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Let $D(0,1)$ be the disk of center 0 and radius 1 and call $A_{D(0,1)}= \{ f:\overline{D(0,1)} \rightarrow \mathbb{C} : f \text{ is continuous and } f|_{D(0,1)} \text{ is holomorphic} \}$.

Can somebody help me in proving that the polynomials (in the variable z) are uniformly dense in $A_{D(0,1)}$? I have tried to write a function $f$ as a power series (as it is holomorphic in $D(0,1)$) and then take the $n$ first elements of the sum so that I get a polynomial that is "near" my function. But then, what happens in the boundary? Is this enough and the result is given by continuity?

Thank you!

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    $\begingroup$ This is Mergelyan's theorem, see the last chapter of Rudin's Real and complex analysis. Maybe the proof is simpler for the disk (Mergelyan's theorem applies to any simply-connected compact subset of $\Bbb{C}$), but it is certainly not trivial. $\endgroup$
    – abx
    Feb 26, 2014 at 9:41
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    $\begingroup$ For those wondering why this is not Stone–Weierstrass, the set of polynomials is not closed under conjugation. $\endgroup$
    – Ben Barber
    Feb 26, 2014 at 10:16
  • $\begingroup$ Thank you both. Mergelyan's theorem seems something far more general than what I need. If there's anyone with a simpler proof for the case of the disk I would be really grateful to read it. $\endgroup$
    – Seasoned
    Feb 26, 2014 at 10:44

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You can simply take the Taylor series of $f$. The partial sums might not converge to $f$, but their Cesaro means will, even uniformly.

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