I have a function $f:\mathbb{R}\to\mathbb{R}$ that is continuous which is differentiable in every point except for a point $a \in \mathbb{R}^+$. We also know that the first differential $f':\mathbb{R}\setminus \{a\}\to\mathbb{R}$ is uniformly continuous. Find out whether or not the function is differentiable in the point $a$.
I don't really know how to get started on this. I believe that it's true, but I don't know how to begin to prove it. Can anyone provide a hint to get me going in the right direction?
Best Answer
For any sequence $(x_{n})$ such that $x_{n}\ne a$, $x_{n}\rightarrow a$, we claim that the sequence $(f'(x_{n}))$ is Cauchy and hence convergent. Indeed, given any $\epsilon>0$, since $f':\mathbb{R}\setminus\{a\}$ is uniformly continuous, there exists a $\delta>0$ such that $|x-y|<\delta$, $x,y\ne a$ implies $|f'(x)-f'(y)|<\epsilon$.
Since $x_{n}-x_{m}\rightarrow 0$ as $n,m\rightarrow\infty$, $|x_{n}-x_{m}|<\delta$ for sufficiently large $n,m$, and so $|f'(x_{n})-f'(x_{m})|<\epsilon$ for all such $n,m$, this shows that $(f'(x_{n}))$ is Cauchy.
Let $L=\lim_{n\rightarrow\infty}f'(x_{n})$. We show that $L$ is independent of any choice of such sequence $(x_{n})$. Indeed, for any other $(y_{n})$ such that $y_{n}\ne a$ and $y_{n}\rightarrow a$, consider the sequence $(z_{n})$ defined by $z_{2n}=x_{n}$ and $z_{2n+1}=y_{n}$, a subsequence argument shows that $L=\lim_{n\rightarrow\infty}f'(y_{n})$.
Now we look at the difference quotient \begin{align*} \dfrac{f(a+h)-f(a)}{h} \end{align*} for $h\ne 0$. By Mean Value Theorem, one has \begin{align*} \dfrac{f(a+h)-f(a)}{h}=f'(\eta_{h}) \end{align*} for some $\eta_{h}$ in between $a$ and $a+h$, in particular $\eta_{h}\ne a$. As $h\rightarrow 0$, $\eta_{h}\rightarrow a$ and so $f'(\eta_{h})\rightarrow L$. The derivative exists and the value is simply $L$.
Indeed, one shall realize $h\rightarrow 0$ to arbitrary sequence $(h_{n})$ that $h_{n}\rightarrow 0$, and the $\eta_{h}$ becomes $\eta_{n}$ which lies in between $a$ and $a+h_{n}$ with $\eta_{n}\ne a$.