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I have been asked to prove the following inequality, given that $f:\mathbb{D}\to\mathbb{C}$ is an analytic function on the open unit disc $\mathbb{D}\subset\mathbb{C}$.

$|f(w)-f(0)-wf~^\prime(0)|\leq|w|^2\sup|f(z)-f(0)-zf~^\prime(0)|$ where the supremum is taken over all $z\in\mathbb{D}$ and $w$ is any element of $\mathbb{D}$.

Since $f$ is analytic on $\mathbb{D}$, it is equal to its power series representation at $z=0$ there. So $f(z)=\displaystyle\sum\limits_{n=0}^\infty a_nz^n$ where $a_n=\frac{f^{(n)}(0)}{n!}$.

We see that the expression $f(z)-f(0)-zf~^\prime(0)$ is just the power series expansion for $f$ without the first two terms. This is another analytic function $g(z)=\displaystyle\sum\limits_{n=2}^\infty a_nz^n$ which satisfies $g(0)=g~^\prime(0)=0$. So equivalently we want to show $|g(w)|\leq|w|^2\sup|g(z)|$ or $|\frac{g(w)}{w^2}|\leq\sup|g(z)|$ (the supremum still over $z\in\mathbb{D}$).

Here is where I am getting hung up. I have $\frac{g(w)}{w^2}=\displaystyle\sum\limits_{n=2}^\infty a_nw^{n-2}$ is still an analytic function (right?). But I am not sure what I need to get the inequality.

$\bf{UPDATE:}$ We can use Schwarz's Lemma on $h(w)=\frac{g(w)}{w}$ to obtain $|\frac{g(w)}{w}|\leq|w|$ since $h(0)=0$. Can we show that the supremum is at least $1$?

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up vote 3 down vote accepted

Use the Schwarz lemma on the function $ g(w)/\sup|g|$ to see that $|g(w)|\le|w| \sup |g|$. Then use the lemma again on the function

$$h(w)= \frac{g(w)}{w \sup|g|}$$

in order to find the desired inequality, $|g(w)|\le |w|^2 \sup|g|$.

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Well I'm glad I thought to use Schwarz's Lemma on my own, even if I am a bit upset I didn't see the last step. Thanks! –  RHP Aug 14 '11 at 6:20

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