TITLE: Prove that if a function $f$ has a jump at an interior point of the interval $[a,b]$ then it cannot be the derivative of any function.
QUESTION [8 upvotes]: Prove that if a function $f$ has a jump at an interior point of the interval $[a,b]$ then it cannot be the derivative of any function.
I know that for $f$ is differentiable in $(a,b)$ and that it has one-sided derivative $f_+' (a)≠f_-' (b)$ at the endpoints. If $C$ is a real number between $f_+' (a)$  and $f_-' (b)$, then there exists $c∈(a,b)$ such that $f' (c)=C $. How can I use this to prove the above?

REPLY [2 votes]: Just trying to add some clarity to the tricky part of the proofs presented above.
Let $f:[a,b]\to\mathbb{R}$ be differentiable in $(a,b)$ and $c$ and interior point of the interval, $c\in(a,b)$. We'll prove that
$$\left.\begin{array}{rr}
\displaystyle\lim_{x\to c^+}f'(x)=A\\
\displaystyle\lim_{x\to c^-}f'(x)=B\\
\end{array}\right\}\,\Rightarrow A=B=f'(c)$$
For each $x\in(c,b)$, $f$ is continuous in $[c,x]$ and differentiable in $(c,x)$ so we can apply the mean value theorem to $f$ on the interval $[c,x]$. Thus, for each interval $[c,x]$, there exists a $d(x)\in (c,x)$ such that $$f'\big(d(x)\big)=\frac{f(x)-f(c)}{x-c}$$ 
Since $f$ is differentiable at $c$, making $x\to c^+$ on the RHS of the expression gives:
$$\displaystyle\lim_{x\to c^+}\frac{f(x)-f(c)}{x-c}=f'(c)$$
The LHS of the equality is a bit trickier, we would like to say that
$$\displaystyle\lim_{x\to c^+}f'\big(d(x)\big)=\displaystyle\lim_{x\to c^+}f'(x)=A$$
but that is not obvious, since we can't assume $f'$ to be continuous (on the right) at $c$. 
However we know that $\displaystyle\lim_{x\to c^+}f'(x)=A$, we'll use $\epsilon-\delta$ definition of limit. Let $\epsilon>0$, 
$$\exists\delta>0 \,:\, \text{ if } 0<x-c<\delta\, \text{ then }\, |f'(x)-A|<\epsilon$$
(notice that we write $0<x-c<\delta$ instead of $0<|x-c|<\delta$ because we're taking the right hand limit, that is $x>c$).
The MVT guarantees that $d(x)\in (c,x)$, that is
$$c<d(x)<x\,\Rightarrow\, 0<d(x)-c<x-c$$
But this means
$$0<x-c<\delta\,\Rightarrow\,0<d(x)-c<\delta\,\Rightarrow\,|f'\big(d(x)\big)-A|<\epsilon$$
thus $\displaystyle\lim_{x\to c^+}f'\big(d(x)\big)=A$ as desired.
We've seen 
$$f'\big(d(x)\big)=\frac{f(x)-f(c)}{x-c}\xrightarrow{x\rightarrow c^+} A=f'(c)$$
Similarly, we can prove $B=f'(c)$ and we're done.
The answer to your question follows directly from this result.
Let $f$ be differentiable in $(a,c)\cup(c,b)$. Assume $f'$ has a jump discontinuity at $c$: $$\displaystyle\lim_{x\to c^+}f'(x)=A\not= B=
\displaystyle\lim_{x\to c^-}f'(x)$$
(to say we have a jump we require the limits to \textbf{exist} and to be different!).
By the result, if $f$ was also differentiable at $c$, we'd have $A=B$, that contradicts the hypothesis of having a jump: $A\neq B$. Therefore $f$ can't be differentiable at $c$.