UMD Analysis Qualifying Exam/Aug08 Real

Problem 1

Suppose that $\{f_n\}$ is a sequence of absolutely continuous functions defined on $[0,1]$ such that $f_n(0)=0$ for every $n$ and $\sum _{n=1}^{\mathrm{\infty }}\underset{0}{\overset{1}{\int }}|f_n^{\prime }(x)|dx<+\mathrm{\infty }$ for every $x\in [0,1]$. Prove: the series $\sum _{n=1}^{\mathrm{\infty }}{f}_n(x)$ converges for each $x\in [0,1]$ pointwise to a function $f$ the function $f$ is absolutely continuous on $[0,1]$ $f^{\prime }(x)=\sum _{n=1}^{\mathrm{\infty }}{f}_n^{\prime }(x)a.e.x\in [0,1]$

Solution 1a

Absolutely Continuous <==> Indefinite Integral

$f_n(x)$ is absolutely continuous if and only if $f_n(x)$ can be written as an indefinite integral i.e. for all $x\in [0,1]$

$\begin{array}{cc}{f}_n(x)-\underset{0}{\underbrace{f_n(0)}}& =\underset{0}{\overset{x}{\int }}{f}_n^{\prime }(t)dt\\ f_n(x)& =\underset{0}{\overset{x}{\int }}{f}_n^{\prime }(t)dt\end{array}$

Apply Inequalities,Sum over n, and Use Hypothesis

Let $x_0\in [0,1]$ be given. Then,

$\begin{array}{cc}{f}_n(x_0)& =\underset{0}{\overset{x_0}{\int }}{f}_n^{\prime }(t)dt\\ & \le \underset{0}{\overset{x_0}{\int }}|f_n^{\prime }(t)|dt\\ & \le \underset{0}{\overset{1}{\int }}|f_n^{\prime }(t)|dt\end{array}$

Hence

$f_n(x_0)\le \underset{0}{\overset{1}{\int }}|f_n^{\prime }(t)|dt$

Summing both sides of the inequality over $n$ and applying the hypothesis yields pointwise convergence of the series $f_n$,

$\sum _{n=1}^{\mathrm{\infty }}{f}_n(x_0)\le \sum _{n=1}^{\mathrm{\infty }}\underset{0}{\overset{1}{\int }}|f_n^{\prime }(t)|dt<+\mathrm{\infty }$

Solution 1b

Absolutely continuous <==> Indefinite Integral

Let $f(x)=\sum _{n=1}^{\mathrm{\infty }}{f}_n(x)$.

We want to show:

$f(x)=\underset{0}{\overset{x}{\int }}\sum _{n=1}^{\mathrm{\infty }}{f}_n^{\prime }(t)dt$

Rewrite f(x) and Apply Lebesgue Dominated Convergence Theorem

$\begin{array}{cc}f(x)& =\sum _{n=1}^{\mathrm{\infty }}{f}_n(x)\\ & =\sum _{n=1}^{\mathrm{\infty }}\underset{0}{\overset{x}{\int }}{f}_n^{\prime }(t)dt\text{ (since }{f}_n\text{ is absolutely continuous)}\\ & =\underset{n\to \mathrm{\infty }}{\lim }\sum _{k=1}^n\underset{0}{\overset{x}{\int }}{f}_k^{\prime }(t)dt\\ & =\underset{n\to \mathrm{\infty }}{\lim }\underset{0}{\overset{x}{\int }}\sum _{k=1}^n{f}_k^{\prime }(t)dt\\ & =\underset{0}{\overset{x}{\int }}\underset{n\to \mathrm{\infty }}{\lim }\sum _{k=1}^n{f}_k^{\prime }(t)dt\text{ (by LDCT)}\\ & =\underset{0}{\overset{x}{\int }}\sum _{n=1}^{\mathrm{\infty }}{f}_n^{\prime }(t)dt\end{array}$

Justification for Lebesgue Dominated Convergence Theorem

$\begin{array}{cc}|\sum _{k=1}^n{f}_k^{\prime }(t)|& \le \sum _{k=1}^n|f_k^{\prime }(t)|\\ & \le \underset{g(t)\text{ dominating function}}{\underbrace{\sum _{n=1}^{\mathrm{\infty }}|f_n^{\prime }(t)|}}\\ \\ \\ \underset{0}{\overset{1}{\int }}\sum _{n=1}^{\mathrm{\infty }}|f_n^{\prime }(x)|dx& =\sum _{n=1}^{\mathrm{\infty }}\underset{0}{\overset{1}{\int }}|f_n^{\prime }(x)|dx\text{ (by Tonelli Theorem)}\\ & <\mathrm{\infty }\text{ (by hypothesis) }\end{array}$

Therefore $g(t)$ is integrable

The above inequality also implies $\sum _{n=1}^{\mathrm{\infty }}|f_n^{\prime }(x)|<\mathrm{\infty }$ a.e on $[0,1]$. Therefore,

$|\sum _{k=1}^n{f}_k^{\prime }(t)|\to |\sum _{k=1}^{\mathrm{\infty }}{f}_k^{\prime }(t)|$

a.e on $[0,1]$ to a finite value.

Solution 1c

Since $f(x)=\underset{0}{\overset{x}{\int }}\sum _{n=1}^{\mathrm{\infty }}f{\text{'}}_n(t)dt$,   by the Fundamental Theorem of Calculus

$f^{\prime }(x)=\sum _{n=1}^{\mathrm{\infty }}f{\text{'}}_n(x)$   a.e. $x\in [0,1]$

Problem 3

Suppose that $\{f_n\}$ is a sequence of nonnegative integrable functions such that $f_n\to f$ a.e., with $f$ integrable, and $\underset{R}{\int }{f}_n\to \underset{R}{\int }f$. Prove that $\underset{R}{\int }|f_n-f|\to 0$

Solution 3

Check Criteria for Lebesgue Dominated Convergence Theorem

Define ${\hat{f}}_n=|f-f_n|$, $g_n=f+f_n$.

g_n dominates hat{f}_n

Since $f_n$ is positive, then so is $f$ , i.e., $|f_n|=f_n$ and $|f|=f$. Hence,

$|{\hat{f}}_n|=|f-f_n|\le |f|+|f_n|=f+f_n=g_n$

g_n converges to g a.e.

Let $g=2f$. Since $f_n\to f$, then

$f+f_n\to 2f$ , i.e.,

$g_n\to g$.

integral of g_n converges to integral of g =

$\begin{array}{cc}\int g& =\int (f+f)\\ & =2\int f\\ \\ \\ \underset{n\to \mathrm{\infty }}{\lim }\int g_n& =\underset{n\to \mathrm{\infty }}{\lim }\int (f+f_n)\\ & =\int f+\underset{n\to \mathrm{\infty }}{\lim }\int f_n\\ & =\int f+\int f\text{ (from hypothesis) }\\ & =2\int f\end{array}$

Hence,

$\int g=\underset{n\to \mathrm{\infty }}{\lim }\int g_n$

hat{f_n} converges to hat{f} a.e.

Note that $f_n\to f$ is equivalent to

$|f_n-f|\to 0$

i.e.

$\widehat{f_n}\to 0=\hat{f}$

Apply LDCT

Since the criteria of the LDCT are fulfilled, we have that

$\underset{n}{\lim }\int \widehat{f_n}=\int \hat{f}=0$ , i.e.,

$\underset{n}{\lim }\int |f-f_n|=0$

Problem 5a

Show that if $f$ is absolutely continuous on $[0,1]$ and $p>1$, then $|f{|}^p$ is absolutely continuous on $[0,1]$

Solution 5a

Show that g(x)=|x|^p is Lipschitz

Consider some interval $I=[\alpha ,\beta ]$ and let $x$ and $y$ be two points in the interval $I$.

Also let $K=\Vert g(x){\Vert }_{\mathrm{\infty }}$ for all $x\in I$

$\begin{array}{cc}||x{|}^p-|y{|}^p|& =||x|-|y||(|x{|}^{p-1}+|x{|}^{p-2}|y|+\dots +|x||y{|}^{p-2}+|y{|}^{p-1})\\ & \le |K^{p-1}+K^{p-2}K+\dots +K{K}^{p-2}+K^{p-1}|||x|-|y||\\ & =\underset{M}{\underbrace{p{K}^{p-1}}}||x|-|y||\\ & \le M|x-y|\end{array}$

Therefore $g(x)$ is Lipschitz in the interval $I$

Apply definitions to g(f(x))

Since $f(x)$ is absolutely continuous on $[0,1]$, given $ϵ>0$, there exists $\delta >0$ such that if $\{(x_i,x_i^{\text{'}}\}$ is a finite collection of nonoverlapping intervals of $[0,1]$ such that

$\sum _{i=1}^n|x_i^{\text{'}}-x_i|<\delta$

then

$\sum _{i=1}^n|f(x_i^{\text{'}})-f(x_i)|<ϵ$

Consider $g\circ f(x)=|f(x){|}^p$. Since $g$ is Lipschitz

$\begin{array}{cc}\sum _{i=1}^n|g(f(x_i^{\text{'}}))-g(f(x_i))|& \le \sum _{i=1}^{n}M|f(x_i^{\text{'}})-f(x_i)|\\ & =M\underset{<ϵ}{\underbrace{\sum _{i=1}^n|f(x_i^{\text{'}})-f(x_i)|}}\\ & <Mϵ\end{array}$

Therefore $g\circ f(x)=|f(x){|}^p$ is absolutely continuous.

Problem 5b

Let $0<p<1$. Give an example of an absolutely continuous function $f$ on $[0,1]$ such that $|f{|}^p$ is not absolutely continuous

Solution 5b

f(x)= x^4sin^2(\frac{1}{x^2}) is Lipschitz (and then AC)

Consider $f(x)=x^4{\sin }^2(\frac{1}{x^2})$. The derivate of f is given by

$f^{\prime }(x)=4x^3{\sin }^2(\frac{1}{x^2})-2x\sin (\frac{2}{x^2})$.

The derivative is bounded (in fact, on any finite interval), so $f$ is Lipschitz.

Hence, f is AC

|f|^{1/2} is not of bounded variation (and then is not AC)

$|f(x){|}^{1/2}=x^2|\sin \left(\frac{1}{x^2}\right)|$

Consider the partition $\left\{\sqrt{\frac{2}{n\pi }}\right\}$. Then,

${\left|f\left(\sqrt{\frac{2}{n\pi }}\right)\right|}^{1/2}=\frac{2}{n\pi }|\sin \left(\frac{n\pi }{2}\right)|$

Then, T(f) goes to $\mathrm{\infty }$ as $n$ goes to $\mathrm{\infty }$.

Then, $|f{|}^{1/2}$ is not of bounded variation and then is not AC