# UMD Analysis Qualifying Exam/Aug08 Real

## Problem 1

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

## Solution 1a

### Absolutely Continuous <==> Indefinite Integral

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

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

{\displaystyle {\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

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

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

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

## Solution 1b

### Absolutely continuous <==> Indefinite Integral

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

We want to show:

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

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

{\displaystyle {\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

{\displaystyle {\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 ${\displaystyle g(t)\!\,}$  is integrable

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

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

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

## Solution 1c

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

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

## Problem 3

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

## Solution 3

### Check Criteria for Lebesgue Dominated Convergence Theorem

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

#### g_n dominates hat{f}_n

Since ${\displaystyle f_{n}\!\,}$  is positive, then so is ${\displaystyle f\!\,}$  , i.e., ${\displaystyle |f_{n}|=f_{n}\!\,}$  and ${\displaystyle |f|=f\!\,}$ . Hence,

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

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

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

${\displaystyle f+f_{n}\rightarrow 2f\!\,}$  , i.e.,

${\displaystyle g_{n}\rightarrow g\!\,}$ .

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

{\displaystyle {\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,

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

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

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

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

i.e.

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

### Apply LDCT

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

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

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

## Problem 5a

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

## Solution 5a

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

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

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

{\displaystyle {\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 ${\displaystyle g(x)\!\,}$  is Lipschitz in the interval ${\displaystyle I\!\,}$

### Apply definitions to g(f(x))

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

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

then

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

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

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

## Problem 5b

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

## Solution 5b

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

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

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

The derivative is bounded (in fact, on any finite interval), so ${\displaystyle f\!\,}$  is Lipschitz.

Hence, f is AC

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

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

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

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

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

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