0
$\begingroup$

Let $g:\mathbb{Q}\to \mathbb{R}$ by $g(q(n))=2^{-n}$ for some bijection $q:\mathbb{N}\to\mathbb{Q}$ and $f:[0,1]\to \mathbb{R}$ by $f(x)=\sum_{r\in \mathbb{Q}:r<x}g(r)$.

Prove that for every rational number $q\in \mathbb{Q}\cap [0,1]$ the function $F:[0,1]\to \mathbb{R}$ defined by $F(x):=\int_0^x f(y)\operatorname{dy}$ is not differentiable at $q$

So I assume for contradiction that $F$ is differentiable at some $q\in \mathbb{Q}$ I've already proven previously that $f$ is discontinuous at rationals and is strictly monotone increasing. Since $f$ is monotone increasing it is riemann integrable.

Thus $F'(x)=(\int_0^x f)'(x)=f(x)$ by fundamental theorem of calculus.

Then $\lim_{x\to q}\frac{F(x)-F(q)}{x-q}=F'(q)$ is finite by assumption.

But $\lim_{x\to q}\frac{F(x)-F(q)}{x-q}=F'(q)=\lim_{x\to q}\frac{\int_0^x f-\int_0^q f}{x-q}=\lim_{x\to q}\frac{\int_x^qf}{x-q}$ which is by L'hopitals rule, $\lim_{x\to q} f(x)-f(q)=f(q)$

From here I'm not really sure what to do. I'm not sure if what I've done is correct, but I am trying to show that $f$ is continuous at a rational number.

$\endgroup$
2
  • $\begingroup$ There is a strange sum over rationals in your question. $\endgroup$
    – markvs
    Jul 12, 2020 at 17:42
  • $\begingroup$ @JCAA It doesn't look strange to me. $f$ is a form of the cumulative distribution function of the measure that puts mass $2^{-n}$ at the rational number $q(n)$. $\endgroup$ Jul 12, 2020 at 17:50

1 Answer 1

0
$\begingroup$

It seems to me that what's going on here is already present in the seemingly simpler case of the function $H(x)=\int_0^x h(t)dt$, where $h(t)=0$ if $t<0$ and $h(t)=1$ for $t\ge 0$. That is, $$ H(x)=\begin{cases}0&x<0\\x&x\ge0.\end{cases}$$ The claim is, that $H$ is not differentiable at $0$. The finite difference quotients $(H(h)-0)/(h-0)$ are equal to $0$ or $1$ according to whether $h<0$ or $h>0$, and so on. Note that $h$ is discontinuous at $x=0$, so a naive application of the fundamental theorem of calculus at $x=0$ is not justified.

The $F$ in the problem at hand is a superposition of this $H$: $F(x)=\sum_n 2^{-n}H(x-q(n))$, and for each rational $q\in[0,1]$ there is a unique $n$ such that $q(n)=q$, and the term $2^{-n}H(x-q(n))$ is not differentiable at $x=q$. The difference quotient $(F(q+h)-F(q))/h$ evaluates to $0$ or $2^{-n}$, depending on whether $h<0$ or $h>0$, and so on.

$\endgroup$
4
  • $\begingroup$ Is what I have done completely wrong? I dont think I used L'hopital rule correctly. I was hoping I could show that $\lim_{x\to q} f(x)=f(q)$ which would contradict discontinuity at $q$. $\endgroup$ Jul 12, 2020 at 19:30
  • $\begingroup$ I get confused by contrapositive arguments, distrust L'hopital's rule, and don't know what version of the fundamental theorem of calculus you are using. I have tried to edit some of this into my answer. $\endgroup$ Jul 12, 2020 at 19:40
  • $\begingroup$ "Distrust l'Hopital rule" Is it a new fobia? $\endgroup$
    – markvs
    Jul 12, 2020 at 19:52
  • $\begingroup$ @JCAA Yeah. Maybe I should have said, "distrust my use of l'H's rule". I remember the conclusion but not the hypotheses, so my anxiety level rises when I see it in use. $\endgroup$ Jul 12, 2020 at 20:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .