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Let $f$ be analytic on the open set $G\subset \mathbb{C}$. What's the best way of showing that the function $\phi:G \times G \to \mathbb{C}$ defined by $\phi(z,w)=[f(z)-f(w)]/(z-w)$ for $z \neq w$ and $\phi(z,z)=f'(z)$ is continuous?

This is obvious for $z\neq w$, but I'm having some trouble on the points $(z,z)$. I did it one way but I'm not convinced this is good.

Edit: Rudin's Proof (for the continuity on the diagonal)

Let $z_0\in G$. Since $f$ is analytic, there exists $r>0$ such that $B(a,r)\subset G$ and $|f'(\zeta)-f'(z_0)|<\epsilon$ for all $\zeta \in B(a,r)$. Getting $z,w\in B(a,r)$, than $\zeta(t)=(1-t)z+tw \in B(a,r)$ for $0\le t\le 1$. Now just use that $\phi(z,w)-\phi(z_0,z_0)=\int_0^1[f'(\zeta(t))-f'(z_0)]dt$ and the $\epsilon$-bound to get the desired continuity.

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    $\begingroup$ The fastest way is to use the definition: just compute the limits and voilà $\endgroup$ – Mariano Suárez-Álvarez Jan 25 '11 at 2:17
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Probably not best, but one approach to showing continuity at $(z_0,z_0)$ is to use the power series expansion $f(z)=\sum_{k=0}^\infty a_k(z-z_0)^k$ to get $$ \begin{align*} \frac{f(w)-f(z)}{w-z}-f'(z_0)&=\sum_{k=1}^\infty a_k\frac{(w-z_0)^k-(z-z_0)^k}{(w-z_0)-(z-z_0)}-f'(z_0)\\ &=\sum_{k=2}^\infty a_k\sum_{j=0}^{k-1}(w-z_0)^j(z-z_0)^{k-1-j}, \end{align*}$$ which converges absolutely and uniformly near $(z_0,z_0)$ and goes to $0$ at $(z_0,z_0)$.

A better candidate for "best" is that given in Rudin's Real and complex analysis, Lemma 10.29 on page 314 of the 3rd Edition.

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  • $\begingroup$ Ok, I did something like this too (sorry, it would be better if I told it). I've also tried other expansion arguments (like expand f(w) around z, but I'm not sure how to prove that this power series (in two variables) is continuous). $\endgroup$ – user1971 Jan 25 '11 at 3:59
  • $\begingroup$ @John: Here's a hint. Assume that $|w-z_0|<r$ and $|z-z_0|<r$ where $r$ is less than the radius of convergence, apply absolute values, and use the fact that $\sum_{k=2}^\infty ka_k(z-z_0)^k$ has the same radius of convergence as the original series. For example you could apply Weierstrass's M test to justify uniform convergence of the series in this bidisk. $\endgroup$ – Jonas Meyer Jan 25 '11 at 4:07
  • $\begingroup$ Yes, that ends the problem. Thanks. I've just seen Rudin's proof, it's fantastic. $\endgroup$ – user1971 Jan 25 '11 at 4:34
  • $\begingroup$ @JonasMeyer Can you please clarify how you went from $\sum_{k=1}^\infty a_k\frac{(w-z_0)^k-(z-z_0)^k}{(w-z_0)-(z-z_0)}-f'(z_0)$ to $\sum_{k=2}^\infty a_k\sum_{j=0}^{k-1}(w-z_0)^j(z-z_0)^{k-1-j}$? $\endgroup$ – sequence Mar 16 '17 at 4:51
  • $\begingroup$ The first term in the sum is $f'(z_0)$, which is why $f'(z_0)$ was cancelled and the index starts at $2$. Then, use the identity $a^k - b^k = (a-b)(b^{k-1}+ab^{k-2}+\cdots+a^{k-2}b+a^{k-1})$. $\endgroup$ – Jonas Meyer Mar 16 '17 at 4:57
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I just want to note that existence of the limit of $\phi$ at $(z,z)$ is the definition of the strong (or strict) differentiability of $f$ az $z$. It is a key strengthened concept of differentiability, which already makes the inverse, and thus the implicit function theorem work. It implies that $f'$ is continuous at $z$, and if $f'$ exists on a neighbourhood of $z$ the reverse is also true, which also anwers the question. This is a consequence of the mean value inequality: $$|f(u)-f(w)-f'(z)(u-w)|\le\sup_{0<t<1}|f'(u+t(w-u))-f'(z)||u-w|.$$

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