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There is a corollary in Rudin analysis. But I am not able to understand it. Can someone help to understand it?

The Corollary is:

Let $f$ be a real differential function on $[a,b]$, then $f'$ cannot have any simple discontinuity.

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Fixing the accept rate will pave the way to your answer soon. :) –  B. S. Nov 24 '12 at 13:06
    
Did you understand the definition of simple discontinuity? Did you understand the statement to which this is a corollary? –  Siminore Nov 24 '12 at 13:27
    
YES. It means both left and right hand limit exist, for simple discontinuity. Theorem proves the intermediate value theorem for derivative. –  user38764 Nov 24 '12 at 13:32
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Does "differential" mean "differentiable"? –  Chris Eagle Nov 24 '12 at 13:46
    
yes............ –  user38764 Nov 24 '12 at 13:49
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2 Answers

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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.

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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 $a$.

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