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Let $f$ be a continuous function on $\mathbb{T}=\{|z|=1\}$ so that there exist a series of polynomials that converges to $f$ uniformly on $\mathbb{T}$. Prove that there is a function $F$ that is continuos on $\{|z|\leq 1\}$, analytic on $\{ |z|<1 \}$ and $F=f$ on $\mathbb{T}$.

I thought to define $F=\lim_{n\rightarrow \infty}P_n(z)$ but I dont know to explainwhy should it converge in $\{ |z|<1 \}$, and if it does, why need $\{ P_n \}$ converge uniformly in $\mathbb{T}$.

Thanks.

catch22
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1 Answers1

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By the maximum modulus principle,

$$\sup_{z \in \overline{\mathbb{D}}} \left\lvert P_n(z) - P_m(z) \right\rvert = \sup_{z \in \mathbb{T}} \left\lvert P_n(z) - P_m(z) \right\rvert,$$

therefore uniform convergence of $(P_n)_{n\in\mathbb{N}}$ on $\mathbb{T}$ implies uniform convergence on $\overline{\mathbb{D}}$, thus you know that $F$ exists, is continuous on $\overline{\mathbb{D}}$, and coincides with $f$ on the boundary.

What remains is the analyticity in $\mathbb{D}$. There are various ways to deduce that, by Morera's theorem for example, or by the fact that uniform convergence on the boundary and pointwise convergence in the interior imply that $F$ is the Cauchy integral of its boundary values.

Daniel Fischer
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    If the $P_n$ converge to $f$ on $\mathbb{T}$ only pointwise, it could be that nevertheless the $P_n$ are unbounded on $\mathbb{T}$ (not sure if that can actually happen with polynomials, but well, at least you don't have a guarantee that the sequence is well-behaved). If you don't know that the convergence on $\mathbb{T}$ is uniform, you can't deduce that there is even a pointwise limit in the interior of the disk. – Daniel Fischer Jul 29 '13 at 15:16