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 $x \in [0,1] \!\,$

$\begin{align} f_n(x)-\underbrace{f_n(0)}_0 &= \int_0^{x} f_n^\prime(t) dt \\ f_n(x) &= \int_0^{x} f_n^\prime(t) dt \end{align}$

Apply Inequalities,Sum over n, and Use Hypothesis

Let $x_0 \in [0,1] \!\,$ be given. Then,

$\begin{align} f_n(x_0) &= \int_0^{x_0} f_n^\prime(t) dt \\ &\leq \int_0^{x_0} |f_n^\prime(t)| dt \\ &\leq \int_0^1 |f_n^\prime(t)| dt \end{align}$

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:

$f(x)=\int_0^x \sum_{n=1}^\infty f^\prime_n (t)dt \!\,$

Rewrite f(x) and Apply Lebesgue Dominated Convergence Theorem

$\begin{align} f(x) &= \sum_{n=1}^\infty f_n(x) \\ &= \sum_{n=1}^\infty \int_0^x f_n^\prime(t) dt \quad \mbox{ (since } f_n \mbox{ is absolutely continuous)} \\ &=\lim_{n \rightarrow \infty} \sum_{k=1}^n \int_0^x f_k^\prime(t) dt \\ &=\lim_{n \rightarrow \infty} \int_0^x \sum_{k=1}^n f_k^\prime(t) dt \\ &=\int_0^x \lim_{n \rightarrow \infty} \sum_{k=1}^n f_k^\prime(t) dt \quad \mbox{ (by LDCT)} \\ &=\int_0^x \sum_{n=1}^\infty f_n^\prime(t) dt \end{align}$

Justification for Lebesgue Dominated Convergence Theorem

$\begin{align} | \sum_{k=1}^n f^\prime_k(t) | &\leq \sum_{k=1}^n |f^\prime_k(t)| \\ &\leq \underbrace{\sum_{n=1}^\infty |f_n^\prime(t)|}_{g(t) \mbox{ dominating function} }\\ \\ \\ \int_0^1 \sum_{n=1}^\infty |f^\prime_n(x)|dx &= \sum_{n=1}^\infty \int_0^1 |f^\prime_n(x)|dx \quad \mbox{ (by Tonelli Theorem)} \\ &< \infty \quad \mbox{ (by hypothesis) } \end{align}$

Therefore $g(t) \!\,$ is integrable

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

$| \sum_{k=1}^n f^\prime_k(t) | \rightarrow | \sum_{k=1}^\infty f^\prime_k(t) | \!\,$

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 =

$\begin{align} \int g &= \int(f+f)\\ &=2 \int f \\ \\ \\ \lim_{n \rightarrow \infty} \int g_n &= \lim_{n \rightarrow \infty} \int (f+f_n) \\ &= \int f + \lim_{n\rightarrow \infty} \int f_n \\ &= \int f + \int f \mbox{ (from hypothesis) } \\ &= 2 \int f \end{align}$

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 $x \in I \!\,$

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

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

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

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

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

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