Absolutely Continuous <==> Indefinite Integral

${\displaystyle f_{n}(x)\!\,}$ is absolutely continuous if and only if ${\displaystyle f_{n}(x)\!\,}$ can be written as an indefinite integral i.e. for all ${\displaystyle x\in [0,1]\!\,}$

${\displaystyle {\begin{aligned}f_{n}(x)-\underbrace {f_{n}(0)} _{0}&=\int _{0}^{x}f_{n}^{\prime }(t)dt\\f_{n}(x)&=\int _{0}^{x}f_{n}^{\prime }(t)dt\end{aligned}}}$

Apply Inequalities,Sum over n, and Use Hypothesis

Let ${\displaystyle x_{0}\in [0,1]\!\,}$ be given. Then,

${\displaystyle {\begin{aligned}f_{n}(x_{0})&=\int _{0}^{x_{0}}f_{n}^{\prime }(t)dt\\&\leq \int _{0}^{x_{0}}|f_{n}^{\prime }(t)|dt\\&\leq \int _{0}^{1}|f_{n}^{\prime }(t)|dt\end{aligned}}}$

Hence

${\displaystyle f_{n}(x_{0})\leq \int _{0}^{1}|f_{n}^{\prime }(t)|dt\!\,}$

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

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

Absolutely continuous <==> Indefinite Integral

Let ${\displaystyle f(x)=\sum _{n=1}^{\infty }f_{n}(x)\!\,}$.

We want to show:

${\displaystyle f(x)=\int _{0}^{x}\sum _{n=1}^{\infty }f_{n}^{\prime }(t)dt\!\,}$

Rewrite f(x) and Apply Lebesgue Dominated Convergence Theorem

${\displaystyle {\begin{aligned}f(x)&=\sum _{n=1}^{\infty }f_{n}(x)\\&=\sum _{n=1}^{\infty }\int _{0}^{x}f_{n}^{\prime }(t)dt\quad {\mbox{ (since }}f_{n}{\mbox{ is absolutely continuous)}}\\&=\lim _{n\rightarrow \infty }\sum _{k=1}^{n}\int _{0}^{x}f_{k}^{\prime }(t)dt\\&=\lim _{n\rightarrow \infty }\int _{0}^{x}\sum _{k=1}^{n}f_{k}^{\prime }(t)dt\\&=\int _{0}^{x}\lim _{n\rightarrow \infty }\sum _{k=1}^{n}f_{k}^{\prime }(t)dt\quad {\mbox{ (by LDCT)}}\\&=\int _{0}^{x}\sum _{n=1}^{\infty }f_{n}^{\prime }(t)dt\end{aligned}}}$

Justification for Lebesgue Dominated Convergence Theorem

${\displaystyle {\begin{aligned}|\sum _{k=1}^{n}f_{k}^{\prime }(t)|&\leq \sum _{k=1}^{n}|f_{k}^{\prime }(t)|\\&\leq \underbrace {\sum _{n=1}^{\infty }|f_{n}^{\prime }(t)|} _{g(t){\mbox{ dominating function}}}\\\\\\\int _{0}^{1}\sum _{n=1}^{\infty }|f_{n}^{\prime }(x)|dx&=\sum _{n=1}^{\infty }\int _{0}^{1}|f_{n}^{\prime }(x)|dx\quad {\mbox{ (by Tonelli Theorem)}}\\&<\infty \quad {\mbox{ (by hypothesis) }}\end{aligned}}}$

Therefore ${\displaystyle g(t)\!\,}$ is integrable

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

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

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

Check Criteria for Lebesgue Dominated Convergence Theorem

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

Apply LDCT

Since the criteria of the LDCT are fulfilled, we have that

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

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

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

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

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

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

Therefore ${\displaystyle g(x)\!\,}$ is Lipschitz in the interval ${\displaystyle I\!\,}$

Apply definitions to g(f(x))

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

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

then

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

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

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

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

f(x)= x^4sin^2(\frac{1}{x^2}) is Lipschitz (and then AC)

Consider ${\displaystyle f(x)=x^{4}\sin ^{2}({\frac {1}{x^{2}}})\!\,}$. The derivate of f is given by

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

The derivative is bounded (in fact, on any finite interval), so ${\displaystyle f\!\,}$ is Lipschitz.

Hence, f is AC

|f|^{1/2} is not of bounded variation (and then is not AC)

${\displaystyle |f(x)|^{1/2}=x^{2}\left|\sin \left({\frac {1}{x^{2}}}\right)\right|\!\,}$

Consider the partition ${\displaystyle \left\{{\sqrt {\frac {2}{n\pi }}}\right\}\!\,}$. Then,

${\displaystyle \left|f\left({\sqrt {\frac {2}{n\pi }}}\right)\right|^{1/2}={\frac {2}{n\pi }}\left|\sin \left({\frac {n\pi }{2}}\right)\right|\!\,}$

Then, T(f) goes to ${\displaystyle \infty \!\,}$ as ${\displaystyle n\!\,}$ goes to ${\displaystyle \infty \!\,}$.

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