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Let $\Omega=\{ z\in \mathbb{C}:$ $Im$ $z>0,$ $|z|>1\}\cup\{z \in \mathbb{C}:$ $Im$ $z<0$ $|z|>1\}$

I know that since $\hat{\mathbb{C}}\setminus \Omega$ is connected there's a sequence of polynomials such that $\{P_n\}$ converges uniformly on the compact sets of $\Omega$. My question is:

Given $f\in \mathcal{H}(\Omega)$, can we always find a sequence of holomorphic functions $\{f_n\}$ in some open set $A$ ($\overline{\Omega}\subset A$) such that $\{ f_n \}$ converges uniformly to $f$ in $\Omega$?

I think I'm missing something because each $P_n \in \mathcal{H}({\mathbb{C}})$ and the answer would be positive. I don't really care about the exact solution I'd like to know the differences between these two cases .

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This is not true.

Consider $f=\dfrac{1}{z-2}$ with a singularity at $z=2\in \bar{\Omega}$. Obviously, $f\in H(\Omega)$.

Suppose there exists $\{f_n\}\in H(A)$ with $\bar{\Omega}\subset A$, s.t. $f_n\to f$ uniformly in $\Omega$. Then we pick up the sequence of points $z_m=2+\dfrac{i}{m}\in \Omega$ which converges to $z=2$. Since the converge of $\{f_n\}$ is uniform, which implies that there exists a function $f_k$, s.t. $|f_k(z)-f(z)|\leq 1, \forall z\in \Omega$.

But $|f(z_m)|\to \infty$ tells us that $|f_k(z_m)|\to \infty$, whence $f_k(2)=\infty$, which is a contradiction since $f_k\in H(A)$ and $2\in A$.

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  • $\begingroup$ this seems nice but main question was why my reasoning about the polynomials is wrong? $\endgroup$
    – Abellan
    May 22, 2015 at 21:38
  • $\begingroup$ What is your reasoning about polynomials? $\endgroup$
    – booksee
    May 22, 2015 at 21:42
  • $\begingroup$ $\hat{\mathbb{C}}\setminus \Omega$ is connected so there is a sequence of polynomials $\{P_n\}$ which converges uniformly on compacts $K \subset \Omega$ and each $P_n \in \mathcal{H}(\Omega)$ So why can't I choose $\{f_n\}=\{P_n\}$? $\endgroup$
    – Abellan
    May 22, 2015 at 21:46
  • $\begingroup$ Did you mean $P_n$ converges to $f$ on $K$? This is true because $K$ is compact. However, we can not approximate $f$ uniformly in $\Omega$ with functions in $H(A) (\bar{\Omega}\subset A)$ since $\Omega$ is open and $f$ may have singularities on the boundary, as illustrated in answer. $\endgroup$
    – booksee
    May 22, 2015 at 21:57
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    $\begingroup$ Failure to be compact makes a huge difference. We have result for approximation in compact subsets of $\Omega$(e.g., the Stone–Weierstrass Theorem) but no general result is known for approximation in $\Omega$ and it is usually impossible to do so because otherwise the definition of $f$ may be extended to a larger domain. $\endgroup$
    – booksee
    May 22, 2015 at 22:08

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