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:

Shifting density

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 $N-R_2 > R_1$:

Shifting density

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. $\contradiction$

Integrals: Approximation