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Let $\{P_n\}$ be a series of polynomials which uniformly converges on the unit circle $ T = \{|z| = 1\}$ to $f$.

prove there exists $F$ such that $F$ is analytic in the unit disk $D$ and continuous in the closed unit disk $\bar{D}$ such that $F \equiv f$ on $T$.

I think that f must by analytic on $T$ so the radius of convergence is at least 1. so f actually qualifies to be F. but I'm not sure I'm correct?

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  • $\begingroup$ $f$ does not qualify because $f$ is not defined on $\overline D$... $\endgroup$
    – 5xum
    Aug 31, 2015 at 11:28
  • $\begingroup$ why is it not defined? f is defined on T, and we can write it as a power series, with a radius of convergence = 1 $\endgroup$ Aug 31, 2015 at 11:29
  • $\begingroup$ You are correct in a sense. $f$ is only defined on $T$. However, you can take the power series of $f$ and that power series converges to some function on $D$. The function cannot be $f$ because it has a different domain, however, it is equal to $f$ on $T$. $\endgroup$
    – 5xum
    Aug 31, 2015 at 11:32
  • $\begingroup$ can you please help how do i prove that f must be analytic on T? $\endgroup$ Aug 31, 2015 at 12:23

1 Answer 1

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The restriction of $P_n$ to $T$ converges uniformly, so $(P_n|_T)$ is uniformly Cauchy on $T$. By the maximum modulus principle, $(P_n)$ is also uniformly Cauchy on $\bar D$, and thus converges uniformly to some (continuous) function $F$ on $\bar D$.

On the other hand, Morera's theorem shows that a uniform limit of holomorphic functions is holomorphic, so in fact $F$ is holomorphic on $D$.

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