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this says that f is a constant, and in particular |f(0)|=|f(z)|=1. Since we ask that f(0)=0 for
Schwarz's lemma, this wouldn't count as one of the functions the lemma applies to. To put it another way, it doesn't matter either way whether you let the codomain be the open or closed disc, the set of
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Well this is just the hypothesis, so we can ask anything we like of f (of course that doesn't mean that such an f can exist, but for example f(z)=z, or even f(z)=0, clearly takes D to D and fixes 0). However, what would happen we did allow the
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version, so I got rid of the other, but I also added a note at the bottom about the ambiguity in the theorem statement so hopefully this doesn't happen yet again. I also note Rudin's R&C A uses yet another condition, that
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By the way I know the previous comment is more than a year old, but this might help someone in future (especially since the article let the codomain be closed in a previous revision).
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functions that are included is the same (when you take into account the f(0)=0 requirement). The same applies to the
Schwarz-Pick theorem.
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Just another math undergrad here... Thanks for the page - Schwarz lemma was a result I needed when studying
Hyperbolic geometry.
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D is wrong, because f transforms from open unit disk to closed unit dist (and not from open to open)
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Knowledge. If you would like to participate, please visit the project page, where you can join
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not be contained in D) there would need to be a z such that |z|<1 but |f(z)|=1. By the
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It got changed again so that it included both conditions simultaneously. I favor the
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to be the closed unit disc? For it to be a "new" function (i.e. the
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293:{\displaystyle ||f||_{\infty }\leq 1}
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334:{\displaystyle f:D\to {\bar {D}}}
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394:{\displaystyle |f(z)|\leq 1}
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