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!

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. $\endgroup$