If $U$ is an open set in $\mathbb{C}$ and $z_0\in U$, could there be a function $f:U\rightarrow \mathbb{C}$ which is continuous in $U$, differentiable in $U \setminus \{z_0\}$, but not differentiable at $z_0$?
One possible example I could come up with is $f(z)=\sin(z)/z$ and $z_0=0$, but I’m not sure this is a valid one because if I’m not mistaken the function isn’t defined at $0$.
I’ve also heard of the Riemann removable singularity theorem, but I’m not sure it answers my question given that it’s application here would imply defining another function which holomorphically extends the original one over $z_0$… right?
Best Answer
In fact, Riemann's removable singularity theorem answers the question. It shows that $f|_{U\setminus\{z_0\}}$ can be extended to an analytic function $F\colon U\longrightarrow\Bbb C$. But, since $f$ and $F$ are continuous functions and since $(\forall z\in U\setminus\{z_0\}):f(z)=F(z)$, you have $(\forall z\in U):f(z)=F(z)$. So, $f$ is an analytic function.