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 \({\mathbb{R}}\) 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 \(1\in L^1(0, \alpha)\).

\hfill

\hfill

• 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.

\hfill

\hfill

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_{\mathbb{R}}\chi_{[0, n]} \qty{ 1 + {x^2\over n}}^{-(n+1)} \, dx \\ &= \int_{\mathbb{R}}\lim_{n\to\infty} \chi_{[0, n]} \qty{ 1 + {x^2\over n}}^{-(n+1)} \, dx {\quad \operatorname{by the DCT} \quad} \\ &= \int_{\mathbb{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_{\mathbb{R}}e^{-x^2}\, dx}^2 &= \qty{\int_{\mathbb{R}}e^{-x^2}\,dx} \qty{\int_{\mathbb{R}}e^{-y^2}\,dx} \\ &= \int_{\mathbb{R}}\int_{\mathbb{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 \(\int_0^\infty f = {1\over 2} \int_{\mathbb{R}}f\).

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 \(1 + {x^2 \over n} \geq 1\)

\hfill

\hfill

\(L^2\) 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.

\hfill

\hfill

• 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*}

\hfill

\hfill

\begin{align*} {\frac{\partial }{\partial t}\,} F(t) &= {\frac{\partial }{\partial t}\,} \int_{\mathbb{R}}f(x) \cos(xt) ~dx \\ &\overset{DCT}= \int_{\mathbb{R}}f(x) {\frac{\partial }{\partial t}\,} \cos(xt) ~dx \\ &= \int_{\mathbb{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 \(xf(x) \in L^1({\mathbb{R}})\) by assumption, the DCT applies.

\hfill

\hfill

• 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 \(\lim_{n\to\infty} \int f_n = \int f\).

\hfill

\hfill

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.

\hfill

\hfill

• 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 \(g_n\subset C_c^\infty({\mathbb{R}}^1)\) 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_{\mathbb{R}}{\left\lvert {\tau_x f(y) - \tau_x g(y)} \right\rvert}\, dy \\ &= \int_{\mathbb{R}}{\left\lvert {f(y-x) - g(y-x)} \right\rvert}\, dy \\ &= \int_{\mathbb{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 \(g\in C_c^\infty({\mathbb{R}}^1)\), let \(E = \mathop{\mathrm{supp}}(g)\), and write \begin{align*} {\left\lVert {\tau_x g - g} \right\rVert}_1 &= \int_{\mathbb{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}