1
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ Start with recalling what the maximum modulus principle says. $\endgroup$
    – Daniel Fischer
    Jun 7, 2017 at 18:45
  • 1
    $\begingroup$ 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, 2017 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, 2017 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, 2017 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, 2017 at 18:51

1 Answer 1

Reset to default
1
$\begingroup$

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 $D(0, 1)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.