• Claim: \(f(x) = \frac 1 x\) is uniformly continuous on \((c, \infty)\) for any \(c > 0\).

  • Note that \begin{align*} {\left\lvert {x} \right\rvert}, {\left\lvert {y} \right\rvert} > c > 0 \implies {\left\lvert {xy} \right\rvert} = {\left\lvert {x} \right\rvert}{\left\lvert {y} \right\rvert} > c^2 \implies \frac{1}{{\left\lvert {xy} \right\rvert}} < \frac 1 {c^{2}} .\end{align*}

  • Letting \(\varepsilon\) be arbitrary, choose \(\delta < \varepsilon c^2\).

    • Note that \(\delta\) does not depend on \(x, y\).

  • Then \begin{align*} {\left\lvert {f(x) - f(y)} \right\rvert} &= {\left\lvert {\frac 1 x - \frac 1 y} \right\rvert} \\ &= \frac{{\left\lvert {x-y} \right\rvert}}{xy} \\ &\leq \frac{\delta}{xy} \\ &< \frac{\delta}{c^2} \\ &< \varepsilon .\end{align*}

• Claim: \(f\) is not uniformly continuous when \(c=0\).
  • Take \(\varepsilon < 1\), and let \(\delta = \delta({\varepsilon})\) be arbitrary.
  • Let \(x_n = \frac 1 n\) for \(n\geq 1\).
  • Choose \(n\) large enough such that \({1\over n} < \delta\)
  • Then a computation: \begin{align*} {\left\lvert {x_n - x_{n+1}} \right\rvert} &= \frac 1 n - \frac 1 {n+1} \\ &= {1\over n(n+1) } \\ &< {1\over n} \\ &< \delta ,\end{align*}
  • Why this can be done: by the Archimedean property of \({\mathbf{R}}\), for any \(\delta\in {\mathbf{R}}\), one can choose choose \(n\) such that \(n\delta > 1\). We’ve also used that \(n+1 > 1\) so \({1\over n+1}< 1\)
  • Note that \(f(x_n) = n\), so \begin{align*} {\left\lvert {f(x_{n+1}) - f(x_{n})} \right\rvert} = (n+1) - n = 1 > \varepsilon .\end{align*}

• Set \(f_N(x) = \sum_{n=1}^N {x^n \over n!}\).

  • Then by definition, \(f_N(x) \to f(x)\) pointwise on \({\mathbf{R}}\).

• Claim: \(f_N\) converges on compact intervals

  • For any compact interval \([-M, M]\), we have \begin{align*} {\left\lVert {f_N(x) - f(x)} \right\rVert}_\infty &= \sup_{x\in [-M, M] } ~{\left\lvert {\sum_{n=N+1}^\infty {x^n \over {n!}} } \right\rvert} \\ &\leq \sup_{x\in [-M, M] } ~\sum_{n=N+1}^\infty {\left\lvert { {x^n \over {n!}} } \right\rvert} \\ &\leq \sum_{n=N+1}^\infty {M^n \over n!} \\ &\leq \sum_{n=0}^\infty {M^n \over {n!} } \quad\text{since all additional terms are positive} \\ &= e^M \\ &<\infty ,\end{align*} so \(f_N \to f\) uniformly on \([-M, M]\) by the M-test.
    • Note: we’ve used that this power series converges to \(e^x\) pointwise everywhere.

• This argument shows that \(f\) converges on any bounded set.

• Claim: \(f_N\) does not converge uniformly on all of \({\mathbf{R}}\).

  • Uniformly convergent sums have uniformly decaying terms: \begin{align*} \sum_{n\leq N} g_n \overset{N\to\infty}\longrightarrow\sum g_n \text{ uniformly on } A \implies {\left\lVert {g_n} \right\rVert}_{\infty, A} \coloneqq\sup_{x\in A} {\left\lvert {g_n(x)} \right\rvert} \overset{n\to\infty}\longrightarrow 0 .\end{align*}

  • Take \(B_N\) a ball of radius \(N\) about 0, then for \(N>1\), note that \(x=N\) on the boundary and so \begin{align*} {\left\lVert {x^k \over k!} \right\rVert}_{\infty, B_N} = {N^k \over k!} \overset{N\to\infty}\longrightarrow\infty .\end{align*}

• Conclusion: \(f_N\) converges on any bounded \(A\subseteq {\mathbf{R}}\) but not on all of \({\mathbf{R}}\).

Switching to polar coordinates and integrating over the quarter of the unit disc \(D \cap Q_1 \subseteq I^2\) in quadrant 1, we have \begin{align*} \int_{I^2} f \, dA &\geq \int_D f \, dA \\ &= \int_0^{\pi/2} \int_0^1 \frac{r^2 \cos(\theta)\sin(\theta)}{r^4} ~r~\,dr\,d\theta\\ &= \int_0^{\pi/2} \int_0^1 \frac{\cos(\theta)\sin(\theta)}{r} \,dr\,d\theta\\ &= \qty{ \int_0^1 {1\over r } \,dr} \qty{ \int_0^{\pi/2} \cos(\theta)\sin(\theta) \,d\theta} \\ &= \qty{ \int_0^1 {1\over r } \,dr} \qty{ \int_0^{1} u \,du} && u=\sin(\theta)\\ &= {1\over 2}\qty{ \int_0^1 {1\over r } \,dr} \\ &\longrightarrow\infty .\end{align*}

• Now setting \(F_N\coloneqq\sum_{n=1}^N f_n\) yields a finite sum of continuous functions, which is continuous.
• Each \(F_N\) is continuous and \(F_N\to F\) uniformly, so \(F\) is continuous.

If a limit \(L\) exists, we have \(x_n\to L\) for all \(n\), so \begin{align*} L = {1+L\over 2+L} \implies L^2 + L - 1 = 0 \implies L = -{1\over 2}\qty{-1 \pm \sqrt 5} .\end{align*} Noting that \(\sqrt{5} > 1\), the condition \(x_1>0\) and a small induction noting that if \(x_n>0\) then \({1+x_n \over 2+x_n}>0\), the only solution can be \(L = -1 + \sqrt 5\). To see that this does converge, write \(f(z) = 1 - (2+z)^{-1}\) so that \(x_{n+1} = f(x_n)\). The claim is that \(f\) is a contracting map on a metric space, which implies it has a unique fixed point \(z_0\) by the Banach fixed point theorem, and if \(f(z_0) = z_0\) then \(z_0 = L\). This follows from the mean value theorem, since \begin{align*} {\left\lvert {f(z) - f(w)} \right\rvert} = {\left\lvert {f'(\xi)} \right\rvert}{\left\lvert {z-w} \right\rvert} < {\left\lvert {z-w} \right\rvert} && \text{for some } \xi \in (z, w) .\end{align*} Since \(f'(z) = (2+z)^{-2}\) satisfies \(0 < f'(z) < 1\) for all \(z\), we have \begin{align*} {\left\lvert {f(z) - f(w)} \right\rvert} \leq {\left\lvert {z-w} \right\rvert} .\end{align*}

See this MSE post for many solutions: https://math.stackexchange.com/questions/4603/if-a-n-subset0-infty-is-non-increasing-and-sum-a-n-infty-then-lim Note that the “obvious” thing here is fiddly: there are bounds on the slices \begin{align*} (N-M \pm 1) x_N \leq \sum_{M\leq k \leq N} a_k \leq (N-M\pm 1) x_M ,\end{align*} but arranging it so that the constants match the indices in \((N-M \pm 1)x_N \approx Nx_N\) requires something clever.

Fix \({\varepsilon}>0\), we’ll find \(n\gg 1\) so that \(nx_n < {\varepsilon}\). Find \(n, m\) with \(n>m\) large enough so that \begin{align*} {\varepsilon}> \sum_{m+1\leq k \leq n} x_k \geq \sum_{m+1\leq k \leq n}x_n = (m-n)x_n .\end{align*} Then rearrange: \begin{align*} {\varepsilon}> (m-n)x_n \implies nx_n < {\varepsilon}+ mx_n .\end{align*} Now choose \(n\) large enough so that \(x_n < {\varepsilon}\), which holds since \(\sum x_n < \infty\), to obtain \begin{align*} nx_n < {\varepsilon}+ m{\varepsilon}= {\varepsilon}(1+m) \to 0 .\end{align*}

• Suppose \(p\) is a polynomial, then integrate by parts: \begin{align*} \lim_{k\to\infty} \int_0^1 kx^{k-1} p(x) \, dx &= \lim_{k\to\infty} \int_0^1 \qty{ {\frac{\partial }{\partial x}\,}x^k } p(x) \, dx \\ &= \lim_{k\to\infty} \left[ x^k p(x) \Big|_0^1 - \int_0^1 x^k \qty{{\frac{\partial p}{\partial x}\,}(x) } \, dx \right] \quad\text{IBP}\\ &= p(1) - \lim_{k\to\infty} \int_0^1 x^k \qty{{\frac{\partial p}{\partial x}\,}(x) } \, dx ,\end{align*}

• Thus it suffices to show that \begin{align*} \lim_{k\to\infty} \int_0^1 x^k \qty{{\frac{\partial p}{\partial x}\,} (x) } \, dx = 0 .\end{align*}

• Integrating by parts a second time yields \begin{align*} \lim_{k\to\infty} \int_0^1 x^k \qty{{\frac{\partial p}{\partial x}\,}(x) } \, dx &= \lim_{k\to\infty} {x^{k+1} \over k+1} {\frac{\partial p}{\partial x}\,}(x) \Big|_0^1 - \int_0^1 {x^{k+1} \over k+1} \qty{ {\frac{\partial ^2 p}{\partial x^2}\,}(x)} \, dx \\ &= \lim_{k\to\infty} {p'(1) \over k+1} - \lim_{k\to\infty} \int_0^1 {x^{k+1} \over k+1} \qty{ {\frac{\partial ^2p}{\partial x^2}\,}(x)} \, dx \\ &= - \lim_{k\to\infty} \int_0^1 {x^{k+1} \over k+1} \qty{ {\frac{\partial ^2p}{\partial x^2}\,}(x)} \, dx \\ &= - \int_0^1 \lim_{k\to\infty} {x^{k+1} \over k+1} \qty{ {\frac{\partial ^2p}{\partial x^2}\,}(x)} \, dx \quad\text{by DCT} \\ &= - \int_0^1 0 \qty{ {\frac{\partial ^2p}{\partial x^2}\,}(x)} \, dx \\ &= 0 .\end{align*}

  • The DCT can be applied here because polynomials are smooth and \([0, 1]\) is compact, so \({\frac{\partial ^2 p}{\partial x^2}\,}\) is bounded on \([0, 1]\) by some constant \(M\) and \begin{align*} \int_0^1 {\left\lvert {x^k {\frac{\partial ^2 p}{\partial x^2}\,} (x)} \right\rvert} \leq \int_0^1 1\cdot M = M < \infty.\end{align*}

• So the result holds when \(f\) is a polynomial.

• Now use the Weierstrass approximation theorem:

  • If \(f: [a, b] \to {\mathbf{R}}\) is continuous, then for every \({\varepsilon}>0\) there exists a polynomial \(p_{\varepsilon}(x)\) such that \({\left\lVert {f - p_{\varepsilon}} \right\rVert}_\infty < {\varepsilon}\).

• Thus \begin{align*} {\left\lvert { \int_0^1 kx^{k-1} p_{\varepsilon}(x)\,dx - \int_0^1 kx^{k-1}f(x)\,dx } \right\rvert} &= {\left\lvert { \int_0^1 kx^{k-1} \qty{p_{\varepsilon}(x) - f(x)} \,dx } \right\rvert} \\ &\leq {\left\lvert { \int_0^1 kx^{k-1} {\left\lVert {p_{\varepsilon}-f} \right\rVert}_\infty \,dx } \right\rvert} \\ &= {\left\lVert {p_{\varepsilon}-f} \right\rVert}_\infty \cdot {\left\lvert { \int_0^1 kx^{k-1} \,dx } \right\rvert} \\ &= {\left\lVert {p_{\varepsilon}-f} \right\rVert}_\infty \cdot x^k \Big|_0^1 \\ &= {\left\lVert {p_{\varepsilon}-f} \right\rVert}_\infty \\ \\ &\overset{{\varepsilon}\to 0}\to 0 \end{align*}

  and the integrals are equal.

• By the first argument, \begin{align*}\int_0^1 kx^{k-1} p_{\varepsilon}(x) \,dx = p_{\varepsilon}(1) \text{ for each } {\varepsilon}\end{align*}

• Since uniform convergence implies pointwise convergence, \(p_{\varepsilon}(1) \overset{{\varepsilon}\to 0}\to f(1)\).

\todo[inline]{Revisit, not so clear that the last line can be made smaller than ${\varepsilon}$, since $M, N$ both depend on ${\varepsilon}$...}

#todo

Case of a characteristic function of an interval \([a, b]\):

• First suppose \(f(x) = \chi_{[a, b]}(x)\).

• Note that \(\sin(nx)\) has a period of \(2\pi/n\), and thus \({\left\lfloor (b-a) \over (2\pi / n) \right\rfloor} = {\left\lfloor n(b-a)\over 2\pi \right\rfloor}\) full periods in \([a, b]\).

• Taking the absolute value yields a new function with half the period

  • So \({\left\lvert {\sin(nx)} \right\rvert}\) has a period of \(\pi/n\) with \({\left\lfloor n(b-a) \over \pi \right\rfloor}\) full periods in \([a, b]\).

• We can compute the integral over one full period (which is independent of which period is chosen)

  • We can use translation invariance of the integral to compute this over the period \(0\) to \(\pi/n\).
  • Since \(\sin(nx)\) is positive, it equals \({\left\lvert {\sin(nx)} \right\rvert}\) on its first period, so we have \begin{align*} \int_{\text{One Period}} {\left\lvert {\sin(nx)} \right\rvert} \, dx &= \int_0^{\pi/n} \sin(nx)\,dx \\ &= {1\over n} \int_0^\pi \sin(u) \,du \quad u = nx \\ &= {1\over n} \qty{-\cos(u)\mathrel{\Big|}_0^\pi} \\ &= {2 \over n} .\end{align*}

• Then break the integral up into integrals over full periods \(P_1, P_2, \cdots, P_N\) where \(N \coloneqq{\left\lfloor n(b-a)/\pi \right\rfloor}\)

• Noting that each period is of length \(\pi\over n\), so letting \(L_n\) be the regions falling outside of a full period, we have

• Thus \begin{align*} \int_a^b {\left\lvert {\sin(nx)} \right\rvert} \, dx &= \qty{ \sum_{j=1}^{N} \int_{P_j} {\left\lvert {\sin(nx)} \right\rvert} \, dx } + \int_{L_n} {\left\lvert {\sin(nx)} \right\rvert}\,dx \\ &= \qty{ \sum_{j=1}^{N} {2\over n} } + \int_{L_n} {\left\lvert {\sin(nx)} \right\rvert}\,dx \\ &= N \qty{2\over n} + \int_{L_n} {\left\lvert {\sin(nx)} \right\rvert}\,dx \\ &\coloneqq{\left\lfloor (b-a) n \over \pi \right\rfloor} {2\over n} + R_n \\ &\coloneqq(b-a)C_n + R_n \end{align*} where (claim) \(C_n \overset{n\to\infty}\to {2\over \pi}\) and \(R(n) \overset{n\to\infty}\to 0\).

• \(C_n \to {2\over \pi}\): \begin{align*} {n-1 \over n} \qty{2\over \pi} = {n-1 \over \pi} \qty{2\over n} \leq {\left\lfloor n\over \pi \right\rfloor}\qty{2\over n} \leq {n \over \pi}\qty{2\over n} = {2 \over \pi} ,\end{align*} then use the fact that \({n-1 \over n} \to 1\).

  • Then equality follows by the Squeeze theorem.

• \(R_n \to 0\):

  • We use the fact that \(m(L_n) \to 0\), then \(\int_{L_n} {\left\lvert {\sin(nx)} \right\rvert} \leq \int_{L_n} 1 = m(L_n) \to 0\).
  • This follows from the fact that \(L_n\) is the complement of \(\cup_j P_j\), the set of full periods, so \begin{align*} m(L_n) &= m(b-a) - \sum m(P_j) \\ &= \qty{b-a} - {\left\lfloor n(b-a) \over \pi \right\rfloor}\qty{\pi \over n} \\ &\overset{n\to \infty}\to (b-a) - (b-a) \\ &= 0 .\end{align*} where we’ve used the fact that \begin{align*} \qty{\pi \over n} \qty{(b-a)n-1 \over \pi} &\leq {\left\lfloor n(b-a) \over \pi \right\rfloor}\qty{\pi \over n} \\ &\leq \qty{\pi \over n} \qty{(b-a)n\over \pi} \\ &= (b-a) ,\end{align*} then taking \(n\to \infty\) sends the LHS to \(b-a\), forcing the middle term to be \(b-a\) by the Squeeze theorem.

General case:

• By linearity of the integral, the result holds for simple functions:

  • If \(f = \sum c_j \chi_{E_j}\) where \(E_j = [a_j, b_j]\), we have \begin{align*} \int_0^1 f(x) {\left\lvert {\sin(nx)} \right\rvert}\,dx &= \int_0^1 \sum c_j \chi_{E_j}(x) {\left\lvert {\sin(nx)} \right\rvert}\,dx \\ &= \sum c_j \int_0^1 \chi_{E_j}(x) {\left\lvert {\sin(nx)} \right\rvert}\,dx \\ &= \sum c_j (b_j - a_j) {2\over \pi} \\ &= {2\over \pi} \sum c_j (b_j - a_j) \\ &= {2\over \pi} \sum c_j m(E_j) \\ &\coloneqq{2\over \pi} \int_0^1 f .\end{align*}

• Since \(f\in L^1\), where simple functions are dense, choose \(s_n\nearrow f\) where \({\left\lVert {s_N - f} \right\rVert}_1 < {\varepsilon}\), then \begin{align*} {\left\lvert { \int_0^1 f(x) {\left\lvert {\sin(nx)} \right\rvert} \,dx - \int_0^1 s_N(x) {\left\lvert {\sin(nx)} \right\rvert}\,dx } \right\rvert} &= {\left\lvert { \int_0^1 \qty{f(x) - s_N(x)} {\left\lvert {\sin(nx)} \right\rvert} \,dx } \right\rvert} \\ &\leq \int_0^1 {\left\lvert { f(x) - s_N(x)} \right\rvert} {\left\lvert {\sin(nx)} \right\rvert} \,dx \\ &= {\left\lVert { \qty{f - s_N} {\left\lvert {\sin(nx)} \right\rvert} } \right\rVert}_1 \\ &\leq {\left\lVert {f-s_N} \right\rVert}_1 \cdot {\left\lVert {{\left\lvert {\sin(nx)} \right\rvert}} \right\rVert}_\infty \quad\text{by Holder}\\ &\leq {\varepsilon}\cdot 1 ,\end{align*}

• So the integrals involving \(s_N\) converge to the integral involving \(f\), and \begin{align*} \lim_{n\to\infty} \int f(x){\left\lvert {\sin(nx)} \right\rvert} &= \lim_{n\to\infty} \lim_{N\to\infty} \int s_N(x) {\left\lvert {\sin(nx)} \right\rvert} \\ &= \lim_{N\to\infty} \lim_{n\to\infty} \int s_N(x) {\left\lvert {\sin(nx)} \right\rvert} \quad\text{because ?}\\ &= \lim_{N\to \infty} {2\over \pi} \int s_N(x) \\ &= {2\over \pi} \int f ,\end{align*} which is the desired result.

