UMD Analysis Qualifying Exam/Aug08 Real

Problem 1

Suppose that $\{f_{n}\}\!\,$$\{f_{n}\}\!\,$ is a sequence of absolutely continuous functions defined on $[0,1]\!\,$$[0,1]\!\,$ such that $f_{n}(0)=0\!\,$$f_{n}(0)=0\!\,$ for every $n\!\,$$n\!\,$ and $\sum _{n=1}^{\infty }\int _{0}^{1}|f_{n}^{\prime }(x)|dx<+\infty \!\,$$\sum _{n=1}^{\infty }\int _{0}^{1}|f_{n}^{\prime }(x)|dx<+\infty \!\,$ for every $x\in [0,1]\!\,$$x\in [0,1]\!\,$. Prove: the series $\sum _{n=1}^{\infty }f_{n}(x)\!\,$$\sum _{n=1}^{\infty }f_{n}(x)\!\,$ converges for each $x\in [0,1]\!\,$$x\in [0,1]\!\,$ pointwise to a function $f\!\,$$f\!\,$ the function $f\!\,$$f\!\,$ is absolutely continuous on $[0,1]\!\,$$[0,1]\!\,$ $f^{\prime }(x)=\sum _{n=1}^{\infty }f_{n}^{\prime }(x)\quad a.e.\,\,x\in [0,1]\!\,$$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)\!\,$$f_{n}(x)\!\,$ is absolutely continuous if and only if $f_{n}(x)\!\,$$f_{n}(x)\!\,$ can be written as an indefinite integral i.e. for all $x\in [0,1]\!\,$$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}}$${\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]\!\,$$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}}$${\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\!\,$$f_{n}(x_{0})\leq \int _{0}^{1}|f_{n}^{\prime }(t)|dt\!\,$

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

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

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

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

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

Solution 1c

Since $f(x)=\int _{0}^{x}\sum _{n=1}^{\infty }f'_{n}(t)dt\!\,$$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)\!\,$$f'(x)=\sum _{n=1}^{\infty }f'_{n}(x)\!\,$   a.e. $x\in [0,1]\!\,$$x\in [0,1]\!\,$

Problem 3

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

Solution 3

Check Criteria for Lebesgue Dominated Convergence Theorem

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

g_n dominates hat{f}_n

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

$|{\hat {f}}_{n}|=|f-f_{n}|\leq |f|+|f_{n}|=f+f_{n}=g_{n}\!\,$$|{\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\!\,$$g=2f\!\,$. Since $f_{n}\rightarrow f\!\,$$f_{n}\rightarrow f\!\,$, then

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

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

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

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

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

i.e.

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

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

Problem 5a

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

Solution 5a

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

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

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

Apply definitions to g(f(x))

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

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

then

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

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

Problem 5b

Let $0<p<1\!\,$$0<p<1\!\,$. Give an example of an absolutely continuous function $f\!\,$$f\!\,$ on $[0,1]\!\,$$[0,1]\!\,$ such that $|f|^{p}\!\,$$|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}}})\!\,$$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}}})\!\,$$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\!\,$$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|\!\,$$|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\}\!\,$$\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|\!\,$$\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 \!\,$$\infty \!\,$ as $n\!\,$$n\!\,$ goes to $\infty \!\,$$\infty \!\,$.

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