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

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