# Integrals: Approximation

## Fall 2021.2 #real_analysis/qual/completed

• Let $$F \subset \mathbb{R}$$ be closed, and define \begin{align*} \delta_{F}(y):=\inf _{x \in F}|x-y| . \end{align*} For $$y \notin F$$, show that \begin{align*} \int_{F}|x-y|^{-2} d x \leq \frac{2}{\delta_F(y)}, \end{align*}

• Let $$F \subset \mathbb{R}$$ be a closed set whose complement has finite measure, i.e. $$m({\mathbf{R}}\setminus F)< \infty$$. Define the function \begin{align*} I(x):=\int_{\mathbb{R}} \frac{\delta_{F}(y)}{|x-y|^{2}} d y \end{align*} Prove that $$I(x)=\infty$$ if $$x \not\in F$$, however $$I(x)<\infty$$ for almost every $$x \in F$$.

Hint: investigate $$\int_{F} I(x) d x$$.

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*} 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…

## Spring 2018.3 #real_analysis/qual/completed

Let $$f$$ be a non-negative measurable function on $$[0, 1]$$.

Show that \begin{align*} \lim _{p \rightarrow \infty}\left(\int_{[0,1]} f(x)^{p} d x\right)^{\frac{1}{p}}=\|f\|_{\infty}. \end{align*}

• $${\left\lVert {f} \right\rVert}_\infty \coloneqq\inf_t {\left\{{ t{~\mathrel{\Big\vert}~}m\qty{\left\{{x\in {\mathbf{R}}^n {~\mathrel{\Big\vert}~}f(x) > t}\right\}} = 0 }\right\} }$$, i.e. this is the lowest upper bound that holds almost everywhere.

• $${\left\lVert {f} \right\rVert}_p \leq {\left\lVert {f} \right\rVert}_\infty$$:
• 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 $${\left\lVert {f} \right\rVert}_p \geq {\left\lVert {f} \right\rVert}_\infty$$.

## Spring 2018.4 #real_analysis/qual/completed

Let $$f\in L^2([0, 1])$$ and suppose \begin{align*} \int _{[0,1]} f(x) x^{n} d x=0 \text { for all integers } n \geq 0. \end{align*} Show that $$f = 0$$ almost everywhere.

• Weierstrass Approximation: A continuous function on a compact set can be uniformly approximated by polynomials.
• $$C^1([0, 1])$$ is dense in $$L^2([0, 1])$$
• Polynomials are dense in $$L^p(X, \mathcal{M}, \mu)$$ for any $$X\subseteq {\mathbf{R}}^n$$ compact and $$\mu$$ a finite measure, for all $$1\leq p < \infty$$.
• Use Weierstrass Approximation, then uniform convergence implies $$L^p(\mu)$$ convergence by DCT.

• Fix $$k \in {\mathbf{Z}}$$.

• Since $$e^{2\pi i k x}$$ is continuous on the compact interval $$[0, 1]$$, it is uniformly continuous.

• Thus there is a sequence of polynomials $$P_\ell$$ such that \begin{align*} P_{\ell, k} \overset{\ell\to\infty}\longrightarrow e^{2\pi i k x} \text{ uniformly on } [0,1] .\end{align*}

• Note applying linearity to the assumption $$\int f(x) \, x^n$$, we have \begin{align*} \int f(x) x^n \,dx = 0 ~\forall n \implies \int f(x) p(x) \,dx = 0 \end{align*} for any polynomial $$p(x)$$, and in particular for $$P_{\ell, k}(x)$$ for every $$\ell$$ and every $$k$$.

• But then
\begin{align*} {\left\langle {f},~{e_k} \right\rangle} &= \int_0^1 f(x) e^{-2\pi i k x} ~dx \\ &= \int_0^1 f(x) \lim_{\ell \to \infty} P_\ell(x) \\ &= \lim_{\ell \to \infty} \int_0^1 f(x) P_\ell(x) \quad\quad \text{by uniform convergence on a compact interval} \\ &= \lim_{\ell \to \infty} 0 \quad\text{by assumption}\\ &= 0 \quad \forall k\in {\mathbf{Z}} ,\end{align*} so $$f$$ is orthogonal to every $$e_k$$.

