There is a corollary in Rudin's Analysis. But I am not able to understand it. Can someone help to understand it?
The Corollary is:
Let $f$ be a real differentiable function on $[a,b]$, then $f'$ cannot have any simple discontinuity.
There is a corollary in Rudin's Analysis. But I am not able to understand it. Can someone help to understand it?
The Corollary is:
Let $f$ be a real differentiable function on $[a,b]$, then $f'$ cannot have any simple discontinuity.
I'll sketch the argument. If the left and right hand limits $f'(c-)$ and $f'(c+)$ both exist and are not equal, then we're in a situation similar to $f'(c-) < f'(c) < f'(c+)$. So working on the lefthand side, we can find an $\epsilon > 0$ $f'(x) < f'(c) - \epsilon$ for all $x \in (c-\delta, c)$. Applying the theorem, we have a contradiction.
You don't even need the theorem about the intermediate value property of the derivative, because one has the following fact:
If $f$ is continuous at $a$, differentiable for $x>a$, and $\lim_{x\to a+} f'(x)=p$, then one has $ \lim_{x\to a+}{f(x)-f(a)\over x-a}=p\ .$
Proof. Given an $\epsilon>0$ there is a $\delta>0$ such that $|f'(x)-p|<\epsilon\qquad\bigl(x\in\ ]a,a+\delta[\ \bigr)\ .$ Let $x\in\ ]a,a+\delta[\ $. Then by the mean value theorem there is a $\xi\in\ ]a,x[\ \subset \ ]a,a+\delta[\ $ such that $\left|{f(x)-f(a)\over x-a}- p\right|=\bigl|f'(\xi)-p\bigr|<\epsilon\ .\qquad\qquad\square$ It follows that the limits $\lim_{x\to a+} f'(x)$ and $\lim_{x\to a-} f'(x)$ cannot both exist and be different, if $f$ is differentiable at $a$.