\(f_n\to 0\) pointwise:

• Finding the maximum: can check that \({\frac{\partial f_n}{\partial x}\,} = x(1-x)^{n-1} \qty{1 + (n^2-1)x}\)
• This has critical points \(x=0, 1, {-1 \over n^2 + 1}\), and the latter is a global max on \([0, 1]\).
• Set \(x_n \coloneqq{-1 \over n^2 + 1}\)
• Compute \begin{align*} \lim f_n(x_n) = \lim_{n\to \infty } {-n \over n^2 + 1} \qty{1 + x_n}^n = 0\cdot 1 = 0 .\end{align*}
• So \(\sup f_n \to 0\), forcing \(f_n \to 0\) pointwise.

The convergence is not uniform:

• Let \(x_n = \frac 1 n\) and \(\varepsilon > e^{-1}\), then \begin{align*} {\left\lVert {nx(1-x)^n - 0} \right\rVert}_\infty &\geq {\left\lvert {nx_n (1-x_n)^n} \right\rvert} \\ &= {\left\lvert {\left( 1 - \frac 1 n\right)^n} \right\rvert} \\ &> e^{-1} \\ &> \varepsilon .\end{align*}

  • Here we’ve used that \((1 + {x\over n})^n \leq e^x\) for all \(x\in {\mathbf{R}}\) and all \(n\).
  • Follows from \(1+y \leq e^y\) applied to \(y = x/n\).

• Thus \({\left\lVert {f_n - 0} \right\rVert}_\infty = {\left\lVert {f_n} \right\rVert}_\infty > e^{-1} > 0\).

• ?

\todo[inline]{Possible to use part a with $\sin(x) \leq x$ on $[0, \pi/2]$?}

#todo

• Noting that \(\sin(x) \leq 1\), we have \begin{align*} {\left\lvert {\int_0^1 n(1-x)^{n} \sin(x)} \right\rvert} &\leq \int_0^1 {\left\lvert {n(1-x)^n \sin(x)} \right\rvert} \\ &\leq \int_0^1 {\left\lvert {n (1-x)^n} \right\rvert} \\ &= n\int_0^1 (1-x)^n \\ &= -\frac{n(1-x)^{n+1}}{n+1} \\ &\overset{n\to\infty}\longrightarrow 0 .\end{align*}

• \(f_n\) has a root: \begin{align*} ae^{-nax} = be^{-nbx} &\iff {1\over n} = e^{-nbx} e^{nax} = e^{n(b-a)x} \iff x = {\ln\qty{a\over b} \over n(a-b)} \coloneqq x_n .\end{align*}

• Thus \(f_n\) only changes sign at \(x_n\), and is strictly positive on one side of \(x_n\).

• Then \begin{align*} \int_{\mathbf{R}}\sum_n {\left\lvert {f_n(x)} \right\rvert}\,dx &= \sum_n \int_{\mathbf{R}}{\left\lvert {f_n(x)} \right\rvert} \,dx \\ &\geq \sum_n \int_{x_n}^\infty f_n(x) \, dx \\ &= \sum_n {1\over n} \qty{ e^{-bnx} - e^{-anx}\Big|_{x_n}^\infty } \\ &= \sum_n {1\over n} \qty{ e^{-bnx_n} - e^{-anx_n} } .\end{align*}

?

• Set \(f_N(x) \coloneqq\sum_{n=1}^N n^{-x}\), so \(f(x) = \lim_{N\to\infty} f_N(x)\).

• If an interchange of limits is justified, we have \begin{align*} {\frac{\partial }{\partial x}\,} \lim_{N\to\infty} \sum_{n=1}^N n^{-x} &= \lim_{h\to 0} \lim_{N\to\infty} {1\over h} \left[ \qty{\sum_{n=1}^N n^{-x}} - \qty{\sum_{n=1}^N n^{-(x+h)} }\right] \\ &\mathop{\mathrm{=}}_{?} \lim_{N\to\infty} \lim_{h\to 0} {1\over h} \left[ \qty{\sum_{n=1}^N n^{-x}} - \qty{\sum_{n=1}^N n^{-(x+h)} }\right] \\ &= \lim_{N\to\infty} \lim_{h\to 0} {1\over h} \left[ {\sum_{n=1}^N n^{-x}} - {n^{-(x+h)} }\right] \quad\text{(1)} \\ &= \lim_{N\to\infty} \sum_{n=1}^N \lim_{h\to 0} {1\over h} \left[ n^{-x} - n^{-(x+h)} \right] \quad\text{since this is a finite sum} \\ &\coloneqq\lim_{N\to\infty} \sum_{n=1}^N {\frac{\partial }{\partial x}\,}\qty{1 \over n^x} \\ &= \lim_{N\to\infty} \sum_{n=1}^N -{\ln(n) \over n^x} ,\end{align*} where the combining of sums in (1) is valid because \(\sum n^{-x}\) is absolutely convergent for \(x>1\) by the \(p{\hbox{-}}\)test.

• Thus it suffices to justify the interchange of limits and show that the last sum converges on \((1, \infty)\).

• Claim: \(\sum n^{-x}\ln(n)\) converges.

  • Use the fact that for any fixed \({\varepsilon}>0\), \begin{align*} \lim_{n\to\infty} {\ln(n) \over n^{\varepsilon}} \mathop{\mathrm{=}}^{L.H.} \lim_{n\to\infty}{1/n \over {\varepsilon}n^{{\varepsilon}-1}} = \lim_{n\to\infty} {1\over {\varepsilon}n^{\varepsilon}} = 0 ,\end{align*}

  • This implies that for a fixed \({\varepsilon}>0\) and for any constant \(c>0\) there exists an \(N\) large enough such that \(n\geq N\) implies \(\ln(n)/n^{\varepsilon}< c\), i.e. \(\ln(n) < c n^{{\varepsilon}}\).

  • Taking \(c=1\), we have \(n\geq N \implies \ln(n) < n^{\varepsilon}\)

  • We thus break up the sum: \begin{align*} \sum_{n\in {\mathbb{N}}} {\ln(n) \over n^x} &= \sum_{n=1}^{N-1} { \ln(n) \over n^x} + \sum_{n=N}^\infty {\ln(n) \over n^x} \\ &\leq \sum_{n=1}^{N-1} { \ln(n) \over n^x} + \sum_{n=N}^\infty {n^{\varepsilon}\over n^x} \\ &\coloneqq C_{\varepsilon}+ \sum_{n=N}^\infty {n^{\varepsilon}\over n^x} \quad \text{with $C_{\varepsilon}<\infty$ a constant}\\ &= C_{\varepsilon}+ \sum_{n=N}^\infty {1 \over n^{x-{\varepsilon}}} ,\end{align*} where the last term converges by the \(p{\hbox{-}}\)test if \(x-{\varepsilon}> 1\).

  • But \({\varepsilon}\) can depend on \(x\), and if \(x\in (1, \infty)\) is fixed we can choose \({\varepsilon}< {\left\lvert {x-1} \right\rvert}\) to ensure this.

• Claim: the interchange of limits is justified.

  \todo[inline]{?}

Let \(L\) be the LHS and \(R\) be the RHS.

Claim: \(L\leq R\).

• Since \({\left\lvert {\phi } \right\rvert}\leq {\left\lVert {\phi} \right\rVert}_\infty\) a.e., we can write `

Claim: \(R\leq L\).

• We will show that \(R\leq L + {\varepsilon}\) for every \({\varepsilon}>0\).
• Set \begin{align*} S_{\varepsilon}\coloneqq\left\{{x\in {\mathbf{R}}^n{~\mathrel{\Big\vert}~}{\left\lvert {\phi(x)} \right\rvert} \geq {\left\lVert {\phi} \right\rVert}_\infty - {\varepsilon}}\right\} .\end{align*}
• Then we have \begin{align*} \int_{\mathbf{R}}{{\left\lvert {\phi(x)} \right\rvert}^n \over 1 +x^2}\,dx &\geq \int_{S_{\varepsilon}} {{\left\lvert {\phi(x)} \right\rvert}^n \over 1 +x^2}\,dx \quad S_{\varepsilon}\subset {\mathbf{R}}\\ &\geq \int_{S_{\varepsilon}} { \qty{{\left\lVert {\phi} \right\rVert}_\infty - {\varepsilon}}^n \over 1 +x^2}\,dx \qquad\text{by definition of }S_{\varepsilon}\\ &= \qty{{\left\lVert {\phi} \right\rVert}_\infty - {\varepsilon}}^n \int_{S_{\varepsilon}} { 1 \over 1 +x^2}\,dx \\ &= \qty{{\left\lVert {\phi} \right\rVert}_\infty - {\varepsilon}}^n C_{\varepsilon}\qquad\text{where $C_{\varepsilon}$ is some constant} \\ \\ \implies \qty{ \int_{\mathbf{R}}{{\left\lvert {\phi(x)} \right\rvert}^n \over 1 +x^2}\,dx }^{1\over n} &\geq \qty{{\left\lVert {\phi} \right\rVert}_\infty - {\varepsilon}} C_{\varepsilon}^{1 \over n} \\ &\overset{n\to\infty}\to \qty{{\left\lVert {\phi} \right\rVert}_\infty - {\varepsilon}} \cdot 1 \\ &\overset{{\varepsilon}\to 0}\to {\left\lVert {\phi} \right\rVert}_\infty ,\end{align*} where we’ve again used the fact that \(c^{1\over n} \to 1\) for any constant.

• Use the fact that \(L^p\) has small tails: if \(h\in L^2({\mathbf{R}})\), then for any \({\varepsilon}> 0\), \begin{align*} \forall {\varepsilon},\, \exists N\in {\mathbb{N}}{\quad \operatorname{such that} \quad}\int_{{\left\lvert {x} \right\rvert} \geq {N}} {\left\lvert {h(x)} \right\rvert}^2 \,dx < {\varepsilon} .\end{align*}

• So choose \(N\) large enough so that \begin{align*} \int_{{\left\lVert {x} \right\rVert} \geq N}{\left\lvert {g(x)} \right\rvert}^2 < {\varepsilon}\\ \int_{{\left\lVert {x} \right\rVert} \geq N}{\left\lvert {f(x)} \right\rvert}^2 < {\varepsilon}\\ .\end{align*}

• Then write \begin{align*} \int_{{\mathbf{R}}^d} f(x) g(x+n) \,dx = \int_{{\left\lVert {x} \right\rVert} \leq N} f(x)g(x+n)\,dx + \int_{{\left\lVert {x} \right\rVert} \geq N} f(x) g(x+n)\,dx .\end{align*}

• Bounding the second term: apply Cauchy-Schwarz \begin{align*} \int_{{\left\lVert {x} \right\rVert} \geq N} f(x) g(x+n)\,dx \leq \qty{ \int_{{\left\lVert {x} \right\rVert} \geq N} {\left\lvert {f(x)} \right\rvert}^2}^{1\over 2} \cdot \qty{ \int_{{\left\lVert {x} \right\rVert} \geq N} {\left\lvert {g(x)} \right\rvert}^2}^{1\over 2} \leq {\varepsilon}^{1\over 2} \cdot {\left\lVert {g} \right\rVert}_2 .\end{align*}

• Bounding the first term: also Cauchy-Schwarz, after variable changes \begin{align*} \int_{{\left\lVert {x} \right\rVert} \leq N} f(x) g(x+n)\,dx &= \int_{-N}^N f(x) g(x+n)\,dx \\ &= \int_{-N+n}^{N+n} f(x-n) g(x)\,dx \\ &\leq \int_{-N+n}^{\infty} f(x-n) g(x)\,dx \\ &\leq \qty{\int_{-N+n}^{\infty} {\left\lvert {f(x-n)} \right\rvert}^2}^{1\over 2}\cdot \qty{\int_{-N+n}^{\infty} {\left\lvert {g(x)} \right\rvert}^2}^{1\over 2} \\ &\leq {\left\lVert {f} \right\rVert}_2 \cdot {\varepsilon}^{1\over 2} .\end{align*}

• Then as long as \(n\geq 2N\), we have \begin{align*} \int {\left\lvert {f(x) g(x+n)} \right\rvert} \leq \qty{{\left\lVert {f} \right\rVert}_2 + {\left\lVert {g} \right\rVert}_2} \cdot {\varepsilon}^{1\over 2} .\end{align*}

• Define a sequence of operators \begin{align*} T_N: \ell^2 &\to \ell^1\\ \left\{{b_n}\right\} &\mapsto \sum_{n=1}^N a_n b_n .\end{align*}

• By assumption, these are well defined: the image is \(\ell^1\) since \({\left\lvert {T_N(\left\{{b_n}\right\})} \right\rvert} < \infty\) for all \(N\) and all \(\left\{{b_n}\right\} \in \ell^2\).

• So each \(T_N \in \qty{\ell^2} {}^{ \vee }\) is a linear functional on \(\ell^2\).

• For each \(x\in \ell^2\), we have \({\left\lVert {T_N(x)} \right\rVert}_{{\mathbf{R}}} = \sum_{n=1}^N a_n b_n < \infty\) by assumption, so each \(T_N\) is pointwise bounded.

• By the Uniform Boundedness Principle, \(\sup_N {\left\lVert {T_N} \right\rVert}_{\text{op}} < \infty\).

• Define \(T = \lim_{N \to\infty } T_N\), then \({\left\lVert {T} \right\rVert}_{\text{op}} < \infty\).

• By the Riesz Representation theorem, \begin{align*} \sqrt{\sum a_n^2} \coloneqq{\left\lVert {\left\{{a_n}\right\}} \right\rVert}_{\ell^2} = {\left\lVert {T} \right\rVert}_{\qty{\ell^2} {}^{ \vee }} = {\left\lVert {T} \right\rVert}_{\text{op}} < \infty .\end{align*}

• So \(\sum a_n^2 < \infty\).

• If \(m_*(E) = \infty\), then take \(B = {\mathbf{R}}^n\) since \(m({\mathbf{R}}^n) = \infty\).

• Suppose \(N \coloneqq m_*(E) < \infty\).

• Since \(m_*(E)\) is an infimum, by definition, for every \({\varepsilon}> 0\) there exists a covering by closed cubes \(\left\{{Q_i({\varepsilon})}\right\}_{i=1}^\infty \rightrightarrows E\) depending on \({\varepsilon}\) such that \begin{align*} \sum_{i=1}^\infty {\left\lvert {Q_i({\varepsilon})} \right\rvert} < N + {\varepsilon} .\end{align*}

• For each fixed \(n\), set \({\varepsilon}_n = {1\over n}\) to produce such a covering \(\left\{{Q_i({\varepsilon}_n)}\right\}_{i=1}^\infty\) and set \(B_n \coloneqq\bigcup_{i=1}^\infty Q_i({\varepsilon}_n)\).

• The outer measure of cubes is equal to the sum of their volumes, so \begin{align*} m_*(B_n) = \sum_{i=1}^\infty {\left\lvert {Q_i({\varepsilon}_n)} \right\rvert} < N + {\varepsilon}_n = N + {1\over n} .\end{align*}

• Now set \(B \coloneqq\bigcap_{n=1}^\infty B_n\).

  • Since \(E\subseteq B_n\) for every \(n\), \(E\subseteq B\)
  • Since \(B\) is a countable intersection of countable unions of closed sets, \(B\) is Borel.
  • Since \(B_n \subseteq B\) for every \(n\), we can apply subadditivity to obtain the inequality \begin{align*} E \subseteq B \subseteq B_n \implies N \leq m_*(B) \leq m_*(B_n) < N + {1\over n} {\quad \operatorname{for all} \quad} n\in {\mathbf{Z}}^{\geq 1} .\end{align*}

• This forces \(m_*(E) = m_*(B)\).

Suppose \(m_*(E) < \infty\).

• By (a), find a Borel set \(B\supseteq E\) such that \(m_*(B) = m_*(E)\)
• Note that \(E\subseteq B \implies B\bigcap E = E\) and \(B\bigcap E^c = B\setminus E\).
• By assumption, \begin{align*} m_*(B) &= m_*(B\bigcap E) + m_*(B\bigcap E^c) \\ m_*(E) &= m_*(E) + m_*(B\setminus E) \\ m_*(E) - m_*(E) &= m_*(B\setminus E) \qquad\qquad\text{since } m_*(E) < \infty \\ \implies m_*(B\setminus E) &= 0 .\end{align*}
• So take \(N = B\setminus E\); this shows \(m_*(N) = 0\) and \(E = B\setminus (B\setminus E) = B\setminus N\).

If \(m_*(E) = \infty\):

• Apply result to \(E_R\coloneqq E \bigcap[R, R+1)^n \subset {\mathbf{R}}^n\) for \(R\in {\mathbf{Z}}\), so \(E = {\textstyle\coprod}_R E_R\)
• Obtain \(B_R, N_R\) such that \(E_R = B_R \setminus N_R\), \(m_*(E_R) = m_*(B_R)\), and \(m_*(N_R) = 0\).
• Note that
  • \(B\coloneqq\bigcup_R B_R\) is a union of Borel sets and thus still Borel
  • \(E = \bigcup_R E_R\)
  • \(N\coloneqq B\setminus E\)
  • \(N' \coloneqq\bigcup_R N_R\) is a union of null sets and thus still null
• Since \(E_R \subset B_R\) for every \(R\), we have \(E\subset B\)
• We can compute \begin{align*} N = B\setminus E = \qty{ \bigcup_R B_R } \setminus \qty{\bigcup_R E_R } \subseteq \bigcup_R \qty{B_R\setminus E_R} = \bigcup_R N_R \coloneqq N' \end{align*} where \(m_*(N') = 0\) since \(N'\) is null, and thus subadditivity forces \(m_*(N) = 0\).

• Since \(E\) is measurable, \(E^c\) is measurable.
• Since \(E^c\) is measurable exists an open \(O\supseteq E^c\) such that \(m(O\setminus E^c) = 0\).
• Set \(B \coloneqq O^c\), then \(O\supseteq E^c \iff {\mathcal{O}}^c \subseteq E \iff B\subseteq E\).
• Computing measures yields \begin{align*} E\setminus B \coloneqq E\setminus {\mathcal{O}}^c \coloneqq E\bigcap({\mathcal{O}}^c)^c = E\bigcap{\mathcal{O}}= {\mathcal{O}}\bigcap(E^c)^c \coloneqq{\mathcal{O}}\setminus E^c ,\end{align*} thus \(m(E\setminus B) = m({\mathcal{O}}\setminus E^c) = 0\).
• Since \({\mathcal{O}}\) is open, \(B\) is closed and thus Borel.

d.irect Proof (Todo)

\todo[inline]{Try to construct the set.}

• Strategy: Borel-Cantelli.

• We’ll show that \(m(E) \bigcap[n, n+1] = 0\) for all \(n\in {\mathbf{Z}}\); then the result follows from \begin{align*} m(E) = m \qty{\bigcup_{n\in {\mathbf{Z}}} E \bigcap[n, n+1]} \leq \sum_{n=1}^\infty m(E \bigcap[n, n+1]) = 0 .\end{align*}

• By translation invariance of measure, it suffices to show \(m(E \bigcap[0, 1]) = 0\).

  • So WLOG, replace \(E\) with \(E\bigcap[0, 1]\).

• Define \begin{align*} E_j \coloneqq\left\{{x\in [0, 1] {~\mathrel{\Big\vert}~}\ \exists p\in {\mathbf{Z}}^{\geq 0} \text{ s.t. } {\left\lvert {x - \frac{p}{j} } \right\rvert} < \frac 1 {j^3}}\right\} .\end{align*}

  • Note that \(E_j \subseteq {\textstyle\coprod}_{p\in {\mathbf{Z}}^{\geq 0}} B_{j^{-3}}\qty{p\over j}\), i.e. a union over integers \(p\) of intervals of radius \(1/j^3\) around the points \(p/j\). Since \(1/j^3 < 1/j\), this union is in fact disjoint.

• Importantly, note that \begin{align*} \limsup_{j\to\infty} E_j \coloneqq\bigcap_{n=1}^\infty \bigcup_{j=n}^\infty E_j = E \end{align*}

  since

  \begin{align*} x \in \limsup_j E_j &\iff x \in E_j \text{ for infinitely many } j \\ &\iff \text{ there are infinitely many $j$ for which there exist a $p$ such that } {\left\lvert {x - {p\over j}} \right\rvert} < j^{-3} \\ &\iff \text{ there are infinitely many such pairs $p, j$} \\ &\iff x\in E .\end{align*}

• Intersecting with \([0, 1]\), we can write \(E_j\) as a union of intervals: \begin{align*} E_j =& \qty{0, {j^{-3}}} \quad {\textstyle\coprod}\quad B_{j^{-3}}\qty{1\over j} {\textstyle\coprod} B_{j^{-3}}\qty{2\over j} {\textstyle\coprod} \cdots {\textstyle\coprod} B_{j^{-3}}\qty{j-1\over j} \quad {\textstyle\coprod}\quad (1 - {j^{-3}}, 1) ,\end{align*} where we’ve separated out the “boundary” terms to emphasize that they are balls about \(0\) and \(1\) intersected with \([0, 1]\).

• Since \(E_j\) is a union of open sets, it is Borel and thus Lebesgue measurable.

• Computing the measure of \(E_j\):

  • For a fixed \(j\), there are exactly \(j+1\) possible choices for a numerator (\(0, 1, \cdots, j\)), thus there are exactly \(j+1\) sets appearing in the above decomposition.

  • The first and last intervals are length \(1 \over j^3\)

  • The remaining \((j+1)-2 = j-1\) intervals are twice this length, \(2 \over j^3\)

  • Thus \begin{align*} m(E_j) = 2 \qty{1 \over j^3} + (j-1) \qty{2 \over j^3} = {2 \over j^2} \end{align*}

• Note that \begin{align*} \sum_{j\in {\mathbb{N}}} m(E_j) = 2\sum_{j\in {\mathbb{N}}} \frac 1 {j^2} < \infty ,\end{align*} which converges by the \(p{\hbox{-}}\)test for sums.

• But then \begin{align*} m(E) &= m(\limsup_j E_j) \\ &= m(\bigcap_{n\in {\mathbb{N}}} \bigcup_{j\geq n} E_j) \\ &\leq m(\bigcup_{j\geq N} E_j) \quad\text{for every } N \\ &\leq \sum_{j\geq N} m(E_j) \\ &\overset{N\to\infty}\to 0 \quad\text{} .\end{align*}

• Thus \(E\) is measurable as a subset of a null set and \(m(E) = 0\).

It suffices to consider the bounded case, i.e. \(E \subseteq B_M(0)\) for some \(M\). Then write \(E_n = B_n(0) \bigcap E\) and apply the theorem to \(E_n\), and by subadditivity, \(m^*(E) = m^*(\bigcup_n E_n) \leq \sum_n m^*(E_n) = 0\).

Lemma: \(f(x) = x^2, f^{-1}(x) = \sqrt{x}\) are Lipschitz on any compact subset of \([0, \infty)\).

Proof: Let \(g = f\) or \(f^{-1}\). Then \(g\in C^1([0, M])\) for any \(M\), so \(g\) is differentiable and \(g'\) is continuous. Since \(g'\) is continuous on a compact interval, it is bounded, so \({\left\lvert {g'(x)} \right\rvert} \leq L\) for all \(x\). Applying the MVT, \begin{align*} {\left\lvert {f(x) - f(y)} \right\rvert} = f'(c) {\left\lvert {x-y} \right\rvert} \leq L {\left\lvert {x-y} \right\rvert} .\end{align*}

Lemma: If \(g\) is Lipschitz on \({\mathbf{R}}^n\), then \(m(E) = 0 \implies m(g(E)) = 0\).

Proof: If \(g\) is Lipschitz, then \begin{align*} g(B_r(x)) \subseteq B_{Lr}(x) ,\end{align*} which is a dilated ball/cube, and so \begin{align*} m^*(B_{Lr}(x)) \leq L^n \cdot m^*(B_{r}(x)) .\end{align*}

Now choose \(\left\{{Q_j}\right\} \rightrightarrows E\); then \(\left\{{g(Q_j)}\right\} \rightrightarrows g(E)\).

By the above observation, \begin{align*} {\left\lvert {g(Q_j)} \right\rvert} \leq L^n {\left\lvert {Q_j} \right\rvert} ,\end{align*}

and so \begin{align*} m^*(g(E)) \leq \sum_j {\left\lvert {g(Q_j)} \right\rvert} \leq \sum_j L^n {\left\lvert {Q_j} \right\rvert} = L^n \sum_j {\left\lvert {Q_j} \right\rvert} \to 0 .\end{align*}

Now just take \(g(x) = x^2\) for one direction, and \(g(x) = f^{-1}(x) = \sqrt{x}\) for the other.

Lemma: \(E\) is measurable iff \(E = K {\textstyle\coprod}N\) for some \(K\) compact, \(N\) null.

Write \(E = K {\textstyle\coprod}N\) where \(K\) is compact and \(N\) is null.

Then \(\phi^{-1}(E) = \phi^{-1}(K {\textstyle\coprod}N) = \phi^{-1}(K) {\textstyle\coprod}\phi^{-1}(N)\).

Since \(\phi^{-1}(N)\) is null by part (a) and \(\phi^{-1}(K)\) is the preimage of a compact set under a continuous map and thus compact, \(\phi^{-1}(E) = K' {\textstyle\coprod}N'\) where \(K'\) is compact and \(N'\) is null, so \(\phi^{-1}(E)\) is measurable.

So \(\phi\) is a measurable function, and thus yields a well-defined map \(\mathcal L({\mathbf{R}}) \to \mathcal L({\mathbf{R}})\) since it preserves measurable sets. Restricting to \([0, \infty)\), \(f\) is bijection, and thus so is \(\phi\).

Claim: \(K\) is compact.

• It suffices to show that \(K^c \coloneqq[0, 1]\setminus K\) is open; Then \(K\) will be a closed and bounded subset of \({\mathbf{R}}\) and thus compact by Heine-Borel.

• Strategy: write \(K^c\) as the union of open balls (since these form a basis for the Euclidean topology on \({\mathbf{R}}\)).

  • Do this by showing every point \(x\in K^c\) is an interior point, i.e. \(x\) admits a neighborhood \(N_x\) such that \(N_x \subseteq K^c\).

• Identify \(K^c\) as the set of real numbers in \([0, 1]\) whose decimal expansion does contain a 4.

  • We will show that there exists a neighborhood small enough such that all points in it contain a \(4\) in their decimal expansions.

• Let \(x\in K^c\), suppose a 4 occurs as the \(k\)th digit, and write \begin{align*} x = 0.d_1 d_2 \cdots d_{k-1}~ 4 ~d_{k+1}\cdots = \qty{\sum_{j=1}^k d_j 10^{-j}} + \qty{4\cdot 10^{-k}} + \qty{\sum_{j=k+1}^\infty d_j 10^{-j}} .\end{align*}

• Set \(r_x < 10^{-k}\) and let \(y \in [0, 1] \bigcap B_{r_x}(x)\) be arbitrary and write \begin{align*} y = \sum_{j=1}^\infty c_j 10^{-j} .\end{align*}

• Thus \({\left\lvert {x-y} \right\rvert} < r_x < 10^{-k}\), and the first \(k\) digits of \(x\) and \(y\) must agree:

  • We first compute the difference: \begin{align*} x - y &= \sum_{i=1}^\infty d_j 10^{-j} - \sum_{i=1}^\infty c_j 10^{-j} = \sum_{i=1}^\infty \qty{d_j - c_j} 10^{-j} \\ \end{align*}
  • Thus (claim) \begin{align*} {\left\lvert {x-y} \right\rvert} &\leq \sum_{j=1}^\infty {\left\lvert {d_j - c_j} \right\rvert} 10^j < 10^{-k} \iff {\left\lvert {d_j - c_j} \right\rvert} = 0 \quad \forall j\leq k .\end{align*}
  • Otherwise we can note that any term \({\left\lvert {d_j - c_j} \right\rvert}\geq 1\) and there is a contribution to \({\left\lvert {x-y} \right\rvert}\) of at least \(1\cdot 10^{-j}\) for some \(j < k\), whereas \begin{align*} j < k \iff 10^{-j} > 10^{-k} ,\end{align*} a contradiction.

• This means that for all \(j \leq k\) we have \(d_j = c_j\), and in particular \(d_k = 4 = c_k\), so \(y\) has a 4 in its decimal expansion.

• But then \(K^c = \bigcup_x B_{r_x}(x)\) is a union of open sets and thus open.

Claim: \(K\) is nowhere dense and \(m(K) = 0\):

• Strategy: Show \(\qty{\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}^\circ = \emptyset\).

• Since \(K\) is closed, \(\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu = K\), so it suffices to show that \(K\) does not properly contain any interval.

• It suffices to show \(m(K^c) = 1\), since this implies \(m(K) = 0\) and since any interval has strictly positive measure, this will mean \(K\) can not contain an interval.

• As in the construction of the Cantor set, let

  • \(K_0\) denote \([0, 1]\) with 1 interval \(\left({4 \over 10}, {5 \over 10} \right)\) of length \(1 \over 10\) deleted, so \begin{align*}m(K_0^c) = {1\over 10}.\end{align*}
  • \(K_1\) denote \(K_0\) with 9 intervals \(\left({1 \over 100}, {5\over 100}\right), ~\left({14 \over 100}, {15 \over 100}\right), \cdots \left({94\over 100}, {95 \over 100}\right)\) of length \({1 \over 100}\) deleted, so \begin{align*}m(K_1^c) = {1\over 10} + {9 \over 100}.\end{align*}
  • \(K_n\) denote \(K_{n-1}\) with \(9^{n}\) such intervals of length \(1 \over 10^{n+1}\) deleted, so \begin{align*}m(K_n^c) = {1\over 10} + {9 \over 100} + \cdots + {9^{n} \over 10^{n+1}}.\end{align*}

• Then compute \begin{align*} m(K^c) = \sum_{j=0}^\infty {9^n \over 10^{n+1} } = {1\over 10} \sum_{j=0}^\infty \qty{9\over 10}^n = {1 \over 10} \qty{ {1 \over 1 - {9 \over 10 } } } = 1. \end{align*}

Claim: \(K\) has no isolated points:

• A point \(x\in K\) is isolated iff there there is an open ball \(B_r(x)\) containing \(x\) such that \(B_r(x) \subsetneq K^c\).

  • So every point in this ball should have a 4 in its decimal expansion.

• Strategy: show that if \(x\in K\), every neighborhood of \(x\) intersects \(K\).

• Note that \(m(K_n) = \left( \frac 9 {10} \right)^n \overset{n\to\infty}\to 0\)

• Also note that we deleted open intervals, and the endpoints of these intervals are never deleted.

  • Thus endpoints of deleted intervals are elements of \(K\).

• Fix \(x\). Then for every \(\varepsilon\), by the Archimedean property of \({\mathbf{R}}\), choose \(n\) such that \(\left( \frac 9 {10} \right)^n < \varepsilon\).

• Then there is an endpoint \(x_n\) of some deleted interval \(I_n\) satisfying \begin{align*}{\left\lvert {x - x_n} \right\rvert} \leq \left( \frac 9 {10} \right)^n < {\varepsilon}.\end{align*}

• So every ball containing \(x\) contains some endpoint of a removed interval, and thus an element of \(K\).

• Strategy: use approximation by simple functions to show absolute continuity and apply Radon-Nikodym

• Claim: \(\lambda \ll \mu\), i.e. \(\mu(E) = 0 \implies \lambda(E) = 0\).

  • Note that if this holds, by Radon-Nikodym, \(f = {\frac{\partial \lambda}{\partial \mu}\,} \implies d\lambda = f d\mu\), which would yield \begin{align*} \int g ~d\lambda = \int g f ~d\mu .\end{align*}

• So let \(E\) be measurable and suppose \(\mu(E) = 0\).

• Then \begin{align*} \lambda(E) \coloneqq\int_E f ~d\mu &= \lim_{n\to\infty} \left\{{\int_E s_n \,d\mu {~\mathrel{\Big\vert}~}s_n \coloneqq\sum_{j=1}^\infty c_j \mu(E_j),\, s_n \nearrow f}\right\} \end{align*} where we take a sequence of simple functions increasing to \(f\).

• But since each \(E_j \subseteq E\), we must have \(\mu(E_j) = 0\) for any such \(E_j\), so every such \(s_n\) must be zero and thus \(\lambda(E) = 0\).

\todo[inline]{What is the final step in this approximation?}

• Set \(g(x) = x^2\), note that \(g\) is positive and measurable.

• By part (a), there exists a positive \(f\) such that for any \(E\subseteq {\mathbf{R}}\), \begin{align*} \int_E g ~dm = \int_E gf ~d\mu \end{align*}

  • The LHS is zero by assumption and thus so is the RHS.

  • \(m \ll \mu\) by construction.

  • Note that \(gf\) is positive.

• Define \(A_k = \left\{{x\in X {~\mathrel{\Big\vert}~}gf \cdot \chi_E > {1 \over k} }\right\}\), for \(k\in {\mathbf{Z}}^{\geq 0}\)

• Then by Chebyshev with \(p=1\), for every \(k\) we have

\begin{align*} \mu(A_k) \leq k \int_E gf ~d\mu = 0 \end{align*}

• Then noting that \(A_k \searrow A \coloneqq\left\{{x\in X {~\mathrel{\Big\vert}~}gf\cdot \chi_E(x) > 0}\right\}\), we have \(\mu(A) = 0\).

• Since \(gf\) is positive, we have \begin{align*} x\in E \iff gf\chi_E(x) > 0 \iff x\in A \end{align*} so \(E = A\) and \(\mu(E) = \mu(A)\).

• But \(m \ll \mu\) and \(\mu(E) = 0\), so we can conclude that \(m(E) = 0\).

• Claim: \(G\in {\mathcal{M}}\).

  • Claim: \begin{align*} G = \qty{ \bigcap_{N=1}^\infty \bigcup_{n=N}^\infty E_n}^c = \bigcup_{N=1}^\infty \bigcap_{n=N}^\infty E_n^c .\end{align*}

    • This follows because \(x\) is in the RHS \(\iff\) \(x\in E_n^c\) for all but finitely many \(n\) \(\iff\) \(x\in E_n\) for at most finitely many \(n\).

  • But \({\mathcal{M}}\) is a \(\sigma{\hbox{-}}\)algebra, and this shows \(G\) is obtained by countable unions/intersections/complements of measurable sets, so \(G\in {\mathcal{M}}\).

• Claim: \(\mu(G) = 0\).

  • We have \begin{align*} \mu(G) &= \mu\qty{\bigcup_{N=1}^\infty \bigcap_{n=N}^\infty E_n^c} \\ &\leq \sum_{N=1}^\infty \mu \qty{\bigcap_{n=N}^\infty E_n^c} \\ &\leq \sum_{N=1}^\infty \mu(E_M^c) \\ &\coloneqq\sum_{N=1}^\infty \mu(X\setminus E_N) \\ &\overset{N\to\infty}\to 0 .\end{align*}

\todo[inline]{Last step seems wrong!}

Part a:

\(\implies\):

• Suppose \(\chi_{E_n}\to 1\) uniformly, we want to produce an \(N\) such that \(n\geq N \implies x\in E_n\) for all \(x\in X\).
• Take \({\varepsilon}\coloneqq 1/2\). By uniform convergence, for \(N\) large enough, \begin{align*} & \forall n\geq N \quad {\left\lvert {\chi_{E_n}(x) - 1} \right\rvert} < 1/2 && \forall x\in X\\ &\iff \forall n\geq N \quad \chi_{E_n}(x) = 1 && \forall x\in X \\ &\iff \forall n\geq N \quad x\in E_n && \forall x\in X &\iff \forall n\geq N \quad E_n = X ,\end{align*} where we’ve used that \(E_n \subseteq X\) by definition and this shows \(X \subseteq E_n\). So this \(N\) suffices.

\(\impliedby\):

• Let \({\varepsilon}> 0\) be arbitrary.
• Choose \(N\) such that \(n\geq N \implies X = E_n\). Then \begin{align*} &\forall n\geq N \quad x\in E_n && \forall x\in X \\ &\forall n\geq N \quad \chi_{E_n}(x) = 1 && \forall x\in X \\ &\forall n\geq N \quad {\left\lvert {\chi_{E_n}(x) - 1} \right\rvert} = 0 < {\varepsilon}&& \forall x\in X ,\end{align*} so \(\chi_{E_n} \to 1\) uniformly.

Part b:

• Define \begin{align*} S &\coloneqq\left\{{x\in X {~\mathrel{\Big\vert}~}\chi_{E_k}(x) \to 1}\right\}\\ &\coloneqq\left\{{x\in X {~\mathrel{\Big\vert}~}\forall {\varepsilon},\, \exists N\, \text{ s.t. } {\left\lvert {\chi_{E_k}(x) - 1 } \right\rvert} < {\varepsilon},\forall k\geq N}\right\}\\ L &\coloneqq\bigcap_{n\geq 0} \bigcup_{k\geq n} \qty{X\setminus E_k} ,\end{align*} so \(S\) is the set where \(f_n\to f\) and \(X\setminus S\) is the exceptional set where \(f_n\not\to f\) doesn’t converge pointwise.

• Claim: \(L = X\setminus S\), so if \(x\in S \iff x\in X\setminus L\).