• Thus $$f\in S^\perp \coloneqq\mathop{\mathrm{span}}_{\mathbf{C}}\left\{{e_k}\right\}_{k\in {\mathbf{Z}}}^\perp \subseteq L^2([0, 1])$$, but since this is a basis, $$S$$ is dense and thus $$S^\perp = \left\{{0}\right\}$$ in $$L^2([0, 1])$$.

• Thus $$f\equiv 0$$ in $$L^2([0, 1])$$, which implies that $$f$$ is zero almost everywhere.

• By density of polynomials, for $$f\in L^2([0, 1])$$ choose $$p_{\varepsilon}(x)$$ such that $${\left\lVert {f - p_{\varepsilon}} \right\rVert} < {\varepsilon}$$ by Weierstrass approximation.

• Then on one hand, \begin{align*} {\left\lVert {f(f-p_{\varepsilon})} \right\rVert}_1 &= {\left\lVert {f^2} \right\rVert}_1 - {\left\lVert {f\cdot p_{\varepsilon}} \right\rVert}_1 \\ &= {\left\lVert {f^2} \right\rVert}_1 - 0 \quad\text{by assumption} \\ &= {\left\lVert {f} \right\rVert}_2^2 .\end{align*}

• Where we’ve used that $${\left\lVert {f^2} \right\rVert}_1 = \int {\left\lvert {f^2} \right\rvert} = \int {\left\lvert {f} \right\rvert}^2 = {\left\lVert {f} \right\rVert}_2^2$$.
• On the other hand \begin{align*} {\left\lVert {f(f-p_{\varepsilon})} \right\rVert} &\leq {\left\lVert {f} \right\rVert}_1 {\left\lVert {f-p_{\varepsilon}} \right\rVert}_\infty \quad\text{by Holder} \\ &\leq {\varepsilon}{\left\lVert {f} \right\rVert}_1 \\ &\leq {\varepsilon}{\left\lVert {f} \right\rVert}_2 \sqrt{m(X)} \\ &= {\varepsilon}{\left\lVert {f} \right\rVert}_2 \quad\text{since } m(X)= 1 .\end{align*}

• Where we’ve used that $${\left\lVert {fg} \right\rVert}_1 = \int {\left\lvert {fg} \right\rvert} = \int {\left\lvert {f} \right\rvert}{\left\lvert {g} \right\rvert} \leq \int {\left\lVert {f} \right\rVert}_\infty {\left\lvert {g} \right\rvert} = {\left\lVert {f} \right\rVert}_\infty {\left\lVert {g} \right\rVert}_1$$.
• Combining these, \begin{align*} {\left\lVert {f} \right\rVert}_2^2 \leq {\left\lVert {f} \right\rVert}_2 {\varepsilon}\implies {\left\lVert {f} \right\rVert}_2 < {\varepsilon}\to 0, .\end{align*} so $${\left\lVert {f} \right\rVert}_2 = 0$$, which implies $$f=0$$ almost everywhere.

## Spring 2015.2 #real_analysis/qual/work

Let $$f: {\mathbf{R}}\to {\mathbf{C}}$$ be continuous with period 1. Prove that \begin{align*} \lim _{N \rightarrow \infty} \frac{1}{N} \sum_{n=1}^{N} f(n \alpha)=\int_{0}^{1} f(t) d t \quad \forall \alpha \in {\mathbf{R}}\setminus{\mathbf{Q}}. \end{align*}

Hint: show this first for the functions $$f(t) = e^{2\pi i k t}$$ for $$k\in {\mathbf{Z}}$$.

## Fall 2014.4 #real_analysis/qual/completed

Let $$g\in L^\infty([0, 1])$$ Prove that \begin{align*} \int _{[0,1]} f(x) g(x)\, dx = 0 \quad\text{for all continuous } f:[0, 1] \to {\mathbf{R}} \implies g(x) = 0 \text{ almost everywhere. } \end{align*}

• Polar decomposition: $$f = \operatorname{sign}(f) \cdot {\left\lvert {f} \right\rvert}$$.
• $$L^\infty[0, 1] \subseteq L^1[0, 1]$$.

