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Suppose $f$ is bounded and analytic on the open unit disk $\mathbb{D}$. Say that $f$ extends continuously to one point $z_0$ on $\partial \mathbb{D}$, the boundary of $\mathbb{D}$. Now does the maximum principle apply here, i.e., is it true that $|f(z_0)| \le M$ implies $|f(z)| \le M$ for $z \in \mathbb{D}$? (Equivalently, $|f(z)| \le |f(z_0)|$.)

I know it sounds strange to talk about continuous extensions to just one point on the boundary. This question is derived from Gamelin's Complex Analysis, p. 89, #7. In that problem, there are finitely many points on the boundary such that $f$ extends continuously to the arcs between them. I'm wondering if this is true when there's only one such point.

(I understand that analytic continuation along the whole boundary might be possible, but I haven't learned about that yet.)

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  • $\begingroup$ Would that apply to $f(z) = z-1$ and $z_0=1$, say? $\endgroup$ – WimC Feb 26 '12 at 9:31
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The answer to the question such as you have stated it is clearly no. Consider $f(z)=z-1$ and $z_0=1$.

I believe that what you wanted to ask is the following: If $f$ can be extended as a continuous function to all of the boundary except one point $z_0$, does the maximum principle hold true? This is just an instance of the problem in Gamelin's book, where you have only one point and only one arc between "the points", the whole circle with the point excluded.

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  • $\begingroup$ Yes, you're quite right. I misconstrued the question -- I didn't see how you could have an arc "between" just one point. Thanks very much. $\endgroup$ – WP- Feb 26 '12 at 16:14

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