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The following is exercise IV.2.3. from Analysis I by Amann and Escher.

Let $-\infty < a < b < \infty$ and $f \in C([a, b],\mathbb{R})$ be differentiable on $(a, b]$. Show that, if $\lim_{x\to a} f'(x)$ exists, then $f$ is in $C^1([a, b],\mathbb{R})$ and $f'(a) = \lim_{x\to a} f'(x)$. (Hint: Use the mean value theorem.)

I think this is false and found a counterexample: Set $b=0$ and define $f:[a,0]\to\mathbb{R}$ by $$ f(x):= \begin{cases} x^2\sin(1/x),&x\in[-a,0);\\ 0,&x=0. \end{cases} $$ The function $f$ satifies all the hypotheses but $f'$ is not continuous at $x=0$.

Am I right?

Edit: My point is that $\lim_{x\to a} f'(x)$ does not imply $f\in C^1([a, b],\mathbb{R})$, or, roughly speaking, the continuity of $f'$ at $x=a$ does not imply its continuity on the whole interval. However, the second conclusion, $f'(a) = \lim_{x\to a} f'(x)$, is correct.

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  • $\begingroup$ @Thomas But I think the hypothesis did not say that... $\endgroup$ – Colescu Jan 21 '18 at 13:06
  • $\begingroup$ @Colescu That's exactly what the hypothesis says. $\endgroup$ – José Carlos Santos Jan 21 '18 at 13:07
  • $\begingroup$ @Thomas If you'll post your comment as an answer, I will upvote it. $\endgroup$ – José Carlos Santos Jan 21 '18 at 13:08
  • $\begingroup$ @JoséCarlosSantos thanks, but no. $\endgroup$ – Thomas Jan 21 '18 at 13:10
  • $\begingroup$ @Colescu The statement asked to be prove, as stated, is clearly wrong by the very reason you mentioned. It should be either assumed that $f \in C^1((a,b],\mathbb{R})$, or the conclusion should be changed to then $f$ is differentiable at $a$ and (...). This is what the author probably had in mind. $\endgroup$ – Aloizio Macedo Jan 21 '18 at 15:33
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As I said in the comments, the statement asked to be proved, as stated, is clearly wrong by the very reason you mentioned in your edit. It should be either assumed that $f \in C^1\left((a,b] \right)$, or the conclusion should be changed to then $f$ is differentiable at $a$ and (...).

Assuming this was an oversight by the authors (which is imho very likely true), let's try to answer the proper question. The method will be the same for both of the possible fixes, so I'll assume the second which does not require continuity of the derivative:

Let $-\infty < a < b < \infty$ and $f \in C([a, b],\mathbb{R})$ be differentiable on $(a, b]$. Show that, if $\lim_{x\to a} f'(x)$ exists, then $f$ is differentiable at $a$ and $f'(a) = \lim_{x\to a} f'(x)$. (Hint: Use the mean value theorem.)

Proof: Let $L:= \lim\limits_{x\to a} f'(x)$. It follows from the hypotheses that $f$ is continuous at $[a,c]$ and differentiable at $(a,c)$ for every $c \leq b$.

Take $\epsilon>0$. By the definition of limit, there exists $\delta>0$ such that if $x< a+\delta$, then $|f'(x)-L|<\epsilon$. From the mean-value theorem we have that if $x \in (a,a+\delta),$ $$\left|\frac{f(x)-f(a)}{x-a}-L\right|=|f'(\xi_x)-L|<\epsilon,$$ since $\xi_x \in (a,x)$, hence in $(a,a+\delta)$. But this proves precisely that $f'(a)=L$.

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  • $\begingroup$ The "$\;C^1\;$" in the book is likely a typo..............+1 $\endgroup$ – DanielWainfleet Jan 21 '18 at 23:49

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