Proving a function is continuous for a fixed variable

This is from Conway's book:

Suppse $f:G \rightarrow \mathbb{C}$ is analytic and define $\phi: G \times G \rightarrow \mathbb{C}$ by $\phi(z,w)=\frac{f(z)-f(w)}{(z-w)}$ if $z \neq w$ and $\phi(z,z)=f'(z)$. Prove that $\phi$ is continuous...

Because $f$ is analytic, $f'$ exists for all points $z$ in the region $G \times G$.

I think I get for free that $f'$ is continuous. But I'm not sure where to go from here.

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You should be able to explain why $\phi$ is continuous wherever $z\ne w$, so you're left with showing $\phi$ is continuous at any point on the line $z=w$. Show that for arbitrary $a\in\mathbb{C}$, $$\lim_{(z,w)\to(a,a)}\phi(z,w)=f\,'(a).$$ You can split evaluation of the limit up into two cases: directions towards $(a,a)$ that do not lie on the line $z=w$ (use the definition of the derivative) and the two directions that are on it (use continuity of $f\;'$). –  anon Oct 11 '11 at 3:56

Let me show you why there is a subtlety here. Let's switch to the world of real functions. Let $$f(x) = \begin{cases} x^2 \sin(x^{-1}) & x \neq 0 \\ 0 & x=0 \end{cases}$$ Then $f$ is differentiable. (Obvious at $x \neq 0$, and $\lim_{h \to 0} (f(h)-f(0))/h = \lim_{h \to 0} h \sin h^{-1}=0$.)

Now, let $x = ( (2N+1/2)\pi)^{-1}$ and $y = ((2N-1/2)\pi)^{-1}$ for $N$ an integer. So $$\frac{f(x) - f(y)}{x-y} = \frac{( (2N+1/2)\pi)^{-2} + ((2N-1/2)\pi)^{-2}}{( (2N+1/2)\pi)^{-1} - ((2N-1/2)\pi)^{-1}}$$ $$\approx \frac{1}{\pi} \frac{2 (2N)^{-2}}{- (2N)^{-2}} = - \frac{2}{\pi}.$$

So, as $(x,y) \to (0,0)$ along the sequence $(( (2N+1/2)\pi)^{-1}, ( (2N-1/2)\pi)^{-1})$, the ratio $(f(x)-f(y))/(x-y)$ does not go to $0$.

Your job is to show that this doesn't happen in the complex world.

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The statement follows immediately from the equality $$f(z)-f(w)=(z-w)\ \int_0^1\ f'(w+tz-tw)\ dt.\tag1$$

EDIT 1. More precisely, the formula makes sense for $z$ and $w$ belonging to any given convex subset of $G$. But this is sufficient, because the continuity is obvious off the diagonal.

EDIT 2.

• The above argument shows that $$\frac{f(z)-f(w)}{z-w}$$ is holomorophic.

• To prove $(1)$, set $g(t):=f(w+tz-tw)$ and observe $g(1)-g(0)=\int_0^1\,g'(t)\,dt$.

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Could you tell me what is going on here? I do not understand your solution from the first line to last line. –  Une Femme Douce Apr 30 '13 at 12:55
Dear @Tsotsi: Thanks for your interest. Could you tell me the first thing that is unclear? –  Pierre-Yves Gaillard Apr 30 '13 at 15:11
equation $1$ in your answer –  Une Femme Douce Apr 30 '13 at 18:06
Dear @Tsotsi: This is explained on the last line of the answer. (Sorry for answering your comment so late; I've been busy.) –  Pierre-Yves Gaillard May 1 '13 at 4:52