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I was given a hw question that I just cant seem to solve. It goes:

Let $f$ be defined and differentiable in $(a,b)$. Prove that $f'$ can't have a jump discontinuity point.

I think that by using Darboux's theorem you could get to that conclusion but we were instructed to use Lagrange's theorem.

What I did do:

Suppose $x_0\in(a,b)$ is a jump discontinuity point. We can look at $(x_0,x_0+\delta)\subset(a,b)$ all the conditions of Lagrange's theorem are met so we have a point $c\in(x_0,x_0+\delta)$ such that $f'(c)=\frac{f(x_0+\delta)-f(x_0)}{\delta}$

And this is where I'm stuck. I understand that the right side reminds the derivative definition of $f(x_0)$ but I don't understand how to connect it all.

José Carlos Santos
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1 Answers1

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I will prove that $\lim_{x\to x_0}f'(x)$, if it exists, is equal to $f'(x_0)$ (this is the meaning of $f'$ not having a jump discontinuity at $x_0$). Take $\varepsilon>0$. Since the limit $\lim_{x\to x_0}f'(x)$ exists, there is a $\delta>0$ such that$$|x-x_0|<\delta\implies\left|f'(x)-\lim_{x\to x_0}f'(x)\right|<\varepsilon.$$So, if $|x-x_0|<\delta$ and $x\neq x_0$, $\frac{f(x)-f(x_0)}{x-x_0}=f'(c)$ for some $c$ between $x_0$ and $x$ and therefore $|c-x_0|<\delta$ too. Therefore$$\left|\frac{f(x)-f(x_0)}{x-x_0}-\lim_{x\to x_0}f'(x)\right|=\left|f'(c)-\lim_{x\to x_0}f'(x)\right|<\varepsilon.$$

José Carlos Santos
  • 440,053