• Proof of claim: Suppose there exists an \(N\) such that the first line below is true. Then for a fixed \(x\), there are equivalent statements: \begin{align*} &\qquad x \in S \\ &\iff \exists N \text{ s.t. } \forall {\varepsilon}>0,\quad {\left\lvert {\chi_{E_k}(x) - 1 } \right\rvert} < {\varepsilon}&& \forall k\geq N \\ &\iff \exists N \text{ s.t. } {\left\lvert {\chi_{E_k}(x) - 1 } \right\rvert} = 0 && \forall k\geq N \\ &\iff \exists N \text{ s.t. } \chi_{E_k}(x) = 1 && \forall k\geq N \\ &\iff \exists N \text{ s.t. } x\in E_k && \forall k\geq N \\ &\iff \exists N \text{ s.t. } x\not\in X\setminus E_k &&\forall k\geq N \\ &\iff \exists N \text{ s.t. } x\not\in \bigcup_{k\geq N} X\setminus E_k \\ &{\color{blue} \iff} x\not\in \bigcap_{n\geq 0}\bigcup_{k\geq n} X\setminus E_k \\ &\iff x\not\in L \\ &\iff x\in X\setminus L .\end{align*}

• Proving the iff: \(f_n\to f\) almost everywhere \(\iff \mu(X\setminus S) = 0 \iff \mu(L) = 0\).

• Write \(X\coloneqq[a, b]\),
• Suppose it is not the case that \(f=g\) almost everywhere; then letting \(A\coloneqq\left\{{x\in X {~\mathrel{\Big\vert}~}f(x) \neq g(x)}\right\}\), we have \(m(A) > 0\).
• Write \begin{align*} A = A_1 {\textstyle\coprod}A_2 \coloneqq\left\{{f > g}\right\} {\textstyle\coprod}\left\{{f < g}\right\} .\end{align*}
• Both \(A_i\) are measurable:
  • Since \(f,g\) are measurable functions, so is \(h\coloneqq f-g\).
  • We can write \begin{align*} A_1 &\coloneqq\left\{{ x\in X {~\mathrel{\Big\vert}~}h > 0 }\right\} = h^{-1}((0, \infty)) \\ A_2 &\coloneqq\left\{{ x\in X {~\mathrel{\Big\vert}~}h < 0 }\right\} = h^{-1}((-\infty, 0)) ,\end{align*} and pullbacks of Borel sets by measurable functions are measurable.
• Then on \(E\), we have \(f(x)>g(x)\) pointwise. This is preserved by monotonicity of the integral, thus \begin{align*} f(x) > g(x) \text{ on } E \implies \int_{E} f(x)\,dx > \int_{E} g(x)\, dx .\end{align*}

Part 1: By Holder with \(p=q=2\) on \(L_1[a, b]\), \begin{align*} (b-a)^2 = {\left\lVert {\operatorname{id}} \right\rVert}_1^2 = {\left\lVert {f^{1\over 2}f^{- {1\over 2} } } \right\rVert}_1^2 \leq {\left\lVert {f^{1\over 2}} \right\rVert}_2^2 \cdot {\left\lVert {f^{-{1\over 2}}} \right\rVert}_2^2 = \int_a^b f(x)\,dx\cdot \int_a^b {1\over f(x)}\,dx .\end{align*}

Part 2: It suffices to show \begin{align*} \int_{2^k}^{2^{k+1}}{1\over f} > c_k \text{ where } \sum_{k\geq 0} c_k = \infty .\end{align*} Manipulate the given inequality a bit: \begin{align*} \int_a^b f \leq \int_1^b f \leq b^2 \log(b) \implies \qty{\int_a^b f}^{-1}\geq {1\over b^2\log(b)}\\ \implies .\end{align*} Rewrite the bound in part 1: \begin{align*} \int_a^b {1\over f} \geq \qty{\int_a^b f}^{-1}(b-a)^2 \geq {(b-a)^2 \over b^2 \log(b) } .\end{align*} Now set \(a=2^k, b=2^{k+1}\): \begin{align*} \int_{2^k}^{2^{k+1}} {1\over f(x)} \,dx \geq {(2^{k+1} - 2^k )^2 \over 2^{2(k+1)} (k+1)\log(2) } = {2^{2k} \over 2^{2k} \cdot 4(k+1)\log(2)} = { \mathsf{O}}(1/k) ,\end{align*} and \(\sum 1/k = \infty\).

• If interchanging a limit and integral is justified, we have \begin{align*} L &\coloneqq\lim_{n\to \infty} \int_{(0, n)} {\cos\qty{x\over n} \over x^2 + \cos\qty{x\over n} } \,dx\\ &= \lim_{n\to \infty} \int_{(0, \infty)} \chi_{(0, n)}(x) {\cos\qty{x\over n} \over x^2 + \cos\qty{x\over n} } \,dx\\ &\overset{\text{DCT}}{=} \int_{(0, \infty)} \lim_{n\to \infty} \chi_{(0, n)}(x) {\cos\qty{x\over n} \over x^2 + \cos\qty{x\over n} } \,dx\\ &= \int_{(0, \infty)} \chi_{(0, \infty)}(x) \lim_{n\to \infty} {\cos\qty{x\over n} \over x^2 + \cos\qty{x\over n} } \,dx\\ &= \int_{(0, \infty)} {\lim_{n\to \infty} \cos\qty{x\over n} \over \lim_{n\to \infty} x^2 + \cos\qty{x\over n} } \,dx\\ &= \int_{(0, \infty)} {\cos\qty{\lim_{n\to \infty} {x\over n} } \over x^2 + \cos\qty{\lim_{n\to \infty} {x\over n} } } \,dx\\ &= \int_{(0, \infty)} {1\over x^2 + 1}\,dx\\ &= \arctan(x)\Big|_0^\infty \\ &= {\pi \over 2} ,\end{align*} where we’ve used that \(\cos(\theta)\) is continuous on \({\mathbf{R}}\) to pass a limit inside, noting that \(x\) is fixed in the integrand.

• Justifying the interchange: DCT. Write \(f_n(x) \coloneqq\cos(x/n) / (x^2 + \cos(x/n))\).

• On \((\alpha, \infty)\) for any \(\alpha > 1\):

  • We have \begin{align*} {\left\lvert {f_n(x)} \right\rvert} \leq {\left\lvert {1\over x^2 + \cos(x/n)} \right\rvert} \leq {1\over x^2-1} ,\end{align*} where we’ve used that \(-1\leq \cos(x/n) \leq 1\) for every \(x\), and so the denominator is minimized when \(\cos(x/n) = -1\), and this maximizes the quantity.
  • Setting \(g(x) \coloneqq 1/(x^2-1)\), we have \(g\in L^1(\alpha, \infty)\) by the limit comparison test with \(h(x) \coloneqq x^2\): \begin{align*} {g(x) \over h(x) } \coloneqq{x^2 -1 \over x^2 } = 1 - {1\over x^2} \overset{x\to \infty}\longrightarrow 1 < \infty ,\end{align*} and so \(g, h\) either both converge or both diverge. But \(\int_\alpha^\infty {1\over x^2}\,dx< \infty\) by the \(p{\hbox{-}}\)test for integrals since \(\alpha>1\).

• On \((0, \alpha)\):

  • Just use that \(f_n(x)\) is bounded by a constant: \begin{align*} {\left\lvert {f_n(x)} \right\rvert} = {\left\lvert {\cos(x/n) \over x^2 + \cos(x/n)} \right\rvert} \leq {\left\lvert {\cos(x/n) \over \cos(x/n)} \right\rvert} = 1 ,\end{align*} where we’ve used that \(x^2\) is positive, and removing it from the denominator only makes the quantity larger.
  • Then check that \(\int_0^\alpha 1 \,dx= \alpha < \infty\), so \(1\in L^1(0, \alpha)\).

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• By induction on the number of limits we can pass through the integral.

• For \(n=1\) we first pass one derivative into the integral: let \(x_n \to x\) be any sequence converging to \(x\), then \begin{align*} {\frac{\partial }{\partial x}\,} {\sin(x) \over x} &= {\frac{\partial }{\partial x}\,} \int_0^1 \cos(tx)\,dt \\ &= \lim_{x_n\to x} {1\over x_n - x} \qty{ \int_0^1 \cos(t x_n)\,dt - \int_0^1 \cos(tx) \,dt} \\ &= \lim_{x_n\to x} \qty{ \int_0^1 { \cos(tx_n) - \cos(tx) \over x_n - x} \,dt} \\ &= \lim_{x_n\to x} \qty{ \int_0^1 \qty{t\sin(tx)\mathrel{\Big|}_{x=\xi_n}} \,dt} {\quad \operatorname{where} \quad} \xi_n \in [x_n, x] \text{ by MVT}, \xi_n\to x \\ &= \lim_{\xi_n\to x} \qty{ \int_0^1 t \sin(t \xi_n) \,dt} \\ &=_{\text{DCT}} \int_0^1 \lim_{\xi_n \to x} t \sin(t \xi_n) \,dt \\ &= \int_0^1 t\sin(tx) \,dt \\ .\end{align*}

• Taking absolute values we obtain an upper bound \begin{align*} {\left\lvert { {\frac{\partial }{\partial x}\,} {\sin(x) \over x} } \right\rvert} &= {\left\lvert { \int_0^1 t\sin(tx) \,dt } \right\rvert} \\ &\leq \int_0^1 {\left\lvert {t\sin(tx)} \right\rvert} \,dt \\ &\leq \int_0^1 1 \, dt = 1 ,\end{align*} since \(t\in [0, 1] \implies {\left\lvert {t} \right\rvert} < 1\), and \({\left\lvert {\sin(xt)} \right\rvert} \leq 1\) for any \(x\) and \(t\).

• Note that this bound also justifies the DCT, since the functions \(f_n(t) = t\sin(t \xi_n )\) are uniformly dominated by \(g(t) = 1\) on \(L^1([0, 1])\).

Note: integrating by parts here yields the actual formula: \begin{align*} \int_0^1 t\sin(tx) \,dt &=_{\text{IBP}} \qty{-t\cos(tx) \over x}\mathrel{\Big|}_{t=0}^{t=1} - \int_0^1 {\cos(tx) \over x} \,dt \\ &= {-\cos(x) \over x} - {\sin(x) \over x^2} \\ &= {x\cos(x) - \sin(x) \over x^2} .\end{align*}

• For the inductive step, we assume that we can pass \(n-1\) limits through the integral and show we can pass the \(n\)th through as well. \begin{align*} {\frac{\partial ^n}{\partial x^n}\,} {\sin(x) \over x} &= {\frac{\partial ^n}{\partial x^n}\,} \int_0^1 \cos(tx)\,dt \\ &= {\frac{\partial }{\partial x}\,} \int_0^1 {\frac{\partial ^{n-1}}{\partial x^{n-1}}\,} \cos(tx)\,dt \\ &= {\frac{\partial }{\partial x}\,} \int_0^1 t^{n-1} f_{n-1}(x, t) \,dt \end{align*}
  • Note that \(f_n(x, t) = \pm \sin(tx)\) when \(n\) is odd and \(f_n(x, t) = \pm \cos(tx)\) when \(n\) is even, and a constant factor of \(t\) is multiplied when each derivative is taken.
• We continue as in the base case: \begin{align*} {\frac{\partial }{\partial x}\,} \int_0^1 t^{n-1} f_{n-1}(x, t) \,dt &= \lim_{x_k\to x} \int_0^1 t^{n-1} \qty{ f_{n-1}(x_n, t) - f_{n-1}(x, t) \over x_n - x} \,dt \\ &=_{\text{IVT}} \lim_{x_k\to x} \int_0^1 t^{n-1} {\frac{\partial f_{n-1}}{\partial x}\,}(\xi_k, t) \,dt \quad\text{where } \xi_k\in [x_k, x],\, \xi_k \to x \\ &=_{\text{DCT}} \int_0^1 \lim_{x_k\to x} t^{n-1} {\frac{\partial f_{n-1}}{\partial x}\,}(\xi_k, t) \,dt \\ &\coloneqq\int_0^1 \lim_{x_k\to x} t^{n} f_n(\xi_k, t) \,dt \\ &\coloneqq\int_0^1 t^{n} f_n(x, t) \,dt .\end{align*}
  • We’ve used the fact that \(f_0(x) = \cos(tx)\) is smooth as a function of \(x\), and in particular continuous
  • The DCT is justified because the functions \(h_{n, k}(x, t) = t^n f_n(\xi_k, t)\) are again uniformly (in \(k\)) bounded by 1 since \(t\leq 1 \implies t^n \leq 1\) and each \(f_n\) is a sin or cosine.
• Now take absolute values \begin{align*} {\left\lvert {{\frac{\partial ^n}{\partial x^n}\,} {\sin(x) \over x} } \right\rvert} &= {\left\lvert { \int_0^1 -t^n f_n(x, t) ~dt } \right\rvert} \\ &\leq \int_0^1 {\left\lvert {t^n f_n(x, t)} \right\rvert} ~dt \\ &\leq \int_0^1 {\left\lvert {t^n} \right\rvert} {\left\lvert {f_n(x, t)} \right\rvert} ~dt \\ &\leq \int_0^1 {\left\lvert { t^n} \right\rvert} \cdot 1 ~dt \\ &\leq \int_0^1 t^n ~dt \quad\text{since $t$ is positive} \\ &= \frac{1}{n+1} \\ &< \frac{1}{n} .\end{align*}
  • We’ve again used the fact that \(f_n(x, t)\) is of the form \(\pm \cos(tx)\) or \(\pm \sin(tx)\), both of which are bounded by 1.

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Note that \begin{align*} \lim_{n} \qty{1 + {x^2 \over n}}^{-(n+1)} &= {1 \over \lim_{n} \qty{1 + {x^2 \over n}}^1 \qty{1 + {x^2 \over n}}^n } \\ &= {1 \over 1 \cdot e^{x^2}} \\ &= e^{-x^2} .\end{align*}

If passing the limit through the integral is justified, we will have \begin{align*} \lim_{n\to\infty} \int_0^n \qty{ 1 + {x^2\over n}}^{-(n+1)}\, dx &= \lim_{n\to\infty} \int_{\mathbf{R}}\chi_{[0, n]} \qty{ 1 + {x^2\over n}}^{-(n+1)} \, dx \\ &= \int_{\mathbf{R}}\lim_{n\to\infty} \chi_{[0, n]} \qty{ 1 + {x^2\over n}}^{-(n+1)} \, dx {\quad \operatorname{by the DCT} \quad} \\ &= \int_{\mathbf{R}}\lim_{n\to\infty} \qty{ 1 + {x^2\over n}}^{-(n+1)} \, dx \\ &= \int_0^\infty e^{-x^2} \\ &= {\sqrt \pi \over 2} .\end{align*}

Computing the last integral:

\begin{align*} \qty{\int_{\mathbf{R}}e^{-x^2}\, dx}^2 &= \qty{\int_{\mathbf{R}}e^{-x^2}\,dx} \qty{\int_{\mathbf{R}}e^{-y^2}\,dx} \\ &= \int_{\mathbf{R}}\int_{\mathbf{R}}e^{-(x+y)^2}\, dx \\ &= \int_0^{2\pi} \int_0^\infty e^{-r^2} r\, dr \, d\theta \qquad u=r^2 \\ &= {1\over 2} \int_0^{2\pi } \int_0^\infty e^{-u}\, du \, d\theta \\ &= {1\over 2} \int_0^{2\pi} 1 \\ &= \pi ,\end{align*} and now use the fact that the function is even so \(\int_0^\infty f = {1\over 2} \int_{\mathbf{R}}f\).

Justifying the DCT:

• Apply Bernoulli’s inequality: \begin{align*} {1 + {x^2\over n}}^{n+1} \geq {1 + {x^2\over n}}\qty{1 + x^2} \geq {1 + x^2} ,\end{align*} where the last inequality follows from the fact that \(1 + {x^2 \over n} \geq 1\)

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\(L^2\) bound:

• Since \(f_k \to f\) almost everywhere, \(\liminf_n f_n(x) = f(x)\) a.e.
• \({\left\lVert {f_n} \right\rVert}_2 < \infty\) implies each \(f_n\) is measurable and thus \({\left\lvert {f_n} \right\rvert}^2 \in L^+\), so we can apply Fatou:

\begin{align*} {\left\lVert {f} \right\rVert}_2^2 &= \int {\left\lvert {f(x)} \right\rvert}^2 \\ &= \int \liminf_n {\left\lvert {f_n(x)} \right\rvert}^2 \\ &\underset{\text{Fatou}}\leq\liminf_n \int {\left\lvert {f_n(x)} \right\rvert}^2 \\ &\leq \liminf_n M \\ &= M .\end{align*}

• Thus \({\left\lVert {f} \right\rVert}_2 \leq \sqrt{M} < \infty\) implying \(f\in L^2\).

\todo[inline]{What is the "right" proof here that uses the first part?}

Equality of Integrals:

• Take the sequence \({\varepsilon}_n = {1\over n}\)

• Apply Egorov’s theorem: obtain a set \(F_{\varepsilon}\) such that \(f_n \to f\) uniformly on \(F_{\varepsilon}\) and \(m(I\setminus F_{\varepsilon}) < {\varepsilon}\). \begin{align*} \lim_{n\to \infty} {\left\lvert { \int_0^1 f_n - f } \right\rvert} &\leq \lim_{n\to\infty} \int_0^1 {\left\lvert {f_n - f} \right\rvert} \\ &= \lim_{n\to\infty} \qty{ \int_{F_{\varepsilon}} {\left\lvert {f_n - f} \right\rvert} + \int_{I\setminus F_{\varepsilon}} {\left\lvert {f_n - f} \right\rvert} } \\ &= \int_{F_{\varepsilon}} \lim_{n\to\infty} {\left\lvert {f_n - f} \right\rvert} + \lim_{n\to\infty} \int_{I\setminus F_{\varepsilon}} {\left\lvert {f_n - f} \right\rvert} \quad\text{by uniform convergence} \\ &= 0 + \lim_{n\to\infty} \int_{I\setminus F_{\varepsilon}} {\left\lvert {f_n - f} \right\rvert} ,\end{align*}

  so it suffices to show \(\int_{I\setminus F_{\varepsilon}} {\left\lvert {f_n - f} \right\rvert} \overset{n\to\infty}\to 0\).

• We can obtain a bound using Holder’s inequality with \(p=q=2\): \begin{align*} \int_{I\setminus F_{\varepsilon}} {\left\lvert {f_n - f} \right\rvert} &\leq \qty{ \int_{I\setminus F_{\varepsilon}} {\left\lvert {f_n - f} \right\rvert}^2 } \qty{ \int_{I\setminus F_{\varepsilon}} {\left\lvert {1} \right\rvert}^2 } \\ &= \qty{ \int_{I\setminus F_{\varepsilon}} {\left\lvert {f_n - f} \right\rvert}^2 } \mu(F_{\varepsilon}) \\ &\leq {\left\lVert {f_n - f} \right\rVert}_2 \mu(F_{\varepsilon}) \\ &\leq \qty{ {\left\lVert {f_n} \right\rVert}_2 + {\left\lVert {f} \right\rVert}_2 } \mu(F_{\varepsilon}) \\ &\leq 2M \cdot \mu(F_{\varepsilon}) \end{align*} where \(M\) is now a constant not depending on \({\varepsilon}\) or \(n\).

• Now take a nested sequence of sets \(F_{{\varepsilon}}\) with \(\mu(F_{\varepsilon}) \to 0\) and applying continuity of measure yields the desired statement.

