Disk algebra norm clousre I have a trouble with a question and i need help to solve it.
Define
$A_1$={$f$ $\in  C(\overline{\mathbb{D}})$ | f is analytic in  $\mathbb{D}\}$
$A_2$=the norm closure of polynomials in $C(\overline{\mathbb{D}})$
i need to show that $A_1$=$A_2$
i have already shown that $A_2 \subseteq A_1$
but i dont know how to do  the other direction.
thanks!
 A: Let $f \in A_1$, and $\varepsilon > 0$ arbitrary. Since $f$ is uniformly continuous ($\overline{\mathbb{D}}$ is compact), there is an $r \in (0,1)$ with
$$\max \left\{ \lvert f(z) - f(rz)\rvert : z \in \overline{\mathbb{D}}\right\} < \frac{\varepsilon}{2}.$$
Now for such an $r$, the function $g(z) := f(rz)$ is analytic on a neighbourhood of the closed unit disk, namely on the disk $D_{1/r}(0) = \left\{ z : \lvert z\rvert < \frac{1}{r}\right\}$. The Taylor series of $g$ about $0$ converges uniformly on every compact subset of $D_{1/r}(0)$, in particular on $\overline{\mathbb{D}}$. That means there is an $n\in\mathbb{N}$ such that
$$\left\lvert g(z) - \sum_{k=0}^n \frac{g^{(k)}(0)}{k!}z^k \right\rvert < \frac{\varepsilon}{2}$$
for all $z\in \overline{\mathbb{D}}$. Then
$$T_n(z) = \sum_{k=0}^n \frac{g^{(k)}(0)}{k!}z^k$$
is a polynomial with $\lVert f - T_n\rVert < \varepsilon$. Since $\varepsilon > 0$ was arbitrary, that shows $f \in A_2$.
A: Let $\overline{\mathbb D}$ denote the closed unit disk and let $f: \overline{\mathbb D} \to \mathbb C$ be continuous on $\overline{\mathbb D}$ and analytic on $\mathbb D$. Let $p_n : \overline{\mathbb D} \to \mathbb C$ denote the $n$-th Taylor polynomial of $f$ at $0$. Since $|p_n-f|$ is continuous it is uniformly continuous on $\overline{\mathbb D}$. Let $g_n = p_n -f$ for convenience. From complex analysis we know that for every $r \in (0,1)$ on $B(0,r)$ we have $\|g_n\|_\infty \to 0$. 
The goal now is to show that this also holds for $r=1$. To this end, let $x \in \partial \overline{\mathbb D}$ and $\varepsilon >0$. Let $\delta$ be such that $\|g_n(x)-g_n(y)\|<{\varepsilon \over 2}$ if $|x-y|<\delta$. Then $\|g_n\|_\infty \to 0$ on $B(0,1-{\delta \over 2})$ hence we may choose $n$ so large that $\|g_n\|\infty < {\varepsilon \over 2}$ on $B(0,1-{\delta \over 2})$.
Now pick $y_x \in B(0,1-{\delta \over 2}) \cap B(x,\delta)$. Then
$$ |g_n(x)| \le |g_n(x) -g_n(y_x)| + |g_n(y_x)| < {\varepsilon \over 2} + {\varepsilon \over 2} $$
