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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!

david
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  • What is you norm? – AlexR Jan 20 '14 at 10:16
  • Hint: Every function that is analytic in $\mathbb{D}$ is represented by its Taylor series about $0$ in the entire open unit disk. – Daniel Fischer Jan 20 '14 at 10:20
  • the norm is the sup norm edit: and i try to use poisson kernel and i feel it the right direction but i dont know how to prove it. i read in some book yesterday that i should look at the function f(rz) when r<1 and i know that the possion kernal converges uniformly to f when r tends to 1. – david Jan 20 '14 at 10:42

2 Answers2

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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 $f \in A_2$.

Daniel Fischer
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  • In the meantime I proved it. More or less like you do except instead of $rz$ I use $y \in \overline{\mathbb D}$ and I don't mention the disk of radius $1/r$. – Student Feb 03 '14 at 07:59
  • ''The Taylor series of $g$ about $0$ converges uniformly on every compact subset of $D_{1/r}(0)$'' -- doesn't it converge uniformly on every subset that doesn't contain any points of the boundary? Like for example, it converges uniformly on the open disk around $0$ of radius $1/2$, no? – Student Feb 10 '14 at 09:26
  • It converges uniformly on every subset of $D_{1/r}(0)$ that "has a positive distance from the boundary". But every such subset is contained in a compact subset. If a sequence converges uniformly on $A$, it also converges uniformly on every subset of $A$, so uniform convergence on all compact subsets implies uniform convergence on all relatively compact subsets. – Daniel Fischer Feb 10 '14 at 10:02
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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

$$ |g_n(x)| \le |g_n(x) -g_n(y_x)| + |g_n(y_x)| < {\varepsilon \over 2} + {\varepsilon \over 2} $$

Student
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  • To show that you can pick the $\delta$ independent of $n$, you need to show that the $g_n$ are uniformly equicontinuous on $\overline{\mathbb{D}}$. The easiest method to do that is, I think, to show that the Taylor polynomials converge uniformly on $\overline{\mathbb{D}}$. – Daniel Fischer Feb 03 '14 at 12:58
  • @DanielFischer I see the flaw in my argument. Thank you for your comment! – Student Feb 10 '14 at 09:15