Lipschitz and Absolute Continuity

Question

A problem in Royden & Fitzpatrick:

Define $f$ and $g$ on $[-1,1]$ by

$f(x)=x^{\frac{1}{3}}$ and

$g(x)=x^2cos(\frac{\pi}{2x})$ if $x\neq0$, $x\in[-1,1]$

$g(x)=0$, if $x=0$.

$(i)$ Show that both $f$ and $g$ are absolutely continuous on $[-1,1]$

$(ii)$ For the partition $P_n=<-1,0,\frac{1}{2n},\frac{1}{2n-1}...,\frac{1}{3},\frac{1}{2},1>$ Examine $V(f(g),P_n)$

$(iii)$ Show that $f(g)$ fails to be of bounded variation, and hence also fails to be absolutely continuous on $[-1,1]$

MY ATTEMPT: (EDIT)

$(i)$ $f'(x)$ is bounded $=>f$ is Lipschitz $=>f$ is absolutely continuous. I show the derivatives of both functions are bounded and I'm done. (WRONG - just realized. Is there any way other than directly showing absolute continuity, I have trouble with doing so rigorously)

$(ii)$ I am computing this bruteforce, if you have a better suggestion please let me hear it before I die of boredom.

$(iii)$ Since I made a mistake in $(i)$, I'm guessing this will be an easy consequence of $(ii)$ if I can show the variation for each $n$ is a series that diverges.

Any help is very much appreciated!

Answer 1

I will assume you are familiar with this characterization of absolutely continuous function: $f$ is absolutely continuous $\Leftrightarrow$ the derivative $f'$ of $f$ exists almost everywhere, is in $L^1$, and $$ f(x)-f(a)=\int_a^x f'(s)ds $$ for some $a$.

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1. Using the characterization I just mentioned, note $$ f'(x)=\frac{1}{3}x^{-2/3} $$ and $$ g'(x)=2x\cos\left(\frac{\pi}{2x}\right)+\frac{\pi}{2}\sin\left(\frac{\pi}{2x}\right) $$ Both are well defined, except on $0$, are Lebesgue integrable, and belong to $L^1([-1,1])$.
2. (I think you are looking for a somewhat short and elegant way of making the calculation here, right? I don't know if the argument to do the calculation I will show can be considered elegant, but I don't think this can be done any better). Note $g(-1)=2cos\left(-\frac{\pi}{2}\right)=0$, $g(0)=0$ and $$ g\left(\frac{1}{k}\right)=\frac{1}{k^2}\cos(\frac{k\pi}{2})= \begin{cases} 0 & k \text{ odd}\\ -\frac{1}{k^2} & k = 2j, j \text{ odd}\\ \frac{1}{k^2} & k = 2j, j \text{ even} \end{cases} $$ which implies $$ f\circ g\left(\frac{1}{k}\right)= \begin{cases} f(0)=0 & k \text{ odd}\\ -\frac{1}{k^{2/3}} & k=2j, j \text{ odd}\\ \frac{1}{k^{2/3}} & k=2j, j \text{ even} \end{cases} $$ Then, we deduce, for k even, $$ \left\vert f \circ g\left(\frac{1}{k+1}\right)-f \circ g\left(\frac{1}{k}\right)\right\vert = \frac{1}{k^{2/3}} $$ and for k odd, $$ \left\vert f \circ g\left(\frac{1}{k+1}\right)-f \circ g\left(\frac{1}{k}\right)\right\vert = \frac{1}{(k+1)^{2/3}} $$ With all this, we deduce $$ V(f(g),P_n)=2\sum_{k=1}^{n} \frac{1}{k^{2/3}} $$ What can you say about this series?

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Edit for part 1: Since you are assumed to not know the characterization I just mentioned for absolute continuity, here I will show other ways of proving absolute continuity of $g$ and $f$ .

Notice $\vert g'(x)\vert\leq 2+\pi$ for all $x\in[-1,1]\setminus\{0\}$. This implies $g$ is Lipschitz.

