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Let $\{p_n\}$ be a set of (complex) polynomials such that $$ \displaystyle \lim_{n\to \infty} p_n(z) =1$$, for all $z$ on the unit circle $C(0,1)$ (uniformly). How to show that this also holds (using the maximum modulus principle) that on the unit disc $D(0,1)$?

Approach: I've no idea how to start. I don't see in particular how I can use the max mod principle in this problem, so any hints/solutions are welcome. Thanks in advance!

EDIT: the maximum modulus principle says that a non constant holomorphic function (in a region) cannot attain it's maximum in that region.

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  • $\begingroup$ Start with recalling what the maximum modulus principle says. $\endgroup$ – Daniel Fischer Jun 7 '17 at 18:45
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    $\begingroup$ This may help: It's the same as showing $|p_n(z)-1|\to 0$ uniformly on the closed unit disc. $\endgroup$ – zhw. Jun 7 '17 at 18:47
  • $\begingroup$ Do you know a version of the maximum modulus principle that mentions the boundary of the domain? $\endgroup$ – Daniel Fischer Jun 7 '17 at 18:48
  • $\begingroup$ Yes it says that (with some continuity assumptions) that the maximum is attained on the boundary. So using the comment of @zhw. I can apply this to the function $p_n(z) - 1$, then the result follows immediately because the maximum is 0 on the boundary? $\endgroup$ – bob Jun 7 '17 at 18:51
  • $\begingroup$ Bingo! Well, the maximum isn't necessarily $0$, but it tends to $0$ as $n\to\infty$ by assumption. $\endgroup$ – Daniel Fischer Jun 7 '17 at 18:51
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For the sake of completenes of the answer data base:

If $p_n(z)$ converges uniformly to $1$ on the unit circle $S^1 \subset \Bbb C$, then for any real $\epsilon > 0$ we have a positive integer $N$ such that $m > N$ implies $\vert p_m(z) - 1 \vert < \epsilon$ on $S^1$. But $p_m (z) - 1$ is holomorphic in $D(0, 1)$ for all $m$, and hence, by the maximum modulus principle, $\vert p_m(z) - 1 \vert$ takes is greatest values on $S^1$. Hence,

$\vert p_m(z) - 1\vert_{D(0, 1)} \le \vert p_m(z) - 1 \vert_{S^1} < \epsilon \tag{1}$

for all $m > N$, whence $p_n(z) \to 1$ uniformly on $D(0, 1)$.

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