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• Note that \(x^{1\over n} \overset{n\to\infty}\to 1\) for any \(0 < x < \infty\).
• Thus the integrand converges to \({1\over e^x}\), which is integrable on \((0, \infty)\) and integrates to 1.
• Break the integrand up: \begin{align*} \int_0^\infty {1 \over \qty{ 1 + {x\over n} }^n x^{1\over n}} \,dx = \int_0^1 {1 \over \qty{ 1 + {x\over n} }^n x^{1\over n}} \,dx = \int_1^\infty {1 \over \qty{ 1 + {x\over n} }^n x^{1\over n}} \,dx .\end{align*}

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\begin{align*} {\frac{\partial }{\partial t}\,} F(t) &= {\frac{\partial }{\partial t}\,} \int_{\mathbf{R}}f(x) \cos(xt) ~dx \\ &\overset{DCT}= \int_{\mathbf{R}}f(x) {\frac{\partial }{\partial t}\,} \cos(xt) ~dx \\ &= \int_{\mathbf{R}}xf(x) \cos(xt)~dx ,\end{align*} so it only remains to justify the DCT.

• Fix \(t\), then let \(t_n \to t\) be arbitrary.

• Define \begin{align*} h_n(x, t) = f(x) \left(\frac{\cos(tx) - \cos(t_n x)}{t_n - t}\right) \overset{n\to\infty}\to {\frac{\partial }{\partial t}\,} \qty{f(x) \cos(xt)} \end{align*} since \(\cos(tx)\) is differentiable in \(t\) and this is the limit definition of differentiability.

• Note that \begin{align*} {\frac{\partial }{\partial t}\,} \cos(tx) &\coloneqq\lim_{t_n \to t} \frac{\cos(tx) - \cos(t_n x)}{t_n - t} \\ &\overset{MVT} = {\frac{\partial }{\partial t}\,}\cos(tx)\mathrel{\Big|}_{t = \xi_n} \hspace{6em} \text{for some } \xi_n \in [t, t_n] \text{ or } [t_n, t] \\ &= x\sin(\xi_n x) \end{align*} where \(\xi_n \overset{n\to\infty}\to t\) since wlog \(t_n \leq \xi_n \leq t\) and \(t_n \nearrow t\).

• We then have \begin{align*}{\left\lvert {h_n(x)} \right\rvert} = {\left\lvert {f(x) x\sin(\xi_n x)} \right\rvert} \leq {\left\lvert {xf(x)} \right\rvert}\quad\text{since } {\left\lvert {\sin(\xi_n x)} \right\rvert} \leq 1\end{align*} for every \(x\) and every \(n\).

• Since \(xf(x) \in L^1({\mathbf{R}})\) by assumption, the DCT applies.

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• Since \(\int {\left\lvert {f_n} \right\rvert} \overset{n\to\infty}\to \int {\left\lvert {f} \right\rvert}\), define \begin{align*} h_n &= {\left\lvert {f_n - f} \right\rvert} &\overset{n\to\infty}\to 0 ~a.e.\\ g_n &= {\left\lvert {f_n} \right\rvert} + {\left\lvert {f} \right\rvert} &\overset{n\to\infty}\to 2{\left\lvert {f} \right\rvert} ~a.e. \end{align*}

  • Note that \(g_n - h_n \overset{n\to\infty}\to 2{\left\lvert {f} \right\rvert} - 0 = 2{\left\lvert {f} \right\rvert}\).

• Then \begin{align*} \int 2 {\left\lvert {f} \right\rvert} &= \int \liminf_n (g_n - h_n) \\ &= \int \liminf_n(g_n) + \int \liminf_n(-h_n) \\ &= \int \liminf_n(g_n) - \int \limsup_n(h_n) \\ &= \int 2 {\left\lvert {f} \right\rvert} - \int \limsup_n(h_n) \\ &\leq \int 2{\left\lvert {f} \right\rvert} - \limsup_n \int h_n \quad\text{by Fatou} ,\end{align*}

• Since \(f\in L^1\), \(\int 2{\left\lvert {f} \right\rvert} = 2{\left\lVert {f} \right\rVert}_1 < \infty\) and it makes sense to subtract it from both sides, thus \begin{align*} 0 &\leq - \limsup_n \int h_n \\ &\coloneqq- \limsup_n \int {\left\lvert {f_n - f} \right\rvert} .\end{align*} which forces \(\limsup_n \int {\left\lvert {f_n -f} \right\rvert} = 0\), since

  • The integral of a nonnegative function is nonnegative, so \(\int {\left\lvert {f_n - f} \right\rvert} \geq 0\).
  • So \(\qty{ -\int {\left\lvert {f_n - f} \right\rvert} } \leq 0\).
  • But the above inequality shows \(\qty{ -\int {\left\lvert {f_n - f} \right\rvert} } \geq 0\) as well.

• Since \(\liminf_n \int h_n \leq \limsup_n \int h_n = 0\), \(\lim_n \int h_n\) exists and is equal to zero.

• But then \begin{align*} {\left\lvert {\int f_n - \int f} \right\rvert} &= {\left\lvert {\int f_n -f} \right\rvert} \leq \int {\left\lvert {f_n - f} \right\rvert} ,\end{align*} and taking \(\lim_{n\to\infty}\) on both sides yields \begin{align*} \lim_{n\to\infty} {\left\lvert {\int f_n - \int f} \right\rvert} \leq \lim_{n\to\infty} \int {\left\lvert {f_n - f} \right\rvert} = 0 ,\end{align*} so \(\lim_{n\to\infty} \int f_n = \int f\).

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Part a Claim: \(f_n\) does not converge uniformly to its limit.

• Note each \(f_n(x)\) is clearly continuous on \((0, \infty)\), since it is a quotient of continuous functions where the denominator is never zero.

• Note \begin{align*} x < 1 \implies x^n \overset{n\to\infty}\to 0{\quad \operatorname{and} \quad} x>1 \implies x^n \overset{n\to\infty}\to \infty .\end{align*}

• Thus \begin{align*} f_n(x) = \frac{x}{1+ x^n}\overset{n\to\infty}\longrightarrow f(x) \coloneqq \begin{cases} x, & 0 \leq x < 1 \\ \frac 1 2, & x = 1 \\ 0, & x > 1 \\ \end{cases} .\end{align*}

• If \(f_n \to f\) uniformly on \([0, \infty)\), it would converge uniformly on every subset and thus uniformly on \((0, \infty)\).

  • Then \(f\) would be a uniform limit of continuous functions on \((0, \infty)\) and thus continuous on \((0, \infty)\).
  • By uniqueness of limits, \(f_n\) would converge to the pointwise limit \(f\) above, which is not continuous at \(x=1\), a contradiction.

Part b

• If the DCT applies, interchange the limit and integral: `

• To justify the DCT, write \begin{align*} \int_0^\infty f_n(x) = \int_0^1 f_n(x) + \int_1^\infty f_n(x) .\end{align*}

• \(f_n\) restricted to \((0, 1)\) is uniformly bounded by \(g_0(x) = 1\) in the first integral, since \begin{align*} x \in [0, 1] \implies \frac{x}{1+x^n} < \frac{1}{1+x^n} < 1 \coloneqq g(x) \end{align*} so \begin{align*} \int_0^1 f_n(x)\,dx \leq \int_0^1 1 \,dx = 1 < \infty .\end{align*} Also note that \(g_0\cdot \chi_{(0, 1)} \in L^1((0, \infty))\).

• The \(f_n\) restricted to \((1, \infty)\) are uniformly bounded by \(g_1(x) = {1\over x^{2}}\) on \([1, \infty)\), since \begin{align*} x \in (1, \infty) \implies \frac{x}{1+x^n} \leq {x \over x^n} = {1 \over x^{n-1}} \leq {1\over x^2}\in L^1([1, \infty) \text{ when } n\geq 3 ,\end{align*} by the \(p{\hbox{-}}\)test for integrals.

• So set \begin{align*}g \coloneqq g_0 \cdot \chi_{(0, 1)} + g_1 \cdot \chi_{[1, \infty)},\end{align*} then by the above arguments \(g \in L^1((0, \infty))\) and \(f_n \leq g\) everywhere, so the DCT applies.

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• Fixing notation, set \(\tau_x f(y) \coloneqq f(y-x)\); we then want to show \begin{align*} {\left\lVert {\tau_x f -f} \right\rVert}_{L^1} \overset{x\to 0}\to 0 .\end{align*}
• Claim: by an \({\varepsilon}/3\) argument, it suffices to show this for compactly supported functions:
  • Since \(f\in L^1\), choose \(g_n\subset C_c^\infty({\mathbf{R}}^1)\) smooth and compactly supported so that \begin{align*}{\left\lVert {f-g} \right\rVert}_{L^1} < {\varepsilon}.\end{align*}
  • Claim: \({\left\lVert {\tau_x f - \tau_x g} \right\rVert} < {\varepsilon}\).
    • Proof 1: translation invariance of the integral.
    • Proof 2: Apply a change of variables: \begin{align*} {\left\lVert {\tau_x f - \tau_x g} \right\rVert}_1 &\coloneqq\int_{\mathbf{R}}{\left\lvert {\tau_x f(y) - \tau_x g(y)} \right\rvert}\, dy \\ &= \int_{\mathbf{R}}{\left\lvert {f(y-x) - g(y-x)} \right\rvert}\, dy \\ &= \int_{\mathbf{R}}{\left\lvert {f(u) - g(u)} \right\rvert}\, du \qquad (u=y-x,\, du=dy) \\ &= {\left\lVert {f-g} \right\rVert}_1 \\ &< {\varepsilon} .\end{align*}
  • Then \begin{align*} {\left\lVert {\tau_x f - f} \right\rVert}_1 &= {\left\lVert {\tau_x f - \tau_x g + \tau_x g - g +g - f} \right\rVert}_{1} \\ &\leq {\left\lVert {\tau_x f - \tau_x g} \right\rVert}_1 + {\left\lVert {\tau_x g - g} \right\rVert}_1 + {\left\lVert {g - f} \right\rVert}_{1} \\ &\leq 2{\varepsilon}+ {\left\lVert {\tau_x g - g} \right\rVert}_1 .\end{align*}
• To show this for compactly supported functions:

  • Let \(g\in C_c^\infty({\mathbf{R}}^1)\), let \(E = \mathop{\mathrm{supp}}(g)\), and write \begin{align*} {\left\lVert {\tau_x g - g} \right\rVert}_1 &= \int_{\mathbf{R}}{\left\lvert {g(y-x) - g(y)} \right\rvert}\,dy \\ &= \int_E {\left\lvert {g(y-x) - g(y)} \right\rvert} \,dy + \int_{E^c} {\left\lvert {g(y-x) - g(y)} \right\rvert} \,dy\\ &= \int_E {\left\lvert {g(y-x) - g(y)} \right\rvert} \,dy .\end{align*}

  • But \(g\) is smooth and compactly supported on \(E\), and thus uniformly continuous on \(E\), so \begin{align*} \lim_{x\to 0} \int_E {\left\lvert {g(y-x) - g(y)} \right\rvert} \,dy &= \int_E \lim_{x\to 0} {\left\lvert {g(y-x) - g(y)} \right\rvert} \,dy \\ &= \int_E 0 \,dy \\ &= 0 .\end{align*}

\begin{align*} I = \lim_{n\to\infty} \int_1^\infty {e^{-x} \over {1\over n} + x^2 }\sin\qty{x\over n}\chi_{[1, n]} \,dx = \int_1^\infty{e^{-x}\over x^2}\lim_{n\to\infty }\sin\qty{x\over n } \chi_{[1, n]} \,dx = 0 ,\end{align*} since \(\sin(x/n) \to 0\). Passing the limit through the integral is justified by the DCT: write \begin{align*} f_n(x) \coloneqq{ne^{-x} \over 1 + nx^2}\sin\qty{x\over n}\chi_{[1, n]} .\end{align*} Then \begin{align*} {\left\lvert {f_n(x)} \right\rvert} \leq g(x) \coloneqq{e^{-x}\over x^2}\in L^1(1, \infty) ,\end{align*}

since ` \begin{align*} {\left\lVert {f} \right\rVert}_{L^1(1, \infty)}

\int_1^\infty {\left\lvert {1\over x^2e^x} \right\rvert},dx\leq \int_1^\infty {\left\lvert {1\over x^2} \right\rvert},dx= 1 < \infty .\end{align*} `{=html}

Let \(y\in F^c\) which is open, then one can find an epsilon ball about \(y\) avoiding \(F\). We can take \({\varepsilon}\coloneqq\delta_F(y)\) to define \(A \coloneqq B_{{\varepsilon}}(y)\), and we still have \(A \subseteq F^c\) and \(F \subseteq A^c\). Note that \({\left\lvert {x-y} \right\rvert}^2 = (x-y)^2\) since this is always positive, then \begin{align*} \int_F {\left\lvert {x-y} \right\rvert}^{-2} \,dx &\leq \int_{A^c} {\left\lvert {x-y} \right\rvert}^{-2} \,dx\\ &= \int_{-\infty}^{-{\varepsilon}} \qty{x-y}^{-2} \,dx+ \int_{{\varepsilon}}^{\infty} \qty{x-y}^{-2}\,dx\\ &= \int_{-\infty}^{-{\varepsilon}} u^{-2} \,dx+ \int_{{\varepsilon}}^{\infty} u^{-2} \,dx\\ &= -u^{-1}\Big|_{u=-{\varepsilon}}^{u=-\infty}- u^{-1}\Big|_{u=\infty}^{u={\varepsilon}} \\ &= {2\over {\varepsilon}} \\ &\coloneqq{2\over \delta_F(y)} .\end{align*}

Estimate: ` \begin{align*} \int_F I(x) ,dx &\coloneqq\int_F \int_{\mathbf{R}}{\delta_F(y) \over (x-y)^2 } ,dy,dx\ &= \int_{\mathbf{R}}\delta_F(y) \int_F {1\over (x-y)^2} ,dx,dy\ &= \int_F \delta_F(y) \int_F {1\over (x-y)^2} ,dx,dy

• \int_{F^c} \delta_F(y) \int_F {1\over (x-y)^2} ,dx,dy\ &= 0
• \int_{F^c} \delta_F(y) \int_F {1\over (x-y)^2} ,dx,dy\ &\leq \int_{F^c} 2 ,dy\ &= 2\mu(F^c) \ &<\infty ,\end{align*} `{=html} where we’ve used that \(y\in F\implies \delta_F(y) = 0\) and applied the bound from the first part. We’ve also implicitly used Fubini-Tonelli to change the order of integration, justified by positivity of the integrand and the finite iterated integral. This forces \(I(x) < \infty\) for almost every \(x\in F\), since if \(I(x)\) is unbounded on any positive measure set then this integral would diverge.

If \(x\not\in F\), pick an \({\varepsilon}{\hbox{-}}\)ball \(A\) about \(x\) avoiding \(F\) so that \({\left\lvert {x-y} \right\rvert}> {\varepsilon}\) for any \(y\in A^c\) and thus \((x-y)^{-2} \leq {\varepsilon}^{-2}\). Note that \(\delta_F(y)\geq {\varepsilon}\), so \begin{align*} I(x) &= \int_{\mathbf{R}}\delta_F(y) (x-y)^{-2} \,dy\\ &\geq \int_{A^c} \delta_F(y) (x-y)^{-2} \,dy\\ &\geq \int_{A^c} \delta_F(y) {\varepsilon}^{-2} \,dy\\ &\geq \int_{A^c} {\varepsilon}^{-1} \,dy\\ &= \mu(A^c){\varepsilon}^{-1} ,\end{align*} which can be made arbitrarily large by taking \({\varepsilon}\to 0\).

#todo: Not great, \(A^c\) depends on \({\varepsilon}\) so this ratio has a competing numerator…

• \({\left\lVert {f} \right\rVert}_p \leq {\left\lVert {f} \right\rVert}_\infty\):

  • Note \({\left\lvert {f(x)} \right\rvert} \leq {\left\lVert {f} \right\rVert}_\infty\) almost everywhere and taking \(p\)th powers preserves this inequality.

  • Thus \begin{align*} {\left\lvert {f(x)} \right\rvert} &\leq {\left\lVert {f} \right\rVert}_\infty \quad\text{a.e. by definition} \\ \implies {\left\lvert {f(x)} \right\rvert}^p &\leq {\left\lVert {f} \right\rVert}_\infty^p \quad\text{for } p\geq 0 \\ \implies {\left\lVert {f} \right\rVert}_p^p &= \int_X {\left\lvert {f(x)} \right\rvert}^p ~dx \\ &\leq \int_X {\left\lVert {f} \right\rVert}_\infty^p ~dx \\ &= {\left\lVert {f} \right\rVert}_\infty^p \int_X 1\,dx \\ &= {\left\lVert {f} \right\rVert}_\infty^p \cdot m(X) \quad\text{since the norm doesn't depend on }x \\ &= {\left\lVert {f} \right\rVert}_\infty^p \qquad \text{since } m(X) = 1 .\end{align*}

    • Thus \({\left\lVert {f} \right\rVert}_p \leq {\left\lVert {f} \right\rVert}_\infty\) for all \(p\) and taking \(\lim_{p\to\infty}\) preserves this inequality.
• \({\left\lVert {f} \right\rVert}_p \geq {\left\lVert {f} \right\rVert}_\infty\):

  • Fix \(\varepsilon > 0\).

  • Define \begin{align*} S_\varepsilon \coloneqq\left\{{x\in {\mathbf{R}}^n {~\mathrel{\Big\vert}~}{\left\lvert {f(x)} \right\rvert} \geq {\left\lVert {f} \right\rVert}_\infty - \varepsilon}\right\} .\end{align*}

    • Note that \(m(S_{\varepsilon}) > 0\); otherwise if \(m(S_{\varepsilon}) = 0\), then \(t\coloneqq{\left\lVert {f} \right\rVert}_\infty - {\varepsilon}< {\left\lVert {f} \right\rVert}_{\varepsilon}\). But this produces a smaller upper bound almost everywhere than \({\left\lVert {f} \right\rVert}_{\varepsilon}\), contradicting the definition of \({\left\lVert {f} \right\rVert}_{\varepsilon}\) as an infimum over such bounds.

  • Then \begin{align*} {\left\lVert {f} \right\rVert}_p^p &= \int_X {\left\lvert {f(x)} \right\rvert}^p ~dx \\ &\geq \int_{S_\varepsilon} {\left\lvert {f(x)} \right\rvert}^p ~dx \quad\text{since } S_{\varepsilon}\subseteq X \\ &\geq \int_{S_\varepsilon} {\left\lvert {{\left\lVert {f} \right\rVert}_\infty - \varepsilon} \right\rvert}^p ~dx \quad\text{since on } S_{\varepsilon}, {\left\lvert {f} \right\rvert} \geq {\left\lVert {f} \right\rVert}_\infty - {\varepsilon}\\ &= {\left\lvert {{\left\lVert {f} \right\rVert}_\infty - \varepsilon} \right\rvert}^p \cdot m(S_\varepsilon) \quad\text{since the integrand is independent of }x \\ & \geq 0 \quad\text{since } m(S_{\varepsilon}) > 0 \end{align*}

  • Taking \(p\)th roots for \(p\geq 1\) preserves the inequality, so \begin{align*} \implies {\left\lVert {f} \right\rVert}_p &\geq {\left\lvert {{\left\lVert {f} \right\rVert}_\infty - \varepsilon} \right\rvert} \cdot m(S_\varepsilon)^{\frac 1 p} \overset{p\to\infty}\longrightarrow{\left\lvert {{\left\lVert {f} \right\rVert}_\infty - \varepsilon} \right\rvert} \overset{\varepsilon \to 0}\longrightarrow{\left\lVert {f} \right\rVert}_\infty \end{align*} where we’ve used the fact that above arguments work

  • Thus \({\left\lVert {f} \right\rVert}_p \geq {\left\lVert {f} \right\rVert}_\infty\).

