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Suppose that $\mathcal f$ is continuous on a closed interval and $\mathcal v$ is a point in its interior. $\mathcal f$ is differentiable when $\mathcal x\not=v$, and suppose that $\,\displaystyle{\lim_{x\rightarrow v}f'(x)}\,$ exists. Show that $\mathcal f$ is differentiable at $\mathcal v$.

Attempt at a solution

I want to show that

$$ \lim_{h\rightarrow 0}\frac {f(v+h)-f(v)}{h}$$

exists. Based on the fact that

$$ \lim_{x\rightarrow v}f'(x)=\lim_{x\rightarrow v}\ \lim_{h\rightarrow 0}\frac {f(x+h)-f(x)}{h}$$

exists. Using the fact that $\mathcal f$ is continuous, I think if somehow I could interchange the limits $$\mathcal {\lim_{x\rightarrow v}\ \lim_{h\rightarrow 0}\frac {f(x+h)-f(x)}{h}}={\lim_{h\rightarrow 0}\ \lim_{x\rightarrow v}\frac {f(x+h)-f(x)}{h}}=\lim_{h\rightarrow 0}\frac {f(v+h)-f(v)}{h}$$ then the result comes out. But I don't think that I could do that, based on continuity. Other than that, I am not sure how to approach this.

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How about the mean value theorem? – William Oct 26 '12 at 3:06
L'Hospital is your friend... – DonAntonio Oct 26 '12 at 3:18

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 $(0,1)$

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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 $L$.

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