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).

# $$L^1$$

## Spring 2021.4 #real_analysis/qual/completed

Let $$f, g$$ be Lebesgue integrable on $${\mathbf{R}}$$ and let $$g_n(x) \coloneqq g(x- n)$$. Prove that \begin{align*} \lim_{n\to \infty } {\left\lVert {f + g_n} \right\rVert}_1 = {\left\lVert {f} \right\rVert}_1 + {\left\lVert {g} \right\rVert}_1 .\end{align*}

• For $$f\in L^1(X)$$, $${\left\lVert {f} \right\rVert}_1 \coloneqq\int_X {\left\lvert {f(x)} \right\rvert} \,dx< \infty$$.

• Small tails in $$L_1$$: if $$f\in L^1({\mathbf{R}}^n)$$, then for every $${\varepsilon}>0$$ exists some radius $$R$$ such that \begin{align*} {\left\lVert {f} \right\rVert}_{L^1(B_R^c)} < {\varepsilon} .\end{align*}

• Shift $$g$$ to the right far enough so that the two densities are mostly disjoint:

• Any integral $$\int_a^b f$$ can be written as $${\left\lVert {f} \right\rVert}_1 - O(\text{err})$$.

• Bounding technique: \begin{align*} a-{\varepsilon}\leq b \leq a+{\varepsilon}\implies b=a .\end{align*}

• 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*}

• 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.

## Fall 2020.4 #real_analysis/qual/completed

Prove that if $$xf(x) \in L^1({\mathbf{R}})$$, then \begin{align*} F(y) \coloneqq\int f(x) \cos(yx)\, dx \end{align*} defines a $$C^1$$ function.

• 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.

## Spring 2020.3 #real_analysis/qual/completed

• Prove that if $$g\in L^1({\mathbf{R}})$$ then \begin{align*} \lim_{N\to \infty} \int _{{\left\lvert {x} \right\rvert} \geq N} {\left\lvert {f(x)} \right\rvert} \, dx = 0 ,\end{align*} and demonstrate that it is not necessarily the case that $$f(x) \to 0$$ as $${\left\lvert {x} \right\rvert}\to \infty$$.

• Prove that if $$f\in L^1([1, \infty])$$ and is decreasing, then $$\lim_{x\to\infty}f(x) =0$$ and in fact $$\lim_{x\to \infty} xf(x) = 0$$.

• If $$f: [1, \infty) \to [0, \infty)$$ is decreasing with $$\lim_{x\to \infty} xf(x) = 0$$, does this ensure that $$f\in L^1([1, \infty))$$?

• Limits
• Cauchy Criterion for Integrals: $$\int_a^\infty f(x) \,dx$$ converges iff for every $${\varepsilon}>0$$ there exists an $$M_0$$ such that $$A,B\geq M_0$$ implies $${\left\lvert {\int_A^B f} \right\rvert} < {\varepsilon}$$, i.e. $${\left\lvert {\int_A^B f} \right\rvert} \overset{A\to\infty}\to 0$$.
• Integrals of $$L^1$$ functions have vanishing tails: $$\int_{N}^\infty {\left\lvert {f} \right\rvert} \overset{N\to\infty}\longrightarrow 0$$.
• Mean Value Theorem for Integrals: $$\int_a^b f(t)\, dt = (b-a) f(c)$$ for some $$c\in [a, b]$$.

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 $${\left\lvert {x} \right\rvert}\to\infty$$.

• Since $$f$$ is decreasing on $$[1, \infty)$$, for any $$t\in [x-n, x]$$ we have \begin{align*} x-n \leq t \leq x \implies f(x) \leq f(t) \leq f(x-n) .\end{align*}

• Integrate over $$[x, 2x]$$, using monotonicity of the integral: \begin{align*} \int_x^{2x} f(x) \,dt \leq \int_x^{2x} f(t) \,dt \leq \int_x^{2x} f(x-n) \,dt \\ \implies f(x) \int_x^{2x} \,dt \leq \int_x^{2x} f(t) \,dt \leq f(x-n) \int_x^{2x} \,dt \\ \implies xf(x) \leq \int_x^{2x} f(t) \, dt \leq xf(x-n) .\end{align*}

• By the Cauchy Criterion for integrals, $$\lim_{x\to \infty} \int_x^{2x} f(t)~dt = 0$$.

