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.

  • $\begingroup$ 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
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    $\begingroup$ 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

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


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