# 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 $\sum _{n=1}^{\infty }\int _{0}^{1}|f_{n}^{\prime }(x)|dx<+\infty \!\,$ for every $x\in [0,1]\!\,$ . Prove: the series $\sum _{n=1}^{\infty }f_{n}(x)\!\,$ converges for each $x\in [0,1]\!\,$ pointwise to a function $f\!\,$ the function $f\!\,$ is absolutely continuous on $[0,1]\!\,$ $f^{\prime }(x)=\sum _{n=1}^{\infty }f_{n}^{\prime }(x)\quad a.e.\,\,x\in [0,1]\!\,$ ## 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{aligned}f_{n}(x)-\underbrace {f_{n}(0)} _{0}&=\int _{0}^{x}f_{n}^{\prime }(t)dt\\f_{n}(x)&=\int _{0}^{x}f_{n}^{\prime }(t)dt\end{aligned}}

### Apply Inequalities,Sum over n, and Use Hypothesis

Let $x_{0}\in [0,1]\!\,$  be given. Then,

{\begin{aligned}f_{n}(x_{0})&=\int _{0}^{x_{0}}f_{n}^{\prime }(t)dt\\&\leq \int _{0}^{x_{0}}|f_{n}^{\prime }(t)|dt\\&\leq \int _{0}^{1}|f_{n}^{\prime }(t)|dt\end{aligned}}

Hence

$f_{n}(x_{0})\leq \int _{0}^{1}|f_{n}^{\prime }(t)|dt\!\,$

Summing both sides of the inequality over $n\!\,$  and applying the hypothesis yields pointwise convergence of the series $f_{n}\!\,$ ,

$\sum _{n=1}^{\infty }f_{n}(x_{0})\leq \sum _{n=1}^{\infty }\int _{0}^{1}|f_{n}^{\prime }(t)|dt<+\infty \!\,$

## Solution 1b

### Absolutely continuous <==> Indefinite Integral

Let $f(x)=\sum _{n=1}^{\infty }f_{n}(x)\!\,$ .

We want to show:

$f(x)=\int _{0}^{x}\sum _{n=1}^{\infty }f_{n}^{\prime }(t)dt\!\,$

### Rewrite f(x) and Apply Lebesgue Dominated Convergence Theorem

{\begin{aligned}f(x)&=\sum _{n=1}^{\infty }f_{n}(x)\\&=\sum _{n=1}^{\infty }\int _{0}^{x}f_{n}^{\prime }(t)dt\quad {\mbox{ (since }}f_{n}{\mbox{ is absolutely continuous)}}\\&=\lim _{n\rightarrow \infty }\sum _{k=1}^{n}\int _{0}^{x}f_{k}^{\prime }(t)dt\\&=\lim _{n\rightarrow \infty }\int _{0}^{x}\sum _{k=1}^{n}f_{k}^{\prime }(t)dt\\&=\int _{0}^{x}\lim _{n\rightarrow \infty }\sum _{k=1}^{n}f_{k}^{\prime }(t)dt\quad {\mbox{ (by LDCT)}}\\&=\int _{0}^{x}\sum _{n=1}^{\infty }f_{n}^{\prime }(t)dt\end{aligned}}

### Justification for Lebesgue Dominated Convergence Theorem

{\begin{aligned}|\sum _{k=1}^{n}f_{k}^{\prime }(t)|&\leq \sum _{k=1}^{n}|f_{k}^{\prime }(t)|\\&\leq \underbrace {\sum _{n=1}^{\infty }|f_{n}^{\prime }(t)|} _{g(t){\mbox{ dominating function}}}\\\\\\\int _{0}^{1}\sum _{n=1}^{\infty }|f_{n}^{\prime }(x)|dx&=\sum _{n=1}^{\infty }\int _{0}^{1}|f_{n}^{\prime }(x)|dx\quad {\mbox{ (by Tonelli Theorem)}}\\&<\infty \quad {\mbox{ (by hypothesis) }}\end{aligned}}

Therefore $g(t)\!\,$  is integrable

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

$|\sum _{k=1}^{n}f_{k}^{\prime }(t)|\rightarrow |\sum _{k=1}^{\infty }f_{k}^{\prime }(t)|\!\,$

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

## Solution 1c

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

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

## Problem 3

 Suppose that $\{f_{n}\}\!\,$ is a sequence of nonnegative integrable functions such that $f_{n}\rightarrow f\!\,$ a.e., with $f\!\,$ integrable, and $\int _{R}f_{n}\rightarrow \int _{R}f\!\,$ . Prove that $\int _{R}|f_{n}-f|\rightarrow 0\!\,$ ## 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,

$|{\hat {f}}_{n}|=|f-f_{n}|\leq |f|+|f_{n}|=f+f_{n}=g_{n}\!\,$

#### g_n converges to g a.e.

Let $g=2f\!\,$ . Since $f_{n}\rightarrow f\!\,$ , then

$f+f_{n}\rightarrow 2f\!\,$  , i.e.,

$g_{n}\rightarrow g\!\,$ .

#### integral of g_n converges to integral of g =

{\begin{aligned}\int g&=\int (f+f)\\&=2\int f\\\\\\\lim _{n\rightarrow \infty }\int g_{n}&=\lim _{n\rightarrow \infty }\int (f+f_{n})\\&=\int f+\lim _{n\rightarrow \infty }\int f_{n}\\&=\int f+\int f{\mbox{ (from hypothesis) }}\\&=2\int f\end{aligned}}

Hence,

$\int g=\lim _{n\rightarrow \infty }\int g_{n}\!\,$

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

Note that $f_{n}\rightarrow f\!\,$  is equivalent to

$|f_{n}-f|\rightarrow 0\!\,$

i.e.

${\hat {f_{n}}}\rightarrow 0={\hat {f}}\!\,$

### Apply LDCT

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

$\lim _{n}\int {\hat {f_{n}}}=\int {\hat {f}}=0\!\,$  , i.e.,

$\lim _{n}\int |f-f_{n}|=0\!\,$

## 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 $K=\|g(x)\|_{\infty }\!\,$  for all $x\in I\!\,$

{\begin{aligned}||x|^{p}-|y|^{p}|&=||x|-|y||(|x|^{p-1}+|x|^{p-2}|y|+\ldots +|x||y|^{p-2}+|y|^{p-1})\\&\leq |K^{p-1}+K^{p-2}K+\ldots +KK^{p-2}+K^{p-1}|||x|-|y||\\&=\underbrace {pK^{p-1}} _{M}||x|-|y||\\&\leq M|x-y|\end{aligned}}

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 $\epsilon >0\!\,$ , there exists $\delta >0\!\,$  such that if $\{(x_{i},x_{i}^{'}\}\!\,$  is a finite collection of nonoverlapping intervals of $[0,1]\!\,$  such that

$\sum _{i=1}^{n}|x_{i}^{'}-x_{i}|<\delta \!\,$

then

$\sum _{i=1}^{n}|f(x_{i}^{'})-f(x_{i})|<\epsilon \!\,$

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

{\begin{aligned}\sum _{i=1}^{n}|g(f(x_{i}^{'}))-g(f(x_{i}))|&\leq \sum _{i=1}^{n}M|f(x_{i}^{'})-f(x_{i})|\\&=M\underbrace {\sum _{i=1}^{n}|f(x_{i}^{'})-f(x_{i})|} _{<\epsilon }\\&

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

## Problem 5b

 Let $0 . 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

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

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)

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

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

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

Then, T(f) goes to $\infty \!\,$  as $n\!\,$  goes to $\infty \!\,$ .

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