## how to prove this version of Schwartz lemma ? [closed]

I see this theorem in Hans Grauert & Reinhold Remmert's book 'Theory of stein space' , page 190, in chapter 6 . The classical Schwartz lemma is stated as follows , but i don't know how to prove it. Let $E$ , $E'$ be disks centered at the origin in the $w$-plane with radii $0<r'<r$. Let $a:=r'r^{-1}$ . Suppose $h\in {\mathcal{O}}(E)$ vanishes of order $e$ at the origin . Then $|h|_{E'}\leq a^{e}|h|_E$ , where $|\cdot|_E$ means the super-norm of functions defined on $E$ .

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Surely a natural place to look is in a 1st book on complex analysis? I think that for questions like this, you should explain before asking where else you have tried looking for the answer – Yemon Choi May 19 2011 at 9:11
does there anything wrong in the mathoverflow , why i typed what i want to ask ,but it only demonstrate part of my words ? – HKSHLZW May 19 2011 at 9:14
Sorry , I really can't see what you have fixed ! I think there may be something wrong with my computer ! – HKSHLZW May 19 2011 at 9:16
Nothing wrong with your computer, you just undid (accidentally) some edits I made. I have done them again. – Yemon Choi May 19 2011 at 9:18
BTW, it's (Hermann) Schwarz's lemma and not (Laurent) Schwartz's lemma. In Boas's words : "The Schwarz of inequality / And lemma too, he has no T. / The "distribution" Schwartz, you see / Is French, and so he has a T." – Maxime Bourrigan May 19 2011 at 9:47