For $f$, we will prove it by definition. Let $\epsilon>0$. We want to show there is $\delta>0$ s.t. if $S=\bigcup_{i=1}^{n}(x_{i},y_{i})$ (with $x_{i}<y_{i}$) is s.t. $m(S)<\delta$ then $\sum_{i=1}^{n}\vert f(y_{i})-f(x_{i})\vert<\epsilon$.

When $0 \notin (x_i,y_i)$, we have $f$ is derivable, hence by the fundamental theorem of calculus, $f(y_i)-f(x_i)=\int_{x_i}^{y_i}f'(s)ds$. When $0 \in (x_i,y_i)$, we note, for $a_n\searrow 0$, $$ h_{a_n}(x):=f'(x)1_{[a_n,y_i]}\nearrow h_0(x):=f'(x)1_{(0,y_i]} $$ pointwise, thus by monotone convergence theorem, $$ y_i^{1/3}-a_n^{1/3}=\int h_{a_n}(x)dx\\ =\int_{a_n}^{y_i} f'(x)dx\\ \nearrow \int h_0(x)dx\\ = \int_0^{y_i}f'(x)dx=y_i^{1/3} $$ thus $\int_0^{y_i}f'(x)dx=(y_i)^{1/3}$ is well defined. Likewise $\int_{x_i}^0 f'(x)dx=-(x_i)^{1/3}$, thus $f(y_i)-f(x_i)=\int_{x_i}^{y_i}f'(s)ds$.

Hence, we can say \begin{eqnarray} \sum_{i=1}^{n}\vert f(y_{i})-f(x_{i})\vert & = & \sum_{i=1}^{n}\left\vert\int_{x_{i}}^{y_{i}}\frac{1}{3}s^{-2/3}ds\right\vert\\ & \leq & \sum_{i=1}^{n}\int_{x_{i}}^{y_{i}}\frac{1}{3}\vert s^{-2/3}\vert ds\\ & = & \int_{S}\frac{1}{3}\vert s^{-2/3}\vert ds \end{eqnarray}

Let $\delta$ be s.t. $(\delta/2)^{1/3}-(-\delta/2)^{1/3}<\epsilon$ and $S$ s.t. $m(S)<\delta$. We can show $$ \int_{S}\frac{1}{3}\vert s^{-2/3}\vert ds\leq \int_{-\delta/2}^{\delta/2}\frac{1}{3}\vert s^{-2/3}\vert ds=(\delta/2)^{1/3}-(-\delta/2)^{1/3}<\epsilon $$ (ie, the "worst case" is $S=(-\delta/2,\delta/2)$). This shows $f$ absolutely continuous.

Answer 2

For completeness, I'll add to Nate's answer by showing $g'(0)=0$. By definition, $$g'(0)=\lim_{x \rightarrow 0} \frac{g(x)-g(0)}{x}=\lim_{x \rightarrow 0} \frac{g(x)}{x}=\lim_{x \rightarrow 0} \frac{x^2 \cos(\frac{\pi}{2x})}{x}$$ Since $-1 \leq \cos (\frac{\pi}{2x}) \leq 1 \implies -x \leq x\cos (\frac{\pi}{2x}) \leq x $ and by squeeze theorem, we are done.

For (1), an easier way is presented in Proof that $\sqrt x$ is absolutely continuous. but the proof essentially follows through: if $0<t\leq x < y$, then $(y-t)^{1/3}-(x-t)^{1/3} \geq y^{1/3}-x^{1/3}$ since $f'(t)=\frac{1}{3(x-t)^{1/3}}-\frac{1}{3(y-t)^{1/3}}\geq 0$ for $f(t)=(y-t)^{1/3}+(x-t)^{1/3}$ and now, for a given $\epsilon>0$, choose $\delta = \epsilon^3$ and shift your disjoint intervals, as done in the link.