• So the LHS term $$xf(x) \overset{x\to\infty}\longrightarrow 0$$.

• Since $$x>1$$, $${\left\lvert {f(x)} \right\rvert} \leq {\left\lvert {xf(x)} \right\rvert}$$

• Thus $$f(x) \overset{x\to\infty}\longrightarrow 0$$ as well.

• Use mean value theorem for integrals: \begin{align*} \int_x^{2x} f(t)\, dt = xf(c_x) \quad\text{for some $c_x \in [x, 2x]$ depending on $x$} .\end{align*}

• Since $$f$$ is decreasing, \begin{align*} x\leq c_x \leq 2x &\implies f(2x)\leq f(c_x) \leq f(x) \\ &\implies 2xf(2x)\leq 2xf(c_x) \leq 2xf(x) \\ &\implies 2xf(2x)\leq 2x\int_x^{2x} f(t)\, dt \leq 2xf(x) \\ .\end{align*}

• By Cauchy Criterion, $$\int_x^{2x} f \to 0$$.

• So $$2x f(2x) \to 0$$, which by a change of variables gives $$uf(u) \to 0$$.

• Since $$u\geq 1$$, $$f(u) \leq uf(u)$$ so $$f(u) \to 0$$ as well.

Just showing $$f(x) \overset{x\to \infty}\longrightarrow 0$$:

• Toward a contradiction, suppose not.

• Since $$f$$ is decreasing, it can not diverge to $$+\infty$$

• If $$f(x) \to -\infty$$, then $$f\not\in L^1({\mathbf{R}})$$: choose $$x_0 \gg 1$$ so that $$t\geq x_0 \implies f(t) < -1$$, then

• Then $$t\geq x_0 \implies {\left\lvert {f(t)} \right\rvert} \geq 1$$, so \begin{align*} \int_1^\infty {\left\lvert {f} \right\rvert} \geq \int_{x_0}^\infty {\left\lvert {f(t) } \right\rvert} \, dt \geq \int_{x_0}^\infty 1 =\infty .\end{align*}

• Otherwise $$f(x) \to L\neq 0$$, some finite limit.

• If $$L>0$$:

• Fix $${\varepsilon}>0$$, choose $$x_0\gg 1$$ such that $$t\geq x_0 \implies L-{\varepsilon}\leq f(t) \leq L$$
• Then \begin{align*}\int_1^\infty f \geq \int_{x_0}^\infty f \geq \int_{x_0}^\infty \qty{L-{\varepsilon}}\,dt = \infty\end{align*}
• If $$L<0$$:

• Fix $${\varepsilon}> 0$$, choose $$x_0\gg 1$$ such that $$t\geq x_0 \implies L \leq f(t) \leq L + {\varepsilon}$$.
• Then \begin{align*}\int_1^\infty f \geq \int_{x_0}^\infty f \geq \int_{x_0}^\infty \qty{L}\,dt = \infty\end{align*}

Showing $$xf(x) \overset{x\to \infty}\longrightarrow 0$$.

