Question on differentiability of a continuous function

Solution 1:

Assume that $f$ is continuous on $[a,b], v\in[a,b]$. f is then continuous and differentiable on $(v,b]$. so for all $h$ such that $v+h \in(v,b]$ we can apply the mean value theorem to f and write $f(v+h) = f(v)+hf'(v+\theta h)$ for some $\theta \in (0,1)$. Hence $\frac{f(v+h)-f(v)}{h} = f'(v+\theta h)$ taking the limit as h goes to zero on both sides gives that the $\displaystyle \lim_{h\to0}\frac{f(v+h)-f(v)}{h} = \displaystyle \lim_{h\to0} f'(v+\theta h)$which we know exists by hypothesis. Bear in mind that $\theta$ does vary with h, but this doesn't cause us any problems with what we are showing because it is always in the interval $(0,1)$

Solution 2:

This is true in slightly more general conditions.

Suppose $f:U \to \mathbb{R}$ is continuous on $U$, an open subset of a Banach space, differentiable on $U\setminus \{v\}$, and $\lim_{x\to v} Df(x)$ exists. Then $f$ is (Fréchet) differentiable on $U$.

The proof is essentially the same as Tom's above.

Let $L = \lim_{x\to v} Df(x)$, and note that $L$ is a continuous operator by the Banach Steinhaus theorem. Let $\epsilon>0$ and choose $\delta>0$ such that $\sup_{\|x-v\|< \delta} \|Df(x)-L\| < \epsilon$. Let $V = B(v,\delta)$. As in the real case, if $x\in V$, there exists a $\theta \in (0,1)$ such that $f(x)-f(v) = Df(v+\theta(x-v))(x-v)$. Then for $x\in V$ we have the estimate $$\|f(x)-f(v)-L(x-v)\| = \|Df(v+\theta(x-v))(x-v)-L(x-v)\|\leq \epsilon \|x-v\|$$ Hence $f$ is differentiable at $v$ with derivative $L$.