Use that \(L^\infty[0, 1] \subseteq L^1[0, 1]\), so fixing \(g\), choose a sequence of compactly supported continuous functions \(f_k\) converging to \(\operatorname{sign}(g)\) in \(L^1\). We can arrange so that \({\left\lvert {g_k} \right\rvert} \leq 1\). Then \begin{align*} \int {\left\lvert {g} \right\rvert} &= \int\operatorname{sign}(g)\cdot g \\ &= \int \lim_k g_k\cdot g \\ &\overset{\text{DCT}}{=} \lim_k \int g_k\cdot g \\ &=\lim_k 0 \\ &= 0 ,\end{align*} where the DCT applies since defining \(h_k \coloneqq g_k\cdot g\) we have \({\left\lvert {h_k} \right\rvert} \leq g\in L^1[0, 1]\), and each integral is zero since \(g_k\) is continuous (and we use the hypothesis).

• Fix \({\varepsilon}\).

• Using small tails for \(f, g \in L^1\), choose \(R_1, R_2 \gg 0\) so that \begin{align*} \int_{B_{R_1}(0)^c} {\left\lvert {f} \right\rvert} &< {\varepsilon}\\ \int_{B_{R_2}(0)^c} {\left\lvert {g} \right\rvert} &< {\varepsilon} .\end{align*}

  • Note that this implies \begin{align*} \int_{-R_1}^{R_1} {\left\lvert {f} \right\rvert} &= {\left\lVert {f} \right\rVert}_1 - 2{\varepsilon}\\ \int_{-R_2}^{R_2} {\left\lvert {g_N} \right\rvert} &= {\left\lVert {g_N} \right\rVert} - 2{\varepsilon} .\end{align*}

  • Also note that by translation invariance of the Lebesgue integral, \({\left\lVert {g} \right\rVert}_1 = {\left\lVert {g_N} \right\rVert}_1\).

• Now use \(N\) to make the densities almost disjoint: choose \(N\gg 1\) so that \(N-R_2 > R_1\):

• Consider the change of variables \(x\mapsto x-N\): \begin{align*} \int_{-R_2}^{R_2} {\left\lvert {g(x)} \right\rvert}\,dx = \int_{N-R_2} ^{N+R_2} {\left\lvert {g(x-N)} \right\rvert} \,dx \coloneqq\int_{N-R_2} ^{N+R_2} {\left\lvert {g_N(x)} \right\rvert} \,dx .\end{align*}
  • Use this to conclude that \begin{align*} \int_{N-R_2}^{N+R_2} {\left\lvert {g_N} \right\rvert} = {\left\lVert {g_N} \right\rVert} - 2{\varepsilon} .\end{align*}
• Now split the integral in the problem statement at \(R_1\):

` \begin{align*} {\left\lVert {f + g_N} \right\rVert}1 = \int{\mathbf{R}}{\left\lvert {f+g_N} \right\rvert} = \int_{-\infty}^{R_1} {\left\lvert {f+ g_N} \right\rvert}

• \int_{R_1}^{\infty} {\left\lvert {f+ g_N} \right\rvert} \coloneqq I_1 + I_2 .\end{align*} `{=html}

• Idea: from the picture,

  • On \(I_1\), \(f\) is big and \(g_N\) is small
  • On \(I_2\), \(f\) is small and \(g_N\) is big

• Casework: estimate \(I_1, I_2\) separately, bounding from above and below.

• \(I_1\) upper bound: \begin{align*} I_1 &\coloneqq\int_{-\infty}^{R_1} {\left\lvert {f + g_N} \right\rvert} \\ &\leq \int_{-\infty}^{R_1} {\left\lvert {f} \right\rvert} + {\left\lvert {g_N} \right\rvert} \\ &= \int_{-\infty}^{R_1} {\left\lvert {f} \right\rvert} + \int_{-\infty}^{R_1} {\left\lvert {g_N} \right\rvert} \\ &\leq \int_{-\infty}^{R_1} {\left\lvert {f} \right\rvert} + \int_{-\infty}^{\color{green} N - R_2} {\left\lvert {g_N} \right\rvert} && R_1 < N-R_2 \\ &= {\left\lVert {f} \right\rVert}_1 - \int_{R_1}^{\infty} {\left\lvert {f} \right\rvert} + \int_{-\infty}^{N - R_2} {\left\lvert {g_N} \right\rvert} \\ &\leq {\left\lVert {f} \right\rVert}_1 - \int_{R_1}^{\infty} {\left\lvert {f} \right\rvert} + {\varepsilon}\\ &\leq {\left\lVert {f} \right\rVert}_1 + {\varepsilon} .\end{align*}

  • In the last step we’ve used that we’re subtracting off a positive number, so forgetting it only makes things larger.

  • We’ve also used monotonicity of the Lebesgue integral: if \(A\leq B\), then \((c, A) \subseteq (c, B)\) and \(\int_{c}^A {\left\lvert {f} \right\rvert} \leq \int_c^B {\left\lvert {f} \right\rvert}\) since \({\left\lvert {f} \right\rvert}\) is positive.

• \(I_1\) lower bound: \begin{align*} I_1 &\coloneqq\int_{-\infty}^{R_1} {\left\lvert {f + g_N} \right\rvert} \\ &\geq \int_{-\infty}^{R_1} {\left\lvert {f} \right\rvert} - {\left\lvert {g_N} \right\rvert} \\ &= \int_{-\infty}^{R_1} {\left\lvert {f} \right\rvert} - \int_{-\infty}^{R_1} {\left\lvert {g_N} \right\rvert} \\ &\geq \int_{-\infty}^{R_1} {\left\lvert {f} \right\rvert} - \int_{-\infty}^{\color{green} N-R_2} {\left\lvert {g_N} \right\rvert} && R_1 < N-R_2 \\ &= {\left\lVert {f} \right\rVert}_1 - \int_{R_1}^{ \infty } {\left\lvert {f} \right\rvert} - \int_{- \infty }^{N-R_2} {\left\lvert {g_N} \right\rvert} \\ &\geq {\left\lVert {f} \right\rVert}_1 - {\varepsilon}- {\varepsilon}\\ &= {\left\lVert {f} \right\rVert}_1 - 2{\varepsilon} .\end{align*}

  • Now we’ve used that the integral with \(g_N\) comes in with a negative sign, so extending the range of integration only makes things smaller. We’ve also used the \({\varepsilon}\) bound on both \(f\) and \(g_N\) here, and both are tail estimates.

• Taken together we conclude \begin{align*} {\left\lVert {f} \right\rVert}_1 - 2{\varepsilon} \leq I_1 \leq {\left\lVert {f} \right\rVert}_1 && {\varepsilon}\to 0 \implies I_1 = {\left\lVert {f} \right\rVert}_1 .\end{align*}

• \(I_2\) lower bound: \begin{align*} I_2 &\coloneqq\int_{R_1}^{\infty} {\left\lvert {f + g_N} \right\rvert} \\ &\leq \int_{R_1}^{\infty} {\left\lvert {f} \right\rvert} + \int_{R_1}^{\infty} {g_N} \\ &\leq \int_{R_1}^{\infty} {\left\lvert {f} \right\rvert} + {\left\lVert {g_N} \right\rVert}_1 - \int_{-\infty}^{R_1} {\left\lvert {g_N} \right\rvert} \\ &\leq {\varepsilon}+ {\left\lVert {g_N} \right\rVert}_1 - \int_{-\infty}^{R_1} {\left\lvert {g_N} \right\rvert} \\ &\leq {\varepsilon}+ {\left\lVert {g_N} \right\rVert}_1 \\ &= {\varepsilon}+ {\left\lVert {g} \right\rVert}_1 .\end{align*}

  • Here we’ve again thrown away negative terms, only increasing the bound, and used the tail estimate on \(f\).

• \(I_2\) upper bound:

\begin{align*} I_2 &\coloneqq\int_{R_1}^{\infty} {\left\lvert {f + g_N} \right\rvert} \\ &= \int_{R_1}^{\infty} {\left\lvert {g_N + f} \right\rvert} \\ &\geq \int_{R_1}^{\infty} {\left\lvert {g_N} \right\rvert} - \int_{R_1}^{\infty} {\left\lvert {f} \right\rvert} \\ &= {\left\lVert {g_N} \right\rVert} - \int_{-\infty}^{R_1} {\left\lvert {g_N} \right\rvert} - \int_{R_1}^{\infty} {\left\lvert {f} \right\rvert} \\ &\geq {\left\lVert {g_N} \right\rVert} - 2{\varepsilon} .\end{align*}

• Here we’ve swapped the order under the absolute value, and used the tail estimates on both \(g\) and \(f\).

• Taken together: \begin{align*} {\left\lVert {g} \right\rVert}_1 - {\varepsilon}\leq I_2 \leq {\left\lVert {g} \right\rVert}_1 + 2{\varepsilon} .\end{align*}

• Note that we have two inequalities: \begin{align*} {\left\lVert {f} \right\rVert}_1 - 2{\varepsilon}&\leq \int_{-\infty}^{R_1} {\left\lvert {f -g_N} \right\rvert} \leq {\left\lVert {f} \right\rVert}_1 + {\varepsilon}\\ {\left\lVert {g} \right\rVert}_1 - 2{\varepsilon}&\leq \int^{\infty}_{R_1} {\left\lvert {f -g_N} \right\rvert} \leq {\left\lVert {g} \right\rVert}_1 + {\varepsilon} .\end{align*}

• Add these to obtain \begin{align*} {\left\lVert {f} \right\rVert}_1 + {\left\lVert {g} \right\rVert}_1 - 4{\varepsilon}\leq I_1 + I_2 \coloneqq{\left\lVert {f - g_N} \right\rVert}_1 \leq {\left\lVert {f} \right\rVert} + {\left\lVert {g} \right\rVert}_1 + 2{\varepsilon} .\end{align*}

• Check that as \(N\to \infty\) as \({\varepsilon}\to 0\) to yield the result.

• Fix \(y_0\), we’ll show \(F'\) exists and is continuous at \(y_0\).

• Fix a sequence \(y_n\searrow y_0\) and define \begin{align*} h_n(x) \coloneqq { h(x, y_n) - h(x, y_0) \over y_n - y_0} && h(x, y) \coloneqq f(x) \cos(yx) .\end{align*}

• We can then write \begin{align*} {\frac{\partial h}{\partial y}\,}(x, y_0) = \lim_{n\to \infty} h_n(x) .\end{align*}

• Apply the MVT: \begin{align*} h_n(x) \coloneqq{ h(x, y_n) - h(x, y_0) \over y_n - y_0} &= {\frac{\partial h}{\partial y}\,}(x, \tilde y) && \text{ for some } \tilde y \in [y_0, y_n] .\end{align*}

• Use this to get a bound for DCT: \begin{align*} {\left\lvert {h_n(x)} \right\rvert} &\coloneqq{\left\lvert { h(x, y_n) - h(x, y_0) \over y_n - y_0} \right\rvert} \\ &= {\left\lvert { {\frac{\partial h}{\partial y}\,}(x, \tilde y) } \right\rvert} \\ &\leq \sup_{y\in [y_0, y_n]} {\left\lvert { {\frac{\partial h}{\partial y}\,}(x, y) } \right\rvert} \\ &\leq \sup_{y\in [y_0, y_n]} {\left\lvert { xf(x) \sin(yx) } \right\rvert} \\ &\leq {\left\lvert { xf(x) } \right\rvert} ,\end{align*} and by assumption \(xf(x) \in L^1\).

• So this justifies commuting an integral and a limit: \begin{align*} F'(y_0) &\coloneqq\lim_{y_n\to y_0} { F(y_n) - F(y_0) \over y_n - y_0} \\ &= \lim_{n\to 0} \int {h_n(x) } \,dx\\ &\overset{\text{DCT}}{=} \int \lim_{n\to\infty} h_n(x) \,dx\\ &\coloneqq\int {\frac{\partial h}{\partial y}\,}(x, y_0) \,dx\\ &\coloneqq- \int xf(x) \sin(yx) \,dx ,\end{align*} and since this limit exists and is finite, \(F\) is differentiable at \(y_0\).

• That \(F\) is continuous: \begin{align*} \lim_{y_n \to y_0} F'(y_n) &= \lim_{y_n \to y_0} \int {\frac{\partial h}{\partial y}\,}(x, y_n) \,dx\\ &\overset{\text{DCT}}{=} \int \lim_{y_n \to y_0} {\frac{\partial h}{\partial y}\,}(x, y_n) \,dx\\ &= - \int \lim_{y_n \to y_0} xf(x) \sin(y_n x) \,dx\\ &= - \int xf(x) \sin(y_0x) \,dx ,\end{align*} where we’ve used that \(y\mapsto \sin(yx)\) is clearly continuous.

Stated integral equality:

• Let \({\varepsilon}> 0\)
• \(C_c({\mathbf{R}}^n) \hookrightarrow L^1({\mathbf{R}}^n)\) is dense so choose \(\left\{{f_n}\right\} \to f\) with \({\left\lVert {f_n - f} \right\rVert}_1 \to 0\).
• Since \(\left\{{f_n}\right\}\) are compactly supported, choose \(N_0\gg 1\) such that \(f_n\) is zero outside of \(B_{N_0}(\mathbf{0})\).
• Then \begin{align*} N\geq N_0 \implies \int_{{\left\lvert {x} \right\rvert} > N} {\left\lvert {f} \right\rvert} &= \int_{{\left\lvert {x} \right\rvert} > N} {\left\lvert {f - f_n + f_n} \right\rvert} \\ &\leq \int_{{\left\lvert {x} \right\rvert} > N} {\left\lvert {f-f_n} \right\rvert} + \int_{{\left\lvert {x} \right\rvert} > N} {\left\lvert {f_n} \right\rvert} \\ &= \int_{{\left\lvert {x} \right\rvert} > N} {\left\lvert {f-f_n} \right\rvert} \\ &\leq \int_{{\left\lvert {x} \right\rvert} > N} {\left\lVert {f-f_n} \right\rVert}_1 \\ &= {\left\lVert {f_n-f} \right\rVert}_1 \qty{\int_{{\left\lvert {x} \right\rvert} > N} 1} \\ &\overset{n\to\infty}\longrightarrow 0 \qty{\int_{{\left\lvert {x} \right\rvert} > N} 1} \\ &= 0\\ &\overset{N\to\infty}\longrightarrow 0 .\end{align*}

To see that this doesn’t force \(f(x)\to 0\) as \({\left\lvert {x} \right\rvert} \to \infty\):

• Take \(f(x)\) to be a train of rectangles of height 1 and area \(1/2^j\) centered on even integers.
• Then \begin{align*}\int_{{\left\lvert {x} \right\rvert} > N} {\left\lvert {f} \right\rvert} = \sum_{j=N}^\infty 1/2^j \overset{N\to\infty}\longrightarrow 0\end{align*} as the tail of a convergent sum.
• However \(f(x) = 1\) for infinitely many even integers \(x > N\), so \(f(x) \not\to 0\) as \({\left\lvert {x} \right\rvert}\to\infty\).

• No: take \(f(x) = {1\over x\ln x}\)
• Then by a \(u{\hbox{-}}\)substitution, \begin{align*} \int_0^x f = \ln\qty{\ln (x)} \overset{x\to\infty}\longrightarrow\infty \end{align*} is unbounded, so \(f\not\in L^1([1, \infty))\).
• But \begin{align*} xf(x) = { 1 \over \ln(x)} \overset{x\to\infty} \longrightarrow 0 .\end{align*}

• Idea: first show this for characteristic functions, then simple functions, then for arbitrary \(f\).

• For characteristic functions:

  • Consider \(\chi_{A}\) for \(A\) a measurable set.

  • By regularity of the Lebesgue measure, for every \({\varepsilon}>0\) we can find an \(I_{\varepsilon}\) such that \(m(A\Delta I_{\varepsilon})< {\varepsilon}\) where \(I_{\varepsilon}\) is a finite disjoint union of intervals.

  • Then use \begin{align*} {\varepsilon}> m(A\Delta I{\varepsilon}) = \int_X {\left\lvert {\chi_A - \chi_{I_{\varepsilon}}} \right\rvert} ,\end{align*} so the \(\chi_{I_{\varepsilon}}\) converge to \(\chi_A\) in \(L_1\).

  • Then just note that \(\chi_{I_{\varepsilon}} = \sum_{j\leq N} \chi_{I_j}\) where \(I_{\varepsilon}= \coprod_{j\leq N} I_j\), so \(\chi_{I_{\varepsilon}} \in S\).

• For simple functions:

  • Let \(\psi = \sum_{k\leq N} c_k \chi_{E_k}\).
  • By the argument above, for each \(k\) we can find \(I_{{\varepsilon}, k}\) such that \(\chi_{I_{{\varepsilon}, k}}\) converges to \(\chi_{E_k}\) in \(L^1\).
  • So defining \(\psi_{\varepsilon}= \sum_{k\leq N} c_k \chi_{I_{{\varepsilon}, k}}\), the claim is that this will converge to \(\phi\) in \(L_1\).
  • Note that \begin{align*} \psi_{\varepsilon}= \sum_k c_k \chi_{I_{{\varepsilon}, k}} = \sum_k c_k \sum_j \chi_{I_{j, k} } = \sum_{k, j} c_k \chi_{ I_{j, k} } \in S \end{align*} since now the \(I_{j, k}\) are indicators of intervals.
  • Moreover \begin{align*} {\left\lVert {\psi_{\varepsilon}- \psi} \right\rVert} = {\left\lVert { \sum_k c_k \qty{ \chi_{E_k} - \chi_{I_{{\varepsilon}, k} }} } \right\rVert} \leq \sum_k c_k {\left\lVert { \chi_{E_k} - \chi_{I_{{\varepsilon}, k}} } \right\rVert} ,\end{align*} where the last norm can be bounded by the proof for characteristic functions.

• For arbitrary functions:

  • Now just use that every \(f \in L^1\) can be approximated by simple functions \(\phi_n\) so that \({\left\lVert {f-\phi_n} \right\rVert}_1 < {\varepsilon}\) for \(n \gg 1\).
  • So find \(\phi_n\to f\), and for each \(n\), find \(g_{n, k} \in S\) with \({\left\lVert {g_{n, k} - \phi_n} \right\rVert}_1 \overset{k\to \infty}\longrightarrow 0\), an approximation by functions in \(S\).
  • Then \begin{align*} {\left\lVert {f - g_{n, k}} \right\rVert} \leq {\left\lVert {f - \phi_n} \right\rVert} + {\left\lVert {\phi_n - g_{n, k}} \right\rVert} ,\end{align*} which can be made arbitrarily small.

