• 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 \(\sum a_n^2 < \infty\).