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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 $B(1/2,r)\subseteq\mathbb{C}\setminus V$?

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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).

b) A function whose distance to an unbounded function is less than a constant is unbounded.

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.

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  • $\begingroup$ But the uniform convergence is given (by contradiction) on $\Omega$ not on $\overline{\Omega}$. $\endgroup$ May 2, 2023 at 20:50
  • $\begingroup$ Ok, I just need to use Maximum Modulus Principle for that. $\endgroup$ May 2, 2023 at 21:25

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