Show that $g$ is continuous

Question

This question is from an old Ph.D Qualifying Exam for Complex Analysis.

---

Let $\Omega\subset\mathbb{C}$ be an open set. Suppose that $f$ is holomorphic in $\Omega$. Define $g$ on $\Omega\times\Omega$ by $$g(z,w)= \begin{cases} \dfrac{f(w)-f(z)}{w-z}, & w\neq z \\ f'(z), & w=z \end{cases}$$

Show that $g$ is continuous in $\Omega\times\Omega$.

---

My attempt: If $w\neq z$ then $g$ is clearly continuous, so we just have to consider the case of $w=z$. Clearly, for a fixed $a\in \Omega$, $\lim_{w\to a}(\lim_{z\to a}g(z,w))=f'(a)$, but the double limit need not be identical to the joint limit $\lim_{(z,w)\to (a,a)}$, so I'm not sure this is right. Does anyone have ideas?

Answer 1

Let $(z_n,w_n) \to (a,a)$. If $z_n=w_n$ for infinitely many $n$ then $g(z_n,w_n)=f'(z_n) \to f'(a)$ along this subsequence because $f'$ is continuous. Now consider the limit of $g(z_n,w_n)$ along a subsequence with $z_n \neq w_n$. In this case use the fact that $\frac {f(z_n)-f(w_n)} {z_n-w_n} \to f'(a)$. To see why this is true write $\frac {f(z_n)-f(w_n)} {z_n-w_n}$ as $\frac {\int_{\gamma} f'(\zeta )\, d\zeta } {z_n-w_n}$ so $\frac {f(z_n)-f(w_n)} {z_n-w_n}-f'(a) =\frac {\int_{\gamma} (f'(\zeta )-f'(a))\, d\zeta } {z_n-w_n}$ where $\gamma $ is teh line segment from $z_n$ to $w_n$. Use continuoty of $f'$ to complete the proof.

Answer 2

OK so by definition $f'(z)$ is the limit as $w \rightarrow z$ of $(f(z)-f(w))/(w-z)$

https://en.wikipedia.org/wiki/Holomorphic_function

So this tells us that, by bringing $w$ close to $z$ we can make the difference between $f(z)-f(w)/(w-z)$ and $f^{'}(z)$ as small as we please which is simply to say that $g$ is continuous at $(z,z)$. As you say, comtinuty at $(z,w)$ for $z \neq w$ is obvious so we have shown continuity for all of $\Omega \times \Omega$.

Answer 3

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.