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Let $f:]a,b[\rightarrow \mathbb{R}$ a differentiable function. If the limit $$\lim_{x\rightarrow a^+} f'(x) =\lambda$$ exists and is finite, prove that $f$ is continuously extensible at the point $a$, that $f$ is differentiable in $a$ and that $f'(a)=\lambda$.

My thoughts: I tried reasoning by contradiction by proving that the limit of $f$ has to be finite, but I got nowhere. It seems to me that if the function tends to $\pm \infty$ for $x\rightarrow a^+$ then the limit of the derivative either doesn't exist or is infinite. But then the case "$f$ has no limit" remains. Anyone?

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  • $\begingroup$ I seriously hope that "prolongable with continuity" makes sense, because I honestly have no clue how to say that in english (and I didn't find anything on how to say it either). What I mean is: If the limit of a function $f$ for $x \rightarrow x_0$ exists and is finite, but the function is not defined in $x_0$, you define a new function to make the function continous. Basically what we do with $\sin x/x$ in $0$. $\endgroup$ – Gennaro Marco Devincenzis Apr 23 '14 at 14:47
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    $\begingroup$ I think the usual terminology is "continuously extensible". $\endgroup$ – Daniel Fischer Apr 23 '14 at 14:49
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Let $x_k\rightarrow a$, $x_k > a$ There is $\delta>0$ such that $|f^\prime(x)-\lambda| \le 1$ for $|x-a|\le \delta$. Then for $k,l>K_0$ there is $\xi_{kl}\in(x_k,x_l)$ (wlog $x_k<x_l) $ such that $$|f(x_k)-f(x_l)|= |f^\prime(\xi_{kl})(x_k-x_l)|\le( \lambda+1)|x_k-x_l|$$ Hence $f(x_k)$ is Cauchy and converges. This implies $f$ can be extended to $[a,b[$.

For the statement of differentiability see my answer here

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  • $\begingroup$ I am afraid your answer doesn't prove differentiability, see my comment at the question from your link. $\endgroup$ – M. Van Aug 2 '16 at 7:35
  • $\begingroup$ Oops I am sorry I am incorrect, your answer is completely valid, I did not know about the assumptions of the MVT $\endgroup$ – M. Van Aug 2 '16 at 8:58
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For the differentiability: Let $h_n$ be a positive sequence that converges to $0$ such that $a+h_n \in (a,b)$ for all $n$, and assume without loss of generality that $h_n<1$ for all $n$. Fix some $n \in \mathbb{N}$. Because $f$ is continuous at $a$, we can find some $\delta_n>0$ such that $|f(x)-f(a)|<h_n^2$ whenever $|x-a|<\delta_n$. In particular, we can find some $x_n$ with $|x_n-a|<\min\{h_n^2, \delta_n\}$ and $|f(x_n)-f(a)|<h_n^2$. Now we have $$\frac{f(a+h_n)-f(a)}{h_n}-\lambda< \frac{f(a+h_n)+h_n^2-f(x_n)}{h_n}-\lambda.$$ Now there exists some $\alpha_n$ between $a+h_n$ and $x_n$ such that $f(a+h_n)-f(x_n)=f'(\alpha_n)(a+h_n-x_n)$, here we use $h_n<1$. So we get $$\frac{f(a+h_n)+h_n^2-f(x_n)}{h_n}-\lambda=\frac{f'(\alpha_n)(a+h_n-x_n)}{h_n}+h_n - \lambda=f'(\alpha_n)-\lambda+\frac{f'(\alpha_n)(a-x_n)}{h_n}<f'(\alpha_n)-\lambda \pm f'(\alpha_n)h_n.$$ Now we have $f'(\alpha_n) - \lambda \to 0$, and $\pm f'(\alpha_n)h_n \to 0$, because $f'$ is bounded near $a$. Analogously one constructs a sequence $b_n$ such that $b_n \to 0$ and $$\frac{f(a+h_n)-f(a)}{h_n}-\lambda>b_n.$$ By the squeeze lemma the result follows. Please ask me if you want to know $b_n$, in fact one can use the same $(x_n)$ and $(\alpha_n)$ to make the $b_n$.

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