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The following is exercise 13.2 in Rudin's Real & Complex Analysis, which I'm self-studying.

Let $\Omega = \{z: |z| < 1 \text{ and } |2z - 1| > 1\}$, and suppose $f \in H(\Omega)$. Must there exist a sequence of polynomials which converges to $f$ uniformly in $\Omega$?

I have a solution, but I feel it's simple and the shape of $\Omega$ is suspicious so there might be a trick somewhere.

Assume such a sequence exists. Let $0 < \epsilon < 1$, $f(z) = 1/z$ and $P$ be a polynomial that satisfies: $$|f(z) - P(z)| < \epsilon \ \ \forall z \in \Omega$$ Therefore $$|P(z)| < |f(z)| + \epsilon \ \ \forall z \in \Omega$$ But near the boundary of the unit disc, $|f(z)| < 2$, so $|P(z)| < 3$ near the boundary. By the continuity of $P$, we have $P(z) \le 3$ on the boundary. By the maximum modulus principle, $P(z) < 3$ on the unit disc. But $|f(z)|$ gets arbitrarily large near $0$. Therefore $P$ cannot approximate $f$ on $\Omega$.

What gives me confidence in my argument is that it doesn't work on compact subsets of $\Omega$ (for which the existence of the polynomial sequence is guaranteed by Runge's theorem).

Is my counter-example correct?

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Is that really the question? If $\Omega$ is bounded, and $f$ is an unbounded holomorphic function, you can never approximate it uniformly on $\Omega$ by polynomials for trivial reasons. Are you sure the question doesn't ask for locally uniform convergence? –  mrf Feb 19 '13 at 21:54
    
@mrf Copied word for word. It actually contains three sections, but the other two are immediate applications of Runge's theorem and a counter-example in the book (uniform convergence on compact subsets of $\Omega$ and a compact set that contains the closure of $\Omega$). I guess the purpose of the question is to prove the claim you've just made? Is there a simpler proof? –  PeterM Feb 19 '13 at 21:59
    
The simpler proof is just to note that every polynomial is bounded on $\Omega$, so $\|f-p\|=\infty$ for every polynomial $p$. –  mrf Feb 19 '13 at 22:01
    
@mrf Yeah the boundedness of $P$ on $\Omega$ is immediate and there is no need for the $\epsilon$ dance above. But it's the same argument otherwise, right? –  PeterM Feb 19 '13 at 22:04
    
@mrf I checked: this is indeed what part (b) asks about. Would not be much of a problem on its own, but it is pedagogically placed between (a) convergence on compact subsets and (c) uniform convergence under the assumption that $f$ is holomorphic on some open set that contains $\overline{\Omega}$. // PeterM: Not only $\epsilon$ turns out to be unnecessary, but also the maximum principle is not needed in (b). But it's good that you worked out this argument, because it's exactly what is needed for (c). –  user53153 Feb 20 '13 at 3:25

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