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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!
Thanks!

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  • $\begingroup$ 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
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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.

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