Note that

` \begin{align*} \int_{{\mathbf{R}}^2}{\left\lvert {f} \right\rvert} ,d\mu &= \int_0^\infty \int_x^\infty x^{1\over 3}(1+xy)^{-3\over 2} ,dy,dx\ &= \int_0^\infty -2x^{-{ 2\over 3} }(1+xy)^{-{ 1\over 2} }\Big|_^ ,dx\ &= \int_0^\infty {2\over x^{2\over 3} (1+x^2)^{1\over 2}}\ &= \int_0^1 {2\over x^{2\over 3} (1+x^2)^{1\over 2}}

• \int_1^\infty {2\over x^{2\over 3} (1+x^2)^{1\over 2}} \ &= \int_0^1 {2\over x^{2\over 3} }
• \int_1^\infty {2\over x^{5\over 3} } \ &<\infty ,\end{align*} `{=html} where

• For the first term: We’ve entirely neglected the \(1+x^2\) factor, since neglecting to divide by a positive number can only make the integrand larger,

• For the second term: \begin{align*} 1+x^2\geq 0 \implies {1\over \sqrt{1+x^2}} \leq {1\over \sqrt{x^2}} = {1\over x} \end{align*}

• Both terms converge by the \(p{\hbox{-}}\)tests.

The use of iterated integration is justified by Tonelli’s theorem on \({\left\lvert {f} \right\rvert} = f\), since \(f\) is non-negative and clearly measurable on \({\mathbf{R}}^2\), and if any iterated integral is finite then it is equal to \(\int {\left\lvert {f} \right\rvert}\).

• Note that if \(m(E) = 0\) then \(\int_E f = 0\) for any \(f\).
• Toward a contradiction, suppose there exists an \({\varepsilon}>0\) such that for all \(\delta>0\) there exists a set \(E_\delta \subseteq {\mathbf{R}}\) with \(m(E) < \delta\) but \(\int_{E_\delta} {\left\lvert {f} \right\rvert} > {\varepsilon}\).
• Let \(\delta_n \searrow 0\) be any sequence converging to zero and choose \(E_n\) with \(\int_{E_n} {\left\lvert {f} \right\rvert} > {\varepsilon}\) for every \(n\).
• Define \(E \coloneqq\limsup_n E_n \coloneqq\bigcap_{N\geq 1} \bigcup_{n\geq N} E_n\), then \(m(E) = 0\) by Borel-Cantelli.
• Now estimate using Fatou: \begin{align*} \int_{E} {\left\lvert {f} \right\rvert} &= \int_X \chi_E {\left\lvert {f} \right\rvert} \\ &= \int_X \limsup_n \chi_{E_n} {\left\lvert {f} \right\rvert} \\ &\geq \limsup_n \int_X \chi_{E_n }{\left\lvert {f} \right\rvert} \\ &\geq \limsup_n \int_{E_n} {\left\lvert {f} \right\rvert} \\ &\geq \limsup_n {\varepsilon}\\ &= {\varepsilon} ,\end{align*} however \(\displaystyle\int_E {\left\lvert {f} \right\rvert}\,dm= 0\) since \(m(E) = 0\), a contradiction. \(\contradiction\).

Note that this is clear for simple functions: let \(\phi = \sum_{k\leq n} c_k m(A_k) < \infty\) be simple function. then \(\phi\) is necessarily bounded on \({\mathbf{R}}\), so let \(M\coloneqq\sup_{\mathbf{R}}\phi\) and estimate \begin{align*} \int_E \phi &\coloneqq\sum_k c_k m(A_k \cap E) \\ &\leq \sum_k M\cdot m(E)\\ &= C M m(E) ,\end{align*} for some constant \(C\), so choosing \(\delta < { {\varepsilon}\over C M}\) (and its corresponding \(E\) with \(m(E) < \delta\)) bounds this above by \({\varepsilon}\).

For arbitrary \(f \in L^1\), there is a sequence of simple functions \(\phi_n\) with \(\int \phi_n \nearrow\int f\) and \({\left\lVert {\phi_n - f} \right\rVert}_{L_1} \overset{n\to\infty}\longrightarrow 0\). Choose \(\delta\) and \(E\) as above, and use the triangle inequality to estimate \begin{align*} \int_E {\left\lvert {f} \right\rvert} &= \int_E {\left\lvert {f - \phi_n + \phi_n} \right\rvert} \\ &\leq \int_E {\left\lvert {f - \phi_n} \right\rvert} + \int_E {\left\lvert {\phi_n} \right\rvert} ,\end{align*} choose \(n\gg 1\) to bound the first term by \({\varepsilon}\), noting that the second term is bounded by \({\varepsilon}\) by the case for simple functions.

Part 1: Take a train of triangles with base points at \(k\) and \(k+1\), each of area \(2^{-k}\). Then \(\int {\left\lvert {f} \right\rvert} \approx \sum_{k\geq 0} 2^{-k} <\infty\), but \(f(x)\not\to 0\) since \(f(x) > 0\) infinitely often.

Part 2:

• Idea: use contradiction to produce a sequence with arbitrarily large terms, and bound below an integral in a ball about each point.

• Suppose \(\lim_{{\left\lvert {x} \right\rvert}\to \infty}f(x) = L > 0\).

  • Then for any \({\varepsilon}\) there exists an \(M\) such that \(x\geq M \implies {\left\lvert {f(x) - L} \right\rvert} < {\varepsilon}\), so \(L-{\varepsilon}\leq f(x) \leq L+{\varepsilon}\)
  • Choosing \({\varepsilon}=L/2\) yields \(L/2 \leq f(x) \leq 3L/2\), and so \begin{align*} \int_{\mathbf{R}}{\left\lvert {f} \right\rvert} \geq \int_{{\left\lvert {x} \right\rvert} \geq M} {\left\lvert {f} \right\rvert} \geq \int_{{\left\lvert {x} \right\rvert}\geq M} L/2 \to \infty ,\end{align*} contradicting \(f\in L^1({\mathbf{R}})\). \(\contradiction\).

• So it must be that this limit does not exist. Fix \({\varepsilon}>0\), then there are infinitely many \(x\) such that \(f(x) > {\varepsilon}\), so choose a sequence \(x_n\to \infty\) with \(f(x_n) > {\varepsilon}\) for each \(n\).

• Now use uniform continuity: pick a uniform \(\delta = \delta({\varepsilon})\) such that \(x\in B_\delta(x_n) \implies {\left\lvert {f(x) - f(x_n)} \right\rvert} < {\varepsilon}/4\).

• Now use that \(f(x_n) - {\varepsilon}/4 \leq f(x) \leq f(x_n)+{\varepsilon}/4\) implies that \(f(x) \geq 3{\varepsilon}/4\) whenever \(x\in B_\delta(x_n)\) for any \(n\) to estimate \begin{align*} \int_{B_\delta(x_n)} {\left\lvert {f(x)} \right\rvert}\,dx \geq 2\delta \cdot 3{\varepsilon}/4 \coloneqq C = C_{\delta, {\varepsilon}} > 0 ,\end{align*} where \(C\) is a constant.

• But now we’ve contradicted \(f\in L^1\): \begin{align*} \int_{\mathbf{R}}{\left\lvert {f} \right\rvert} \geq \sum_{n\geq 1} \int_{B_\delta(x_n)} {\left\lvert {f} \right\rvert} \geq \sum_{n\geq 1} C \to \infty ,\end{align*} provided we pass to a further subsequence of \(x_n\) such that the balls \(B_\delta(x_n)\) are disjoint. \(\contradiction\)

Write \begin{align*} \Gamma \coloneqq\left\{{(x, f(x)) {~\mathrel{\Big\vert}~}x\in {\mathbf{R}}}\right\} \subseteq {\mathbf{R}}^d .\end{align*} Then \begin{align*} \mu(\Gamma) &= \int_{{\mathbf{R}}^d} \chi_\Gamma \,d\mu\\ &= \int_{{\mathbf{R}}^{d-1}}\int_{\mathbf{R}}\chi_\Gamma(x, y) \,dy\,dx\\ &= \int_{{\mathbf{R}}^{d-1}} 0 \,dx\\ &= 0 ,\end{align*} using that \(\int_{\mathbf{R}}\chi_\Gamma(x, y) \,dy= 0\) since if \(x\) is fixed then \(\chi_\Gamma(x, y) = \left\{{f(x)}\right\}\) is a point with measure zero. Since \(f\) is measurable, \(\Gamma\) is a measurable set and \(\chi_\Gamma\) is measurable. Since the iterated integral was finite, the equalities are justified by Fubini-Tonelli.

• If these norms can be computed via iterated integrals, we have \begin{align*} {\left\lVert {f\ast g} \right\rVert}_1 &\coloneqq\int_{\mathbf{R}}{\left\lvert {(f\ast g)(x)} \right\rvert} \,dx\\ &\coloneqq\int_{\mathbf{R}}{\left\lvert {\int_{\mathbf{R}}H(x, y) \,dy} \right\rvert} \,dx\\ &\coloneqq\int_{\mathbf{R}}{\left\lvert {\int_{\mathbf{R}}f(y)g(x-y) \,dy} \right\rvert} \,dx\\ &\leq \int_{\mathbf{R}}\int_{\mathbf{R}}{\left\lvert {f(y) g(x-y)} \right\rvert} \,dx\,dy\\ &\coloneqq\int_{\mathbf{R}}\int_{\mathbf{R}}{\left\lvert {H(x ,y)} \right\rvert}\,dx\,dy\\ &\coloneqq\int_{{\mathbf{R}}^2} {\left\lvert {H} \right\rvert} \,d\mu_{{\mathbf{R}}^2} \\ &\coloneqq{\left\lVert {H} \right\rVert}_{L^1({\mathbf{R}}^2)} .\end{align*} So it suffices to show \({\left\lVert {H} \right\rVert}_1 < \infty\).

• A preliminary computation, the validity of which we will show afterward: \begin{align*} {\left\lVert {H} \right\rVert}_1 &\coloneqq{\left\lVert {H} \right\rVert}_{L^1({\mathbf{R}}^2)} \\ &= \int _{\mathbf{R}}\qty{ \int_{\mathbf{R}}{\left\lvert {f(y)g(x-y)} \right\rvert} \, dy } \, dx && \text{Tonelli} \\ &= \int _{\mathbf{R}}\qty{ \int_{\mathbf{R}}{\left\lvert {f(y)g(x-y)} \right\rvert} \, dx} \, dy && \text{Tonelli} \\ &= \int _{\mathbf{R}}\qty{ \int_{\mathbf{R}}{\left\lvert {f(y)g(t)} \right\rvert} \, dt} \, dy && \text{setting } t=x-y, \,dt = - dx \\ &= \int _{\mathbf{R}}\qty{ \int_{\mathbf{R}}{\left\lvert {f(y)} \right\rvert}\cdot {\left\lvert {g(t)} \right\rvert} \, dt}\, dy \\ &= \int _{\mathbf{R}}{\left\lvert {f(y)} \right\rvert} \cdot \qty{ \int_{\mathbf{R}}{\left\lvert {g(t)} \right\rvert} \, dt}\, dy \\ &\coloneqq\int _{\mathbf{R}}{\left\lvert {f(y)} \right\rvert} \cdot {\left\lVert {g} \right\rVert}_1 \,dy \\ &= {\left\lVert {g} \right\rVert}_1 \int _{\mathbf{R}}{\left\lvert {f(y)} \right\rvert} \,dy &&\text{the norm is a constant} \\ &\coloneqq{\left\lVert {g} \right\rVert}_1 {\left\lVert {f} \right\rVert}_1 \\ &< \infty && \text{by assumption} .\end{align*}

• We’ve used Tonelli twice: to equate the integral to the iterated integral, and to switch the order of integration, so it remains to show the hypothesis of Tonelli are fulfilled.

• Apply Tonelli to \({\left\lvert {H} \right\rvert}\)
  • \(H\) measurable implies \({\left\lvert {H} \right\rvert}\) is measurable.
  • \({\left\lvert {H} \right\rvert}\) is non-negative.
  • So the iterated integrals are equal in the extended sense
  • The calculation shows the iterated integral is finite, so \(\int {\left\lvert {H} \right\rvert}\) is finite and \(H\) is thus integrable on \({\mathbf{R}}^2\).

Note: Fubini is not needed, since we’re not calculating the actual integral, just showing \(H\) is integrable.

Note: \(f\) is a function \({\mathbf{R}}\to {\mathbf{R}}\) in the original problem, but here I’ve assumed \(f:{\mathbf{R}}^n\to {\mathbf{R}}\).

• Since \(f\geq 0\), set \begin{align*} E\coloneqq\left\{{(x, t) \in {\mathbf{R}}^{n} \times{\mathbf{R}}{~\mathrel{\Big\vert}~}f(x) > t}\right\} = \left\{{(x, t) \in {\mathbf{R}}^n \times{\mathbf{R}}{~\mathrel{\Big\vert}~}0 \leq t < f(x)}\right\} .\end{align*}

• Claim: since \(f\) is measurable, \(E\) is measurable and thus \(m(E)\) makes sense.

  • Since \(f\) is measurable, \(F(x, t) \coloneqq t - f(x)\) is measurable on \({\mathbf{R}}^n \times{\mathbf{R}}\).
  • Then write \(E = \left\{{F < 0}\right\} \cap\left\{{t\geq 0}\right\}\) as an intersection of measurable sets.

• We have slices \begin{align*} E^t &\coloneqq\left\{{x\in {\mathbf{R}}^n {~\mathrel{\Big\vert}~}(x, t) \in E}\right\} = \left\{{x\in {\mathbf{R}}^n {~\mathrel{\Big\vert}~}0 \leq t < f(x)}\right\} \\ E^x &\coloneqq\left\{{t\in {\mathbf{R}}{~\mathrel{\Big\vert}~}(x, t) \in E}\right\} = \left\{{t\in {\mathbf{R}}{~\mathrel{\Big\vert}~}0 \leq t \leq f(x)}\right\} = [0, f(x)] .\end{align*}

  • \(E_t\) is precisely the set that appears in the original RHS integrand.
  • \(m(E^x) = f(x)\).

• Claim: \(\chi_E\) satisfies the conditions of Tonelli, and thus \(m(E) = \int \chi_E\) is equal to any iterated integral.

  • Non-negative: clear since \(0\leq \chi_E \leq 1\)
  • Measurable: characteristic functions of measurable sets are measurable.

• Conclude:

  • For almost every \(x\), \(E^x\) is a measurable set, \(x\mapsto m(E^x)\) is a measurable function, and \(m(E) = \int_{{\mathbf{R}}^n} m(E^x) \, dx\)
  • For almost every \(t\), \(E^t\) is a measurable set, \(t\mapsto m(E^t)\) is a measurable function, and \(m(E) = \int_{{\mathbf{R}}} m(E^t) \, dt\)

• On one hand, \begin{align*} m(E) &= \int_{{\mathbf{R}}^{n+1}} \chi_E(x, t) \\ &= \int_{{\mathbf{R}}} \int_{{\mathbf{R}}^n} \chi_E(x, t) \,dt \,dx \quad\text{by Tonelli}\\ &= \int_{{\mathbf{R}}^n} m(E^x) \,dx \quad\text{first conclusion}\\ &= \int_{{\mathbf{R}}^n} f(x) \,dx .\end{align*}

• On the other hand, \begin{align*} m(E) &= \int_{{\mathbf{R}}^{n+1}} \chi_E(x, t) \\ &= \int_{\mathbf{R}}\int_{{\mathbf{R}}^n} \chi_E(x, t) \, dx \,dt \quad\text{by Tonelli} \\ &= \int_{\mathbf{R}}m(E^t) \,dt \quad\text{second conclusion} .\end{align*}

• Thus \begin{align*} \int_{{\mathbf{R}}^n} f \,dx = m(E) = \int_{\mathbf{R}}m(E^t) \,dt = \int_{\mathbf{R}}m\qty{\left\{{x{~\mathrel{\Big\vert}~}f(x) > t}\right\}} .\end{align*}

This is a direct application of Fubini-Tonelli: \begin{align*} {\left\lVert {A_\delta f} \right\rVert} &\coloneqq\int {\left\lvert { \int f(x-y)\delta^{-1}\varphi(\delta^{-1}y)\,dy} \right\rvert} \,dx\\ &\leq \int \int {\left\lvert {f(x-y)\delta^{-1}\varphi(\delta^{-1}y)} \right\rvert} \,dy\,dx\\ &\overset{FT}{=} \int \int {\left\lvert { f(x-y) } \right\rvert} \cdot {\left\lvert {\delta^{-1}\varphi(\delta^{-1}y)} \right\rvert} \,dx\,dy\\ &= \int {\left\lvert { \delta^{-1}\varphi(\delta^{-1}y) } \right\rvert} \qty{ \int {\left\lvert { f(x-y) } \right\rvert}\,dx} \,dy\\ &= \int {\left\lvert { \delta^{-1}\varphi(\delta^{-1}y)} \right\rvert}\cdot {\left\lVert {f} \right\rVert} \,dy\\ &= {\left\lVert {f} \right\rVert} \cdot \int {\left\lvert { \delta^{-1}\varphi(\delta^{-1}y) } \right\rvert} \,dy\\ &= {\left\lVert {f} \right\rVert} \cdot {\left\lVert {\varphi} \right\rVert} .\end{align*} Here we’ve used translation and dilation invariance of the Lebesgue integral.

Claim: if \(f\in L^1[0, 1]\) and \(\widehat{f}\in \ell^1({\mathbf{Z}})\), then \(S_Nf \to f\) uniformly.

• Since \(\widehat{f}\in \ell^1({\mathbf{Z}})\), we have \(S_Nf\to g\) uniformly for some continuous \(g\) by the \(M{\hbox{-}}\)test.

• Now consider \(\widehat{g}\). We have \begin{align*} \widehat{g}(n) = \int_0^1 \sum_m \qty{\widehat{f}(m)e_m(x)}e_{-n}(x) \,dx= \widehat{f}(n) ,\end{align*} using that \(\int_0^1 e_n(x)\,dx= \chi_{n=0}\).

• We’ll now show \(f-g= 0\) a.e. by mollifying against an approximate identity \(\varphi\in L^1\), setting \begin{align*} \varphi_{\varepsilon}(x) \coloneqq{\varepsilon}^{-1}\varphi({\varepsilon}^{-1}x) \in L^1[0, 1] .\end{align*}

• A computation: \begin{align*} \widehat{f\ast\varphi_{\varepsilon}}(n) &= \widehat{f}\cdot \widehat{\varphi_{\varepsilon}}(n) \\ &= \widehat{g}\cdot \widehat{\varphi_{\varepsilon}}(n) \\ &= \widehat{g\ast\varphi_{\varepsilon}}(n) ,\end{align*} so \begin{align*} \widehat{(f-g)\ast\varphi_{\varepsilon}} = 0 \quad \forall n \implies (f-g)\ast\varphi_{\varepsilon}\equiv 0 ,\end{align*} using that \((f-g)\ast\varphi_{\varepsilon}\in L^2\) and \(\left\{{e_n}\right\}\) for a complete orthonormal basis of \(L^2\).

