Proving that this function is continuous on $G\times G$ Let $G\subset \mathbb{C}$ be a non-empty open set and $f$ be a function holomorphic on $G$. Let $g: G\times G\to \mathbb{C}$ be a function defined as $$g(z,w)= \begin{cases} 
      \frac{f(z)-f(w)}{z-w}, & z\ne w \\
      f'(z), & z=w 
   \end{cases}$$
I need to prove that $g$ is continuous on $G\times G$.
My approach:
Let $\varepsilon > 0$. Then $\exists \delta_1, \delta_2>0$ such that $\| (z,w)-(z_0,w_0) \|^2\le|z-z_0|+|w-w_0|<\delta_1+\delta_2$ implies that $|f(z)-f(z_0)|<\epsilon$, $|f(w)-f(w_0)|<\epsilon$, since $f$ is continuous on $G$.
Now, $\| g(z,w)-g(z_0,w_0) \|=\left| \frac{f(z)-f(w)}{z-w} - \frac{f(z_0)-f(w_0)}{z_0-w_0} \right|\le \left| \frac{f(z)-f(w)}{z-w}\right| + \left|\frac{f(z_0)-f(w_0)}{z_0-w_0} \right|$
Since $f$ is holomorphic on $G$, it is continuous on $G$, thus $\exists \delta'>0$ such that $| {f(z)-f(w)}|<\varepsilon'|z-w|<\varepsilon'\delta'$ whenever $|z-w|<\delta'$.
And here's where I'm stuck, because it's not clear how to deal with $\left|\frac{f(z_0)-f(w_0)}{z_0-w_0} \right|$, because $z_0, w_0$ are fixed points, so we can't just make them approach each other. Of course, my first thought was to somehow rearrange the inequality in such a way as to obtain $\left| \frac{f(z)-f(z_0)}{z-z_0}\right| + \left|\frac{f(w)-f(w_0)}{w-w_0} \right|$, but I don't see how to do so.
Your help would be appreciated.
 A: On the open set $\Omega = G \setminus \Delta_G$, both $z - w$ and $f(z) - f(w)$ are continuous and $z - w$ has no zeros, so $g(z,w)$ is continuous. So we only need to show that $g(z,w)$ is continuous at points in $\Delta_G$, i.e.
$$\Vert (z,w) - (z_0,z_0) \Vert < \delta \implies \vert g(z,w) - g(z_0,z_0) \vert < \epsilon.$$
If $z = w$ this is easy, as then
$$g(z,w) - g(z_0,z_0) = f'(z) - f'(z_0)$$
and $f'$ is continuous. If $z \neq w$, we have
$$g(z,w) - g(z_0,z_0) = \frac{f(z) - f(w)}{z - w} - f'(z_0).$$
For this case write
$$f(z) = \sum_{n=0}^{\infty}a(n)(z - z_0)^n.$$
Then
\begin{equation*}
\begin{aligned}
\frac{f(z) - f(w)}{z - w} &= \frac{1}{z - w}\sum_{n=0}^{\infty}a(n)[(z - z_0)^n - (w - z_0)^n] \\
&= \frac{1}{(z - z_0) - (w - z_0)}\sum_{n=1}^{\infty}a(n)[(z - z_0)^n + (w - z_0)^n] \\
&= \sum_{n=1}^{\infty}a(n)[Z^{n - 1} + Z^{n-2}W + \cdots + ZW^{n-2} + W^{n-1}] \\
\end{aligned}
\end{equation*}
with $Z = z - z_0, W = w - z_0$. We have
$$\frac{f(z) - f(w)}{z - w} - f'(z_0) = \sum_{n=2}^{\infty}a(n)[Z^{n - 1} + Z^{n-2}W + \cdots + ZW^{n-2} + W^{n-1}].$$
Thus if $\vert Z \vert, \vert W \vert < \delta$, then
$$\left\vert \frac{f(z) - f(w)}{z - w} - f'(z_0) \right\vert < \sum_{n=2}^{\infty}n \vert a(n) \vert\delta^{n - 1} < \epsilon.$$
A: If $z_0\ne w_0$ there is an $\epsilon>0$ such that $U_\epsilon(z_0)$ and $U_\epsilon(w_0)$ are disjoint. One then has
$$g(z,w)={f(w)-f(z)\over w-z}\qquad\forall\>(z,w)\in U_\epsilon(z_0)\times U_\epsilon(w_0)\ ,$$
hence $g$ is continuous at $(z_0,w_0)$.
If $z_0=w_0$ choose a convex neighborhood $U\subset \Omega$ of $z_0$. Fix two points $z$, $w\in U$, and consider the auxiliary function
$$\phi(t):=f\bigl((1-t)z+tw\bigr)\qquad(0\leq t\leq 1)$$with derivative
$$\phi'(t)=f'\bigl((1-t)z+tw\bigr)\>(w-z)\ .$$
One has
$$f(w)-f(z)=\phi(1)-\phi(0)=\int_0^1\phi'(t)\>dt=(w-z)\int_0^1f'\bigl((1-t)z+tw\bigr)\>dt\ .$$
If $z\ne w$ we therefore have
$$g(z,w)=\int_0^1f'\bigl((1-t)z+tw\bigr)\>dt\ ,\tag{1}$$
and if $z=w$ formula $(1)$ is trivially true. Since the right hand side of $(1)$ is continuous on $U\times U$ we are done.