• Toward a contradiction, suppose not.
• (How to show that $$xf(x) \not\to + \infty$$?)
• If $$xf(x)\to -\infty$$
• Choose a sequence $$\Gamma = \left\{{\widehat{x}_i}\right\}$$ such that $$x_i \to \infty$$ and $$x_i f(x_i) \to -\infty$$.
• Choose a subsequence $$\Gamma' = \left\{{x_i}\right\}$$ such that $$x_if(x_i) \leq -1$$ for all $$i$$ and $$x_i \leq x_{i+1}$$.
• Choose a further subsequence $$S = \left\{{x_i \in \Gamma' {~\mathrel{\Big\vert}~}2x_i < x_{i+1}}\right\}$$.
• Then since $$f$$ is always decreasing, for $$t\geq x_0$$, $${\left\lvert {f} \right\rvert}$$ is increasing, and $${\left\lvert {f(x_i)} \right\rvert} \leq {\left\lvert {f(2x_i)} \right\rvert}$$, so \begin{align*} \int_1^{\infty} {\left\lvert {f} \right\rvert} \geq \int_{x_0}^\infty {\left\lvert {f} \right\rvert} \geq \sum_{x_i \in S} \int_{x_i}^{2x_i} {\left\lvert {f(t)} \right\rvert} \, dt \geq \sum_{x_i \in S} \int_{x_i}^{2x_i} {\left\lvert {f(x_i)} \right\rvert} &= \sum_{x_i \in S} x_i f(x_i) \to \infty .\end{align*}
• If $$xf(x) \to L \neq 0$$ for $$0 < L< \infty$$:
• Fix $${\varepsilon}> 0$$, choose an infinite sequence $$\left\{{x_i}\right\}$$ such that $$L-{\varepsilon}\leq x_i f(x_i) \leq L$$ for all $$i$$. \begin{align*} \int_1^\infty {\left\lvert {f} \right\rvert} \geq \sum_S \int_{x_i}^{2x_i} {\left\lvert {f(t)} \right\rvert}\,dt \geq \sum_S \int_{x_i}^{2x_i} f(x_i) \,dt = \sum_S x_i f(x_i) \geq \sum_S \qty{L-{\varepsilon}} \to \infty .\end{align*}
• If $$xf(x) \to L \neq 0$$ for $$-\infty < L < 0$$:
• Fix $${\varepsilon}> 0$$, choose an infinite sequence $$\left\{{x_i}\right\}$$ such that $$L \leq x_i f(x_i) \leq L + {\varepsilon}$$ for all $$i$$. \begin{align*} \int_1^\infty {\left\lvert {f} \right\rvert} \geq \sum_S \int_{x_i}^{2x_i} {\left\lvert {f(t)} \right\rvert}\,dt \geq \sum_S \int_{x_i}^{2x_i} f(x_i) \,dt = \sum_S x_i f(x_i) \geq \sum_S \qty{L} \to \infty .\end{align*}

For $$x\geq 1$$, \begin{align*} {\left\lvert {xf(x)} \right\rvert} = {\left\lvert { \int_x^{2x} f(x) \, dt } \right\rvert} \leq \int_x^{2x} {\left\lvert {f(x)} \right\rvert} \, dt \leq \int_x^{2x} {\left\lvert {f(t)} \right\rvert}\, dt \leq \int_x^{\infty} {\left\lvert {f(t)} \right\rvert} \,dt \overset{x\to\infty}\longrightarrow 0 \end{align*} where we’ve used

• Since $$f$$ is decreasing and $$\lim_{x\to\infty}f(x) =0$$ from part (a), $$f$$ is non-negative.
• Since $$f$$ is positive and decreasing, for every $$t\in[a, b]$$ we have $${\left\lvert {f(a)} \right\rvert} \leq {\left\lvert {f(t)} \right\rvert}$$.
• By part (a), the last integral goes to zero.

• Toward a contradiction, produce a sequence $$x_i\to\infty$$ with $$x_i f(x_i) \to \infty$$ and $$x_if(x_i) > {\varepsilon}> 0$$, then \begin{align*} \int f(x) \, dx &\geq \sum_{i=1}^\infty \int_{x_i}^{x_{i+1}} f(x) \, dx \\ &\geq \sum_{i=1}^\infty \int_{x_i}^{x_{i+1}} f(x_{i+1}) \, dx \\ &= \sum_{i=1}^\infty f(x_{i+1}) \int_{x_i}^{x_{i+1}} \, dx \\ &\geq \sum_{i=1}^\infty (x_{i+1} - x_i) f(x_{i+1}) \\ &\geq \sum_{i=1}^\infty (x_{i+1} - x_i) {{\varepsilon}\over x_{i+1}} \\ &= {\varepsilon}\sum_{i=1}^\infty \qty{ 1 - {x_{i-1} \over x_i}} \to \infty \end{align*} which can be ensured by passing to a subsequence where $$\sum {x_{i-1} \over x_i} < \infty$$.