• Now use that \((f-g)\ast\varphi_{\varepsilon}\to (f-g)\) in \(L^1\) and \((f-g)\ast\varphi_{\varepsilon}\equiv 0\) to conclude \(f-g = 0\) a.e.

Note: this proof shows \(L^2(X) \subseteq L^1(X)\) whenever \(\mu(X) < \infty\).

• First, \(S_Nf\) converges in \({\mathcal{H}}\) to something, say \(g \coloneqq\lim_{n\to\infty} S_n f\), since \begin{align*} {\left\lVert {g - S_Nf} \right\rVert} = {\left\lVert {\sum_{{\left\lvert {n} \right\rvert} \geq N} \widehat{f} (n) e_n(x) } \right\rVert} \leq \sum_{{\left\lvert {n} \right\rvert} \geq N } {\left\lvert {\widehat{f}(n)} \right\rvert} \overset{N\to\infty}\longrightarrow 0 ,\end{align*} where the last term is the tail of a convergent sum since \(\left\{{\widehat{f}(n)}\right\} \in \ell^1\).

• This also shows that \(S_N\to g\) uniformly.

• \(g\) is continuous, as the uniform limit of continuous functions.

• Showing that \(g = f\) a.e.: it suffices to show that \(S_N\) converges to \(f\) in \(L^p\) for some \(p\), since this will provide a subsequence that converges to \(f\) a.e..

• Claim: \(\widehat{f}\in \ell^1 \subseteq \ell^2\) implies that \(f \in L^2\). This follows from Parseval: \begin{align*} \infty > {\left\lVert {\widehat{f}} \right\rVert}_{\ell^2}^2 = \sum_{n\in {\mathbf{Z}}} {\left\lvert {\widehat{f}(n)} \right\rvert}^2 = \int_0^1 {\left\lvert {f(z)} \right\rvert}^2 \,dz = {\left\lVert {f} \right\rVert}_{L^2}^2 .\end{align*}

• Claim: \(S_N\to f\) in \(L^2\).

  • This follows from the fact that \(\left\{{e_n}\right\}_{n\in {\mathbf{Z}}}\) is a complete orthonormal basis, so \(f = \sum {\left\langle {f},~{e_n} \right\rangle}e_n\) uniquely, recognizing \(\widehat{f}(n) = {\left\langle {f},~{e_n} \right\rangle}\), and writing \begin{align*} f = \sum_{n} {\left\langle {f},~{e_n} \right\rangle}e_n = \sum_n \widehat{f}(n) e_n \coloneqq\lim_{N\to\infty }S_N f .\end{align*}

• So a subsequence \(\left\{{S_{N_k}}\right\}_{k\geq 0}\) converges to \(f\) a.e.. Since \(S_N\to g\) a.e., \(f=g\) a.e. by uniqueness of limits.

\hfill

\hfill

Part a

Lemma: If \(\varphi \in C_c^1\), then \((f \ast \varphi)' = f \ast \varphi'\) almost everywhere.

Silly Proof:

\begin{align*} \mathcal{F}( (f \ast \varphi)' ) &= 2\pi i \xi ~\mathcal{F}(f\ast \varphi) \\ &= 2\pi i \xi ~ \mathcal{F}(f) ~ \mathcal{F}(\varphi) \\ &= \mathcal{F}(f) \cdot \left( 2\pi i \xi ~\mathcal{F}(\varphi)\right) \\ &= \mathcal{F}(f) \cdot \mathcal{F}(\varphi') \\ &= \mathcal{F}(f\ast \varphi') .\end{align*}

Actual proof:

\begin{align*} (f\ast \varphi)'(x) &= (\varphi\ast f)'(x) \\ &= \lim_{h\to 0} \frac{(\varphi\ast f)'(x+h) - (\varphi\ast f)'(x)}{h} \\ &= \lim_{h\to 0} \int \frac{\varphi(x + h - y) - \varphi(x - y)}{h} f(y) \\ &\overset{DCT}= \int \lim_{h\to 0} \frac{\varphi(x + h - y) - \varphi(x - y)}{h} f(y) \\ &= \int \varphi'(x-y) f(y) \\ &= (\varphi' \ast f)(x) \\ &= (f \ast \varphi')(x) .\end{align*}

To see that the DCT is justified, we can apply the MVT on the interval \([0, h]\) to \(f\) to obtain

\begin{align*} \frac{\varphi(x + h - y) - \varphi(x - y)}{h} &= \varphi'(c) \quad c\in [0, h] ,\end{align*}

and since \(\varphi'\) is continuous and compactly supported, \(\varphi'\) is bounded by some \(M < \infty\) by the extreme value theorem and thus \begin{align*} \int {\left\lvert {\frac{\varphi(x + h - y) - \varphi(x - y)}{h} f(y)} \right\rvert} &= \int {\left\lvert {\varphi'(c) f(y)} \right\rvert} \\ &\leq \int {\left\lvert {M} \right\rvert}{\left\lvert {f} \right\rvert} \\ &= {\left\lvert {M} \right\rvert} \int {\left\lvert {f} \right\rvert} < \infty ,\end{align*}

since \(f\in L^1\) by assumption, so we can take \(g\coloneqq{\left\lvert {M} \right\rvert} {\left\lvert {f} \right\rvert}\) as the dominating function.

Applying this theorem infinitely many times shows that \(f\ast \varphi\) is smooth.

To see that \(f\ast \varphi\) is compactly supported, approximate \(f\) by a continuous compactly supported function \(h\), so \({\left\lVert {h - f} \right\rVert}_1 \overset{L^1}\to 0\).

Now let \(g_x(y) = \varphi(x-y)\), and note that \(\mathrm{supp}(g) = x - \mathrm{supp}(\varphi)\) which is still compact.

But since \(\mathrm{supp}(h)\) is bounded, there is some \(N\) such that \begin{align*} {\left\lvert {x} \right\rvert} > N \implies A_x\coloneqq\mathrm{supp}(h) \cap\mathrm{supp}(g_x) = \emptyset \end{align*}

and thus \begin{align*} (h\ast \varphi)(x) &= \int_{\mathbf{R}}\varphi(x-y) h(y)~dy \\ &= \int_{A_x} g_x(y) h(y) \\ &= 0 ,\end{align*}

so \(\left\{{x {~\mathrel{\Big\vert}~}f\ast g(x) = 0}\right\}\) is open, and its complement is closed and bounded and thus compact.

Part b

\begin{align*} {\left\lVert {f\ast K_j - f} \right\rVert}_1 &= \int {\left\lvert {\int f(x-y) K_j(y) ~dy - f(x)} \right\rvert}~dx \\ &= \int {\left\lvert {\int f(x-y) K_j(y) ~dy - \int f(x) K_j(y) ~ dy} \right\rvert}~dx \\ &= \int {\left\lvert {\int ( f(x-y) - f(x) ) K_j(y) ~dy } \right\rvert} ~dx \\ &\leq \int \int {\left\lvert {(f(x-y) - f(x))} \right\rvert} \cdot {\left\lvert {K_j(y)} \right\rvert} ~ dy~dx \\ &\overset{FT}= \int \int {\left\lvert {(f(x-y) - f(x))} \right\rvert} \cdot {\left\lvert {K_j(y)} \right\rvert} \mathbf{~ dx~dy}\\ &= \int {\left\lvert {K_j(y)} \right\rvert} \left( \int {\left\lvert {(f(x-y) - f(x))} \right\rvert} ~ dx\right) ~dy \\ &= \int {\left\lvert {K_j(y)} \right\rvert} \cdot {\left\lVert {f - \tau_y f} \right\rVert}_1 ~dy .\end{align*}

We now split the integral up into pieces.

• Chose \(\delta\) small enough such that \({\left\lvert {y} \right\rvert} < \delta \implies {\left\lVert {f - \tau_y f} \right\rVert}_1 < \varepsilon\) by continuity of translation in \(L^1\), and

• Since \(\varphi\) is compactly supported, choose \(J\) large enough such that \begin{align*} j > J \implies \int_{{\left\lvert {y} \right\rvert} \geq \delta} {\left\lvert {K_j(y)} \right\rvert} ~dy = \int_{{\left\lvert {y} \right\rvert} \geq \delta} {\left\lvert {j\varphi(jy)} \right\rvert} = 0 \end{align*}

Then ` \begin{align*} {\left\lVert {f\ast K_j - f} \right\rVert}_1 &\leq \int {\left\lvert {K_j(y)} \right\rvert} \cdot {\left\lVert {f - \tau_y f} \right\rVert}1 ~dy \ &= \int{{\left\lvert {y} \right\rvert} < \delta} {\left\lvert {K_j(y)} \right\rvert} \cdot {\left\lVert {f - \tau_y f} \right\rVert}_1 ~dy

• \int_{{\left\lvert {y} \right\rvert} \geq \delta} {\left\lvert {K_j(y)} \right\rvert} \cdot {\left\lVert {f - \tau_y f} \right\rVert}1 ~dy \ &= \varepsilon \int{{\left\lvert {y} \right\rvert} \geq \delta} {\left\lvert {K_j(y)} \right\rvert} + 0 \ &\leq \varepsilon(1) \to 0 .\end{align*} `{=html}

Let \(\Lambda \in L^1(X) {}^{ \vee }\) be arbitrary.

• Denote this norm \({\left\lVert {{-}} \right\rVert}_u\)

• Let \(f_n\) be a Cauchy sequence in this space, so \({\left\lVert {f_n} \right\rVert}_u < \infty\) for every \(n\) and \({\left\lVert {f_j - f_k} \right\rVert}_u \overset{j, k\to\infty}\to 0\).

and define a candidate limit: for each \(x\in I\), set \begin{align*}f(x) \coloneqq\lim_{n\to\infty} f_n(x).\end{align*}

• Note that \begin{align*} {\left\lVert {f_n} \right\rVert}_\infty &\leq {\left\lVert {f_n} \right\rVert}_u < \infty \\ {\left\lVert {f_n'} \right\rVert}_\infty &\leq {\left\lVert {f_n} \right\rVert}_u < \infty .\end{align*}

  • Thus both \(f_n, f_n'\) are Cauchy sequences in \(C^0([a, b], {\left\lVert {{-}} \right\rVert}_\infty)\), which is a Banach space, so they converge.

• So

  • \(f_n \to f\) uniformly (by uniqueness of limits),
  • \(f_n' \to g\) uniformly for some \(g\), and
  • \(f, g\in C^0([a, b])\).

• Claim: \(g = f'\)

  • For any fixed \(a\in I\), we have \begin{align*} f_n(x) - f_n(a) \quad &\overset{u}\to f(x) - f(a) \\ \int_a^x f'_n \quad &\overset{u}\to \int_a^x g .\end{align*}
  • By the FTC, the left-hand sides are equal.
  • By uniqueness of limits so are the right-hand sides, so \(f' = g\).

• Claim: the limit \(f\) is an element in this space.

  • Since \(f, f'\in C^0([a, b])\), they are bounded, and so \({\left\lVert {f} \right\rVert}_u < \infty\).

• Claim: \({\left\lVert {f_n - f} \right\rVert}_u \overset{n\to\infty}\to 0\)

• Thus the Cauchy sequence \(\left\{{f_n}\right\}\) converges to a function \(f\) in the \(u{\hbox{-}}\)norm where \(f\) is an element of this space, making it complete.

\hfill

Let \(\left\{{f_k}\right\}\) be a Cauchy sequence, so \({\left\lVert {f_k} \right\rVert} < \infty\) for all \(k\). Then for a fixed \(x\), the sequence \(f_k(x)\) is Cauchy in \({\mathbf{R}}\) and thus converges to some \(f(x)\), so define \(f\) by \(f(x) \coloneqq\lim_{k\to\infty} f_k(x)\).

Then \({\left\lVert {f_k - f} \right\rVert} = \max_{x\in X}{\left\lvert {f_k(x) - f(x)} \right\rvert} \overset{k\to\infty}\to 0\), and thus \(f_k \to f\) uniformly and thus \(f\) is continuous. It just remains to show that \(f\) has bounded norm.

Choose \(N\) large enough so that \({\left\lVert {f - f_N} \right\rVert} < \varepsilon\), and write \({\left\lVert {f_N} \right\rVert} \coloneqq M < \infty\)

\begin{align*} {\left\lVert {f} \right\rVert} \leq {\left\lVert {f - f_N} \right\rVert} + {\left\lVert {f_N} \right\rVert} < \varepsilon + M < \infty .\end{align*}

For the first, use that \begin{align*} {\left\lvert { \sum_{k\geq 1} {1\over k^2} \sin^n(k) } \right\rvert} \leq \sum_{k\geq 1} {\left\lvert { {1\over k^2} \sin^n(k) } \right\rvert} \sum_{k\geq 1} {\left\lvert { {1\over k^2}} \right\rvert} < \infty ,\end{align*} since \({\left\lvert {\sin(x)} \right\rvert} \leq 1\) and \(x^n < x\) for \({\left\lvert {x} \right\rvert}\leq 1\). By the dominated convergence theorem, we can pass the limit inside. Using the same fact as above, \(\lim_{n\to\infty}\sin^n(x) = 0\),

For the second, the claim is that it diverges (very slowly). Note that \(\lim_{n\to\infty} e^{-k/n} = 1\) for any \(k\). By Fatou, we have \begin{align*} \liminf_{n\to\infty} \sum_{k\geq 1} {e^{-k/n} \over k} \geq \sum_{k\geq 1} \liminf_{n\to\infty} {e^{-k/n} \over k} = \sum_{k\geq 1} {1 \over k} = \infty .\end{align*}

Apply Fubini-Tonelli to commute two sums: \begin{align*} \mu\qty{\bigcup_{1\leq k \leq M} E_k}\coloneqq &= \sum_{n\geq 1} \mu_n\qty{\bigcup_{1\leq k \leq M} E_k}\\ &= \sum_{n\geq 1} \sum_{1\leq k \leq M} \mu_n\qty{E_k}\\ &= \sum_{1\leq k \leq M}\sum_{n\geq 1} \mu_n\qty{E_k} \text{FT} \\ &\coloneqq\sum_{1\leq k \leq M} \mu(E_k) .\end{align*}

Use a stronger result: a continuous function on a compact metric is uniformly continuous. Fix \({\varepsilon}\). Suppose \(f\) is continuous, then for each \(z\in X\) choose \(\delta_z = \delta({\varepsilon}, z)\) to ensure \(B_\delta(z) \hookrightarrow B_{\varepsilon}(f(z))\) and form the cover \(\left\{{B_{\delta_z} (z)}\right\}_{x\in X}\rightrightarrows X\). By compactness, choose a finite subcover corresponding to \(\left\{{z_1, \cdots, z_m}\right\}\) and choose \(\delta = \min \left\{{\delta_1, \cdots, \delta_m}\right\}\). The claim is that this \(\delta\) works for uniform continuity: if \({\left\lvert {x-y} \right\rvert} < \delta\) then \({\left\lvert {x-y} \right\rvert} < \delta_i\) for all \(i\). Note that \(x\in B_{\delta_z}(z)\) for one of the finitely many \(z\) above, and if we adjust \(\delta\) to \(\delta/2\), we can arrange so that both \(x, y\in B_{\delta_z}(z)\) for some \(z\), since \begin{align*} {\left\lvert {x-y} \right\rvert} = {\left\lvert {x-z+z-y} \right\rvert} \leq {\left\lvert {x-z} \right\rvert} + {\left\lvert {z-y} \right\rvert} < {\delta \over 2 } + {\delta \over 2} = \delta < \delta_{z} ,\end{align*} and similarly \begin{align*} {\left\lvert {f(x)-f(y) } \right\rvert} \leq {\left\lvert {f(x)-f(z)} \right\rvert} + {\left\lvert {f(z)-f(y)} \right\rvert} < {\varepsilon}+ {\varepsilon} ,\end{align*} so just adjust the original \({\varepsilon}\) chosen by the continuity of \(f\) to \({\varepsilon}/2\).

Todo

Todo

Todo

• Take \(\left\{{a_n}\right\} \in \ell^2\), then note that \(\sum {\left\lvert {a_n} \right\rvert}^2 < \infty \implies\) the tails vanish.

• Define \(x \coloneqq\displaystyle\lim_{N\to\infty} S_N\) where \(S_N = \sum_{k=1}^N a_k u_k\)

• \(\left\{{S_N}\right\}\) is Cauchy and \(H\) is complete, so \(x\in H\).

• By construction, \begin{align*} {\left\langle {x},~{u_n} \right\rangle} = {\left\langle {\sum_k a_k u_k},~{u_n} \right\rangle} = \sum_k a_k {\left\langle {u_k},~{u_n} \right\rangle} = a_n \end{align*} since the \(u_k\) are all orthogonal.

• By Pythagoras since the \(u_k\) are normal, \begin{align*} {\left\lVert {x} \right\rVert}^2 = {\left\lVert {\sum_k a_k u_k} \right\rVert}^2 = \sum_k {\left\lVert {a_k u_k} \right\rVert}^2 = \sum_k {\left\lvert {a_k} \right\rvert}^2 .\end{align*}

Let \(x\) and \(u_n\) be arbitrary.

` \begin{align*} {\left\langle {x - \sum_^\infty {\left\langle {x},~{u_k} \right\rangle}u_k },~{u_n} \right\rangle} &= {\left\langle {x},~{u_n} \right\rangle}

{\left\langle {\sum_^\infty {\left\langle {x},~{u_k} \right\rangle}u_k },~{u_n} \right\rangle} \ &= {\left\langle {x},~{u_n} \right\rangle}

\sum_^\infty {\left\langle {{\left\langle {x},~{u_k} \right\rangle}u_k },~{u_n} \right\rangle} \ &= {\left\langle {x},~{u_n} \right\rangle}

\sum_^\infty {\left\langle {x},~{u_k} \right\rangle} {\left\langle {u_k },~{u_n} \right\rangle} \ &= {\left\langle {x},~{u_n} \right\rangle} - {\left\langle {x},~{u_n} \right\rangle} = 0 \ \implies x - \sum_^\infty {\left\langle {x},~{u_k} \right\rangle}u_k &= 0 \quad\text{by completeness} .\end{align*} `{=html}

So \begin{align*} x = \sum_{k=1}^\infty {\left\langle {x},~{u_k} \right\rangle} u_k \implies {\left\lVert {x} \right\rVert}^2 = \sum_{k=1}^\infty {\left\lvert {{\left\langle {x},~{u_k} \right\rangle}} \right\rvert}^2. \hfill\blacksquare .\end{align*}

\begin{align*} \int f^p &= \int_{x < 1} f^p + \int_{x=1}f^p + \int_{x > 1} f^p\\ &= \int_{x < 1} f^p + \int_{x=1}1 + \int_{x > 1} f^p \\ &= \int_{x < 1} f^p + m(\left\{{f = 1}\right\}) + \int_{x > 1} f^p \\ &\overset{p\to\infty}\to 0 + m(\left\{{f = 1}\right\}) + \begin{cases} 0 & m(\left\{{x\geq 1}\right\}) = 0 \\ \infty & m(\left\{{x\geq 1}\right\}) > 0. \end{cases} \end{align*}