1
$\begingroup$

If $h(x)=0$ if $x<0$ and $h(x)=1$ if $x\geq 0$, prove there exists does there does not exist a function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f'(x)=h(x)$.

Proof: We will show that $h$ is not continuous ax $x=0$ which will imply that $h$ is not differentiable. Let $x_n=\frac{(-1)^n}{n}$. We see that the $\lim_{n\to\infty}\frac{(-1)^n}{n}=0$. But if $n$ is even then $\lim_{n\to\infty} h(x_n)=\lim_{n\to\infty}h(\frac{(-1)^n}{n})=\lim_{n\to\infty}h(\frac{1}{n})=1$. But if $n$ is odd then $\lim_{n\to\infty}h(x_n)=\lim_{n\to\infty}h(\frac{-1}{n})=0$. Hence $h$ is discontinous at $x=0$. Would this show there cannot be a function $f'(x)=h(x)$?

New Proof: Suppose that there exist a function $f$ such that $f'(x)=h(x)$. Since $f$ is differentiable over all real numbers it follows that $f$ is differentiable on $[-1,1]$. Then $f'(-1)=h(-1)=0$ and $f'(1)=h(1)=1$. Thus $f([-1,1])=[0,1]$ by the preservation of interval's theorem. Then pick $a\in[0,1]$ where $a=\frac{1}{2}$ and it follows by darboux's theorem that there is a point $c\in[-1,1]$ such that $f'(c)=a=\frac{1}{2}$. But this contradicts my assumption.

$\endgroup$
  • $\begingroup$ It doesn't help at all to show that $h$ is not differentiable. You have to show that $h$ can't be the derivative of any function $f$, which is just about the opposite! $\endgroup$ – TonyK Dec 10 '13 at 17:55
  • $\begingroup$ oh ok then I'll use the hint that is given to me. $\endgroup$ – user60887 Dec 10 '13 at 17:56
  • $\begingroup$ This isn't sufficient: Derivatives do not need to be continuous, but in fact can be very poorly behaved. They do, however, satisfy the intermediate value property - so they can't have jump discontinuities. $\endgroup$ – user61527 Dec 10 '13 at 17:56
  • $\begingroup$ I have a theorem that states if $f$ is differentiable then $f$ must be continuous. So I tried working with the contrapositive. $\endgroup$ – user60887 Dec 10 '13 at 17:57
  • 1
    $\begingroup$ The contrapositive would be "if $f'$ does not exist at all points in the domain of $f$, then $f$ is not continuous over its domain." Again, not much help here. $\endgroup$ – Omnomnomnom Dec 10 '13 at 18:00
6
$\begingroup$

Hint: For any differentiable function $f$, $f'(x)$ must satisfy the intermediate value property.

$\endgroup$

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.