• 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*}

## Fall 2019.5 #real_analysis/qual/completed

• Show that if $$f$$ is continuous with compact support on $${\mathbf{R}}$$, then \begin{align*} \lim _{y \rightarrow 0} \int_{\mathbb{R}}|f(x-y)-f(x)| d x=0 \end{align*}

• Let $$f\in L^1({\mathbf{R}})$$ and for each $$h > 0$$ let \begin{align*} \mathcal{A}_{h} f(x):=\frac{1}{2 h} \int_{|y| \leq h} f(x-y) d y \end{align*}

• Prove that $$\left\|\mathcal{A}_{h} f\right\|_{1} \leq\|f\|_{1}$$ for all $$h > 0$$.

• Prove that $$\mathcal{A}_h f \to f$$ in $$L^1({\mathbf{R}})$$ as $$h \to 0^+$$.

\todo[inline]{Walk through.}


• Continuity in $$L^1$$ (recall that DCT won’t work! Notes 19.4, prove it for a dense subset first).
• Lebesgue differentiation in 1-dimensional case. See HW 5.6.
• Fix $$\varepsilon > 0$$. If we can find a set $$A$$ such that the following calculation holds for $$h$$ small enough, we’re done: \begin{align*} \int_{\mathbf{R}}{\left\lvert {f(x-h) - f(x)} \right\rvert} \,dx &= \int_A {\left\lvert {f(x-h) - f(x)} \right\rvert} \,dx\\ &\leq \int_A {\varepsilon}\\ &= {\varepsilon}\mu(A) \longrightarrow 0 ,\end{align*} provided $$h\to 0$$ as $${\varepsilon}\to 0$$, which we can arrange if $${\left\lvert {h} \right\rvert} < {\varepsilon}$$.

• Choose $$A\supseteq\mathop{\mathrm{supp}}f$$ compact such that $$\mathop{\mathrm{supp}}f \pm 1 \subseteq A$$

• Why this can be done: $$\mathop{\mathrm{supp}}f$$ is compact, so closed and bounded, and contained in some compact interval $$[-M, M]$$. So e.g. $$A\coloneqq[-M-1, M+1]$$ suffices.
• Note that $$f$$ is still continuous on $$A$$, since it is zero on $$A\setminus\mathop{\mathrm{supp}}f$$, and thus uniformly continuous (by Heine-Cantor, for example).

• We can rephrase the usual definition of uniform continuity: \begin{align*} \forall {\varepsilon}\exists \delta = \delta({\varepsilon}) \text{ such that } {\left\lvert {x - y} \right\rvert} < \delta \implies {\left\lvert {f(x) - f(y)} \right\rvert} < {\varepsilon}\quad \forall x, y\in A \end{align*} as \begin{align*} \forall {\varepsilon}\exists \delta = \delta({\varepsilon}) \text{ such that } {\left\lvert {h} \right\rvert} < \delta \implies {\left\lvert {f(x-h) - f(x)} \right\rvert} < {\varepsilon}\quad \forall x \text{ such that }x, x\pm h \in A \end{align*}

• So fix $${\varepsilon}$$ and choose such a $$\delta$$ for $$A$$, and choose $$h$$ such that $${\left\lvert {h} \right\rvert} < \min(1, \delta)$$. Then the desired computation goes through by uniform continuity of $$f$$ on $$A$$.

We have \begin{align*} \int_{\mathbf{R}}{\left\lvert {A_h(f)(x)} \right\rvert} ~dx &= \int_{\mathbf{R}}{\left\lvert {\frac{1}{2h} \int_{x-h}^{x+h} f(y)~dy} \right\rvert} ~dx \\ &\leq \frac{1}{2h} \int_{\mathbf{R}}\int_{x-h}^{x+h} {\left\lvert {f(y)} \right\rvert} ~dy ~dx \\ &=_{FT} \frac{1}{2h} \int_{\mathbf{R}}\int_{y-h}^{y+h} {\left\lvert {f(y)} \right\rvert} ~\mathbf{dx} ~\mathbf{dy} \\ &= \int_{\mathbf{R}}{\left\lvert {f(y)} \right\rvert} ~{dy} \\ &= {\left\lVert {f} \right\rVert}_1 ,\end{align*}

and (rough sketch)

\begin{align*} \int_{\mathbf{R}}{\left\lvert {A_h(f)(x) - f(x)} \right\rvert} ~dx &= \int_{\mathbf{R}}{\left\lvert { \left(\frac{1}{2h} \int_{B(h, x)} f(y)~dy\right) - f(x)} \right\rvert}~dx \\ &= \int_{\mathbf{R}}{\left\lvert { \left(\frac{1}{2h} \int_{B(h, x)} f(y)~dy\right) - \frac{1}{2h}\int_{B(h, x)} f(x) ~dy} \right\rvert}~dx \\ &\leq_{FT} \frac{1}{2h} \int_{\mathbf{R}}\int_{B(h, x)}{\left\lvert { f(y-x) - f(x)} \right\rvert} ~\mathbf{dx} ~\mathbf{dy} \\ &\leq \frac 1 {2h} \int_{\mathbf{R}}{\left\lVert {\tau_x f - f} \right\rVert}_1 ~dy \\ &\to 0 \quad\text{by (a)} .\end{align*}

This works for arbitrary $$f\in L^1$$, using approximation by continuous functions with compact support:

• Choose $$g\in C_c^0$$ such that $${\left\lVert {f- g} \right\rVert}_1 \to 0$$.

• By translation invariance, $${\left\lVert {\tau_h f - \tau_h g} \right\rVert}_1 \to 0$$.

• Write \begin{align*} {\left\lVert {\tau f - f} \right\rVert}_1 &= {\left\lVert {\tau_h f - g + g - \tau_h g + \tau_h g - f} \right\rVert}_1 \\ &\leq {\left\lVert {\tau_h f - \tau_h g} \right\rVert} + {\left\lVert {g - f} \right\rVert} + {\left\lVert {\tau_h g - g} \right\rVert} \\ &\to {\left\lVert {\tau_h g - g} \right\rVert} ,\end{align*}

so it suffices to show that $${\left\lVert {\tau_h g - g} \right\rVert} \to 0$$.

## Fall 2017.3 #real_analysis/qual/completed

Let \begin{align*} S = \mathop{\mathrm{span}}_{\mathbf{C}}\left\{{\chi_{(a, b)} {~\mathrel{\Big\vert}~}a, b \in {\mathbf{R}}}\right\}, \end{align*} the complex linear span of characteristic functions of intervals of the form $$(a, b)$$.

Show that for every $$f\in L^1({\mathbf{R}})$$, there exists a sequence of functions $$\left\{{f_n}\right\} \subset S$$ such that \begin{align*} \lim _{n \rightarrow \infty}\left\|f_{n}-f\right\|_{1}=0 \end{align*}

• From homework: $$E$$ is Lebesgue measurable iff there exists a finite union of closed cubes $$A$$ such that $$m(E\Delta A) < \varepsilon$$.

• 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.

## Spring 2015.4 #real_analysis/qual/completed

Define \begin{align*} f(x, y):=\left\{\begin{array}{ll}{\frac{x^{1 / 3}}{(1+x y)^{3 / 2}}} & {\text { if } 0 \leq x \leq y} \\ {0} & {\text { otherwise }}\end{array}\right. \end{align*}

Carefully show that $$f \in L^1({\mathbf{R}}^2)$$.

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|_{y=x}^{y=\infty} \,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*} 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 $$\int {\left\lvert {f} \right\rvert}$$.

## Fall 2014.3 #real_analysis/qual/completed

Let $$f\in L^1({\mathbf{R}})$$. Show that \begin{align*} \forall\varepsilon > 0 \exists \delta > 0 \text{ such that } \qquad m(E) < \delta \implies \int _{E} |f(x)| \, dx < \varepsilon \end{align*}

• 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.

## Spring 2014.1 #real_analysis/qual/completed

• Give an example of a continuous $$f\in L^1({\mathbf{R}})$$ such that $$f(x) \not\to 0$$ as$${\left\lvert {x} \right\rvert} \to \infty$$.

• Show that if $$f$$ is uniformly continuous, then \begin{align*} \lim_{{\left\lvert {x} \right\rvert} \to \infty} f(x) = 0. \end{align*}

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. $$\contradiction$$

#real_analysis/qual/completed #todo #real_analysis/qual/work