# Is such a function of class $C^1$

Let $X$ be a real Banach space and $f:\mathbb{R} \rightarrow X$ be continuous on $\mathbb{R}$, of class $C^1$ on $\mathbb{R}\setminus \{0\}$ and there is a limit $$g=\lim_{x \rightarrow 0} f'(x).$$ Is then $f$ of class $C^1$ on $\mathbb{R}$?

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Have you tried the case when $X$ is $\mathbb{R}$? – Hui Yu Sep 3 '12 at 9:12
Yes, in this case equality $f'(0)=lim_{x\rightarrow 0} f'(x)$ follows from de L'Hospital rule, continuity of $f$ and definition of derivative. – R.S Sep 3 '12 at 9:22
Then everything you need still holds in the case of real Banach spaces I guess. – Hui Yu Sep 3 '12 at 9:28
It may be more subtle than this. De l'Hospital's theorem is based on some mean value theorem, which is not expected to hold for vector-valued functions. I guess that a proof should follow a different path. – Siminore Sep 3 '12 at 9:39

Theorem 5.20. Suppose $f$ is a continuous mapping of $[a,b]$ into ${\mathbb R}^k$ and $f$ is differentiable on $\ ]a,b[\$. Then there is an $x\in\ ]a,b[\$ such that $$|f(b)-f(a)|\leq(b-a)|f'(x)|\ .$$
I think the proof carries over from ${\mathbb R}^k$ to any Banach space $X$. Using this theorem we can argue as follows:
After replacing $f$ with $x\mapsto f(x)- x g$ we may assume $g=0\in X$. Let an $\epsilon>0$ be given. Then there is a $\delta>0$ such that $$|f'(x)|<\epsilon\qquad\forall x\in\dot U_\delta(0)\ .$$ Consider an arbitrary $t\in\dot U_\delta(0)$. By the above theorem there is a $\tau\in\ ]0,t[\$ (resp. $\in\ ]t,0[\$), such that $$|f(t)-f(0)|\leq|t|\ |f'(\tau)|<\epsilon |t|\ .$$ As $\epsilon>0$ was arbitrary this proves that $f'(0)=0\in X$ and at the same time that $f'$ is continuous at $0$.