# 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
${\begin{aligned}\sum _{i=1}^{n}|g(f(x_{i}^{'}))-g(f(x_{i}))|&\leq \sum _{i=1}^{n}M|f(x_{i}^{'})-f(x_{i})|\\&=M\underbrace {\sum _{i=1}^{n}|f(x_{i}^{'})-f(x_{i})|} _{<\epsilon }\\&<M\epsilon \end{aligned}}$
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
$f'(x)=4x^{3}\sin ^{2}({\frac {1}{x^{2}}})-2x\sin({\frac {2}{x^{2}}})\!\,$.
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