# Uniform convergence of polynomials onto a holomorphic function

I need help with these exercise.

Let $$\Omega=\{z\in\mathbb{C}||z|<1,|z-1/2|>1/2\}$$ and $$f\in\mathcal{H}(\Omega)$$.

(a) Prove that there exists a sequence of polynomials $$\{P_n\}_{n\in\mathbb{N}}$$ converging uniformly to $$f$$ on every compact set of $$\Omega$$.

I know this result which is a consequence of Runge's theorem:

If $$K$$ is compact, $$\mathbb{C}\setminus K$$ is connected and $$f$$ is holomorphic on an open set which contains $$K$$, then $$f$$ is uniform limit on $$K$$ of a sequence of polynomials.

For this part of the exercise, I defined the exhaustive sequence of compact sets $$\{K_m\}_{m\in\mathbb{N}}$$ of $$\Omega$$ as follows:

$$K_m=\{z\in\Omega|\text{dist}(x,\partial\Omega)\geq 1/m\}.$$

It's very easy to check geometrically that $$\mathbb{C}\setminus K_n$$ is connected for every $$m\in\mathbb{N}$$. Applying the result I mentioned above, for each $$m\in\mathbb{N}$$, there exists a sequece of polynomials $$\{P^m_n\}_{n\in\mathbb{N}}$$ converging uniformly to $$f$$ on $$K_m$$.

I know that if $$K\subseteq\Omega$$, there exists $$m\in\mathbb{N}$$ such that $$K\subseteq K_m$$. But how can I simplified $$\{P^m_n\}_{n\in\mathbb{N}}$$ to just one index and be valid for every $$K_m$$?

(b) Does always exist a sequence of polynomials $$\{P_n\}_{n\in\mathbb{N}}$$ which converges uniformly to $$f$$ on $$\Omega$$?

No, in general. Let $$f:z\in\Omega\mapsto\dfrac{1}{z}\in\mathbb{C}$$ and consider a sequence $$\{a_m\}_{m\in\mathbb{N}}\subseteq\Omega$$ converging to $$0$$. Suppose by contradiction that there exists a sequence of polynomials $$\{P_n\}_{n\in\mathbb{N}}$$ which converges uniformly to $$f$$ on $$\Omega$$. Then, there's a constant $$C>0$$ such that $$||P_n||_{\infty,\Omega}+||P_n-f||_{\infty,\Omega}\leq C$$ for all $$n\in\mathbb{N}$$. Therefore, for all $$m\in\mathbb{N}$$, we have the following:

$$|f(a_m)|\leq|f(a_m)-P_n(a_m)|+|P_n(a_m)|\leq||P_n-f||_{\infty,\Omega}+||P_n||_{\infty,\Omega}\leq C.$$

However, $$|f(a_m)|\stackrel{m\to\infty}{\rightarrow}+\infty$$ and we got a contradiction.

(c) Does the answer on (b) change if we ask $$f$$ to be holomorphic on an open set which contains $$\overline{\Omega}$$?

Let $$V\subseteq\mathbb{C}$$ an open set such that $$\overline{\Omega}\subseteq V$$ and $$f\in\mathcal{H}(V)$$. If $$B(0,1)\subseteq V$$, then it is trivially satisfied by the result I mentioned before. If $$1/2\in V'\setminus V$$, reasoning as on (b) with the function $$f(z)=\dfrac{1}{z-1/2}$$ we got a contradiction. What would it happen if there exists $$r>0$$ such that $$B(1/2,r)\subseteq\mathbb{C}\setminus V$$?

a) Take $$p_n = P^n_k$$ where $$k$$ is large enough that, say, $$|P^n_k - f| < 1/n$$ on $$K_n$$ (and therefore on $$K_m$$ for $$m < n$$ too).
c) Use $$K = \overline{\Omega}$$ in Runge. It would be more interesting if you had $$|z| < 1$$ rather than $$1/2$$ in the definition of $$\Omega$$.
EDIT: Now that it's $$|z| < 1$$, consider integrals around the unit circle. Any polynomial gives $$0$$, $$1/(z - 1/2)$$ does not.
• But the uniform convergence is given (by contradiction) on $\Omega$ not on $\overline{\Omega}$. May 2, 2023 at 20:50