UMD Analysis Qualifying Exam/Aug08 Real

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 $$f_n \!\,$$,

$$ \sum_{n=1}^\infty f_n(x_0) \leq \sum_{n=1}^\infty \int_0^1 |f_n^\prime(t)|dt < +\infty \!\,$$

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

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

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.,

$$ \lim_n \int |f-f_n| =0    \!\,$$

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 $$x \in I \!\,$$

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

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.

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