Can someone, help me in this question, please?

Let $U\subset\mathbb{C}$ be open set and $H_\infty(U)=\{f:U\to\mathbb{C}:f\text{ is bounded and homolorphic}\}$. Show that $H_\infty(U)$ is a closed subset of $C(U)=\{f:U\to\mathbb{C}\text{ continuous}\}$, with respect the norm $$ \|f\|=\sup\{f(x):x\in U\} $$

This is a random question, of random studies on function analysis, and I don't have a good background in complex variables.

Given a sequence $(f_n)$ in $H_\infty(U)$ that converges to $f\in C(U)$, i would like to prove that the limit commutes with the derivative (or partial derivatives of the real/imaginary parts), hence $f$ satisfies the Cauchy-Riemman conditions, so $f$ is holomorphic.
But, since the derivative operator is not continuous (or is it in these conditions?), this isn't true, and i have no idea what to do...

Any help will be appreciated!

  • $\begingroup$ The operator taking $f \in H_{\infty}(U)$ to $f'(z_0)$ where $z_0 \in U$ is continuous in the uniform norm on $H_{\infty}(U)$. Do you see how to prove that using the Cauchy integral formula? $\endgroup$ – A Blumenthal Sep 10 '13 at 17:25

This follows from Morera's Theorem:

Suppose $f: U \to \mathbb{C}$ is continuous and for any triangle $T$ contained in $U$ we have $$\int_T \! f(z) \, dz = 0.$$ Then $f$ is holomorphic.

(You can find a proof in Complex Analysis by Stein & Shakarchi.)

Now just note that if $f_n: U \to \mathbb{C}$ is holomorphic and $f_n \to f$ uniformly, then for any triangle $T$ in $U$ we have $$\int_T \! f(z) \, dz = \lim_{n \to \infty} \int_T \! f_n(z) \, dz = \lim_{n \to \infty} 0 = 0$$ by Cauchy's Theorem.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.