4
$\begingroup$

Let $E$ have measure zero contained in the open interval $(a,b)$. In a previous problem I showed that there is a countable collection of open intervals, $\{(c_k,d_k)\}_k$, contained in $(a,b)$ for which each point in $E$ is contained in infinitely many intervals of the collection and $\sum_k |(c_k,d_k)|=\sum_kd_k-c_k<\infty.$ Define $f(x)=\sum_k|(c_k,d_k)\cap(-\infty,x)|$ for $x \in (a,b)$. Show $f$ is increasing and fails to be differentiable for every point in $E$.

The absolute value bars mean measure of the interval. I have already shown $f$ was increasing by showing that if $a \le u <v \le b$ then $$\begin{align} f(v)-f(u) &=\sum_k |(c_k,d_k)\cap [(-\infty,u)\cup[u,v)] |-\sum_k|(c_k,d_k)\cap (-\infty,u)|\\&=\sum_k|(c_k,d_k)\cap [u,v)] |\ge 0 \end{align}$$

Let $D^+f(x)=\lim_{h\rightarrow 0}\left[\sup_{0<|t|\le h}\dfrac{f(x+t)-f(x)}{t} \right]$ and $D^-f(x)=\lim_{h\rightarrow 0}\left[\inf_{0<|t|\le h}\dfrac{f(x+t)-f(x)}{t} \right]$

To show f is not differentiable at any point in $E$ then I need to show that $D^+f(x) \not= D^-f(x)$ for every $x \in E$. This is where I am having trouble. First from using what I did above to show that $f$ is increasing I found that for $t>0$

$$ A_t(f(x))= \dfrac{f(x+t)-f(x)}{t}=\dfrac{1}{t}\sum_{k=1}^{\infty}|(c_k,d_k)\cap[x,x+t)| $$

So at this point what we know is that if $x\in E$ then $x$ belongs to $(c_k,d_k)$ for infinitely many $k$ and $\sum_{k=1}^{\infty}|(c_k,d_k)\cap[x,x+t)|$ converges since $\sum_{k=1}^{\infty}|(c_k,d_k)|<\infty$. But Im not sure how to find the $\sup_{0<|t|\le h}$ and $\inf_{0<|t|\le h}$ of $A_t(f(x))$ I think that $\inf_{0<|t|\le h}A_t(f(x))=0$ and $\sup_{0<|t|\le h}A_t(f(x))>0$ but I dont know how to show it.

Any help is appreciated, thanks

$\endgroup$
  • $\begingroup$ $D^+f(x)$ does not converge, as $x \in E$ belongs to infinite $(c_k, d_k)$, $\endgroup$ – Zhuanghua Liu Jul 5 '17 at 6:38
2
$\begingroup$

I don't know if you are still seeking for the answer (probably not...), but here's my try.

Indeed, I cannot figure out the supremum and the infimum from what you got. Let's think about it differently.

Let $\{k_1,k_2,...,k_n,...\}$ be the collection of natural numbers for which $x\in I_k:=(c_k,d_k)$. Let $N\in \mathbb{N}$ be such that $k_{N}$ is in the latter enumeration. Then, $x\in I_{k_1}\cap I_{k_2}\cap...\cap I_{k_N}$. We know that finite intersections of open sets is open. Therefore, the latter finite intersection is open. This means we can pick an $\epsilon_N>0$ small enough for $x+\epsilon_N$ to remain in the intersection. Consequently, $[x,x+\epsilon_N)\subset I_{k_1}\cap I_{k_2}\cap...\cap I_{k_N}$. In particular, this means that the $N^{th}$-partial sum of $f(x+\epsilon_N)-f(x)$ is

\begin{equation} [f(x+\epsilon_N)-f(x)]_N=\sum_{k=1}^{N}\ell((c_k,d_k)\cap [x,x+\epsilon_N))=\sum_{k=1}^{N}\epsilon_N=N\epsilon_N. \end{equation}

Hence, we observe that $f(x+\epsilon_N)-f(x)\ge N\epsilon_N$. Finally, we see that

\begin{equation} \begin{split} D^+f(x)&=\lim_{h\rightarrow 0}\left[\sup_{0<|t|\le h}\dfrac{f(x+t)-f(x)}{t} \right]\ge\lim_{h\rightarrow 0}\left[\sup_{0<|t|\le h} \dfrac{f(x+\epsilon_N)-f(x)}{t}\right]\\&=\lim_{h\rightarrow 0}\left[\sup_{0<|t|\le h} \dfrac{N\epsilon_N}{t}\right]\ge N. \end{split} \end{equation}

But recall that $N$ can be arbitrarily large based on the fact that $x\in I_k$ for infinitely many $k$. This tells us that $D^+f(x)$ is unbounded. Zhuanghua Liu gave you the correct hint :)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.