Suppose $f : G \longrightarrow \mathbb C$ is analytic and define $\phi : G \times G \longrightarrow \mathbb C$ by $\phi(z,w) = [f(z) - f(w)] (z-w)^{-1}$ if $z \neq w$ and $\phi(z,z) = f'(z)$. Prove that $\phi$ is continuous and for each fixed $w$, $z \mapsto \phi(z,w)$ is analytic.

I have proved the continuity of $\phi$ in $G \times G$ which is pretty easy and analyticity of the map $z \mapsto \phi (z,w)$ in $G \setminus \{w\}$ for each fixed $w$ in $G$. At $w \in G$ I don't find any direct approach to prove that the map is analytic at $z=w$ for each fixed $w \in G$. I first show that $\int_{\partial T} f(z)\ dz =0$ for any closed triangle $T$ in $G$ (by using Goursat's theorem for the function analytic except at one point in a region) and then by Morera's theorem it has been concluded that the map is indeed analytic in $G$.

However I don't like this process anyway. I think there is an elegant way to prove it directly though I have failed to find it inspite of my effort. Please help me by giving some suggestions.

Thank you in advance.

  • $\begingroup$ Hello! how does prove that $φ(z,w)$ is continuous on$G×G$? I don't know how prove $φ(z,w)$ is continuous at $(z_0,z_0)$ $\endgroup$
    – user158796
    Aug 21, 2021 at 9:26
  • 1
    $\begingroup$ @user158796$:$ Observe that $$f(z) - f(w) = \int_0^1 f'(w +t(z-w))(z-w)\ dt$$ for $z,w$ in any $B(a,r)\subset G.$ So if $z,w \in B(a,r)\subset G,$ then $\varphi(z,w) = \int_0^1f'(w+t(z-w))\ dt,$ including the case where $z=w.$ Note that $|w+t(z-w) - z_0| \le |z-z_0|+|w-z_0|$ for $t\in [0,1].$ Thus as $(z,w) \to (z_0,z_0),$ $f'(w+t(z-w)) \to f'(z_0)$ uniformly on $[0,1].$ That gives the desired convergence of the integral to $f'(z_0),$ which is what you want. $\endgroup$ Oct 2, 2021 at 13:55

1 Answer 1


The function $f$ has a power series expansion around $w$:$$f(z)=a_0+a_1(z-w)+a_2(z-w)^2+\cdots,$$with $a_k=\frac{f^{(k)}(w)}{k!}$. Therefore$$\frac{f(z)-f(w)}{z-w}=a_1+a_2(z-w)+a_3(z-w)^2+\cdots$$around $w$.

  • $\begingroup$ Thanks for your help @Jose Carlos Santos.That means if the map is $F$ then $F'(w) = f(w)$ and for $z \neq w$ we have $F'(z) = \frac {1} {z-a} \left [ f'(z) - \frac {f(z) - f(a)} {z - a} \right ]$. Then as $z \rightarrow w$ what will happen? Does $F'(z) \rightarrow f(w)$? As we have to show that $F$ is $C^1$. $\endgroup$ Sep 2, 2017 at 19:57
  • $\begingroup$ Sorry! $F'(z) = \frac {1} {z-a} \left [ f'(z) - \frac {f(z) - f(a)} {z - a} \right ]$ for $z \neq w$. $\endgroup$ Sep 2, 2017 at 20:03
  • $\begingroup$ Sorry! $F'(z) = \frac {1} {w-a} \left [ f'(z) - \frac {f(z) - f(w)} {z - w} \right ]$ for $z \neq w$. $\endgroup$ Sep 2, 2017 at 20:07
  • $\begingroup$ @ArnabChatterjee. No, it's the other way around. If this map is $F$, then $F(w)=f'(w)$. Besides, there is no $a$ here. $\endgroup$ Sep 2, 2017 at 20:22
  • $\begingroup$ Sorry it was a typo which I couldn't avoid in any of my line. Here $F'(w) = \frac {f"(w)} {2!}$ and $$\lim_{z \rightarrow w} F'(z) = \frac {f"(w)} {2!} = F'(w).$$ Another way of looking into it is that since $F$ has power series expansion around $w$ we have $F$ is analytic at $w$ by analyticity of power series. Isn't it? $\endgroup$ Sep 2, 2017 at 20:32

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