polynomial converges uniformily to analytic function Let $D=\{z:|z|<1\}$.   suppose $f(z)$ is analytic on $D$ and  continuous on $\{z:|z|\leqslant1\}$. prove that there exists polynomial sequence  $\{P_n(z)\}$ such that $\{P_n(z)\}$  converges uniformily to $f(z)$ on  $\{z:|z|\leqslant1\}$
 A: Assume that $f \in C(\bar D) \cap \mathcal{O}(D)$. Then the Maclaurin polynomials of $f$ converge locally uniformly on $D$ to $f$, but in general it is not true that the Maclaurin polynomials converge uniformly to $f$ on $\bar D$. In fact, it is even possible to construct examples of such $f$:s where the Maclaurin series diverges at some points of $\partial D$. (See this question.)
If you really want uniform convergence on the whole of $\bar D$, you need to work a little more. Let $r_n$ be an increasing sequence of positive numbers tending to $1$ and let
$$f_n(z) = f(r_n z).$$
Then each $f_n$ is analytic on a neighbourhood of $\bar D$. 
Let $P_n$ be the $d_n$:th Maclaurin polynomial of $f_n$, where $d_n$ is chosen so large that $$\sup_{z\in\bar D} |P_n(z) - f_n(z)| < 2^{-n}$$
(This is possible, since the Maclaurin polynomials converge locally uniformly and $f_n$ is analytic on a neighbourhood of $\bar D$.)
Hence
\begin{align}
\sup_{z\in \bar D} |P_n(z) - f(z)| &\le
\sup_{z\in \bar D} |P_n(z) - f_n(z)| + \sup_{z\in \bar D} |f_n(z) - f(z)| \\[10pt]
&\le 2^{-n} + \sup_{z\in \bar D} |f_n(z) - f(z)|.
\end{align}
By uniform continuity of $f$ on $\bar D$, the second term tends to $0$ as $n\to\infty$ and we are done.

All of this is a (simple) special case of a deep theorem by Mergelyan: If $K$ is a compact subset of $\mathbb{C}$ with connected complement, and $f$ is continuous on $K$, analytic on the interior of $K$, then $f$ can be approximated uniformly on $K$ by polynomials.
