MHB Complex Variables - Max Modulus Inequality

[joypav]

Suppose that f is analytic on the disc $\vert{z}\vert<1$ and satisfies $\vert{f(z)}\vert\le{M}$ if $\vert{z}\vert<1$. If $f(\alpha)=0$ for some $\alpha, \vert{\alpha}\vert<1$. Show that,
$$\vert{f(z)}\vert\le{M\vert{\frac{z-\alpha}{1-\overline{\alpha}z}}\vert}$$

What I have:
Let $g(z)=f((1-\overline{\alpha}z)z+\alpha)$. Then g is analytic on the disc $\vert{z}\vert<1$ and $g(0)=f(\alpha)=0$.
If $\vert{f(z)}\vert\le{M}$ if $\vert{z}\vert<1$, then $\vert{g(z)}\vert\le{M}$ if $\vert{z}\vert<1$

1. $\vert{f((1-\overline{\alpha}z)z+\alpha)}\vert=\vert{g(z)}\vert\le{M\vert{z}\vert} \rightarrow \vert{f(z)}\vert\le{M\vert{\frac{z-\alpha}{1-\overline{\alpha}z}}}\vert$
(By substituting $z=\frac{z-\alpha}{1-\overline{\alpha}z}$)

2. Show that $g$ does satisfy $\vert{g(z)}\vert\le{M\vert{z}\vert}$.
Let $h(z)=\frac{g(z)}{z}$. Then h(z) is analytic on $\vert{z}\vert<1$, (removable singularity at z=0). Then by the Max Modulus Theorem, $\vert{h(z)}\vert$ attains its max somewhere on the boundary of $\vert{z}\vert<1$. Call this max M. Then,
$$\forall{z}, \vert{h(z)}\vert\le{M} \rightarrow \frac{\vert{g(z)}\vert}{\vert{z}\vert}\le{M} \rightarrow \vert{g(z)}\vert\le{M\vert{z}\vert}$$

Then, show that
$$M\vert{\frac{f(z)-f(z_0)}{M^2-\overline{f(z_0)}f(z)}}\vert \le{\vert{\frac{z-z_0}{1-\overline{z_0}z}}}\vert$$
for all $z, z_0$ in $w:\vert{w}\vert<1$.I'm just wondering if my first part is correct. If so, I assume I'll utilize the first to prove the second?

 

[Euge]

Hi joypav,

In order for $g$ to be well-defined, $(1 - \overline{\alpha}z)z + \alpha$ has to be in $\Bbb D$ for all $z\in \Bbb D$, which is not generally true (consider $\alpha = z = \frac{1}{\sqrt{2}}$). Instead, consider that the function $\phi_\alpha : z\mapsto \frac{z - \alpha}{1 - \overline{\alpha}z}$ is an analytic mapping of $\Bbb D$ onto $\Bbb D$, so we may define $g = \frac{1}{M}f\circ \phi_{\alpha}^{-1}$. Then $g$ is an analytic mapping $\Bbb D\to \overline{\Bbb D}$ with $g(0) = 0$, so by the Schwarz lemma $\lvert g(z) \rvert \le \lvert z\rvert$ for all $z\in \Bbb D$. Thus $\lvert Mg(\phi_\alpha(z))\rvert \le M\lvert \phi_\alpha(z)\rvert$, or, $\lvert f(z)\rvert \le M\lvert\frac{z - \alpha}{1 - \overline{\alpha}z}\rvert$ for all $z\in \Bbb D$.

 

[joypav]

For the first part...
Consider the function
$$\Psi_\alpha:z\rightarrow\frac{z-\alpha}{1-\overline{\alpha}z}$$
an analytic map from D onto D, where D is $\vert{z}\vert<1$.
Then define
$$g=\frac{1}{M}(f\circ\Psi_\alpha^{-1})$$
$$g(0)=\frac{1}{M}f(\alpha)=0$$
Then, by Schwarz Lemma,
$$\forall{z}\in{D}, \vert{g(z)}\vert\leq\vert{z}\vert.$$
Take $z=\Psi_\alpha(z)$ and multiply M to both sides.
$$\vert{Mg(\Psi_\alpha(z))}\vert\leq{M}\vert{\Psi_\alpha(z)}\vert$$
$\rightarrow\vert{f(z)}\vert\leq{M}\vert{\frac{z-\alpha}{1-\overline{\alpha}z}}\vert$, for all $z\in{D}$.

For the second part...

Consider the function
$$g(z)=\frac{f(z)}{M}$$
$\rightarrow\vert{g(z)}\vert\leq{1}$ in $\vert{z}\vert<{1}$.
Define
$\Psi_{\alpha_0}:z\rightarrow\frac{z-\alpha_0}{1-\overline{\alpha_0}z}$ and $h(z)=\Psi_{\alpha_0}(g(z))$
Then $\vert{h(z)}\vert\leq1$ and $h(z_0)=\Psi_{\alpha_0}(g(z_0))=\Psi_{\alpha_0}(\alpha_0)=0$.
Now we can apply the previous inequality to the function h.

$$\vert{h(z)}\vert\leq{1\vert{\frac{z-z_0}{1-\overline{z_0}z}}\vert}$$

$$\rightarrow\vert{(\Psi_{\alpha_0}\circ{g})(z)}\vert\leq{\vert{\frac{z-z_0}{1-\overline{z_0}z}}\vert}$$

$$\rightarrow\vert{\Psi_{\alpha_0}(\frac{f(z)}{M})}\vert\leq{\vert{\frac{z-z_0}{1-\overline{z_0}z}}\vert}$$

$$\rightarrow\vert{\frac{\frac{f(z)}{M}-\alpha_0}{1-\overline{\alpha_0}\frac{f(z)}{M}}}\vert\leq{\vert{\frac{z-z_0}{1-\overline{z_0}z}}\vert}$$

Sub in $\alpha_0=g(z_0)\rightarrow\alpha_0=\frac{f(z_0)}{M}$.

$$\rightarrow\vert{\frac{\frac{f(z)}{M}-\frac{f(z_0)}{M}}{1-\frac{\overline{f(z_0)}}{M}\frac{f(z)}{M}}}\vert\leq{\vert{\frac{z-z_0}{1-\overline{z_0}z}}\vert}$$

$$\rightarrow\vert{\frac{\frac{1}{M}(f(z)-f(z_0))}{\frac{1}{M^2}(M^2-\overline{f(z_0)}f(z))}}\vert\leq{\vert{\frac{z-z_0}{1-\overline{z_0}z}}\vert}$$

$$\rightarrow{M\vert{\frac{f(z)-f(z_0)}{M^2-\overline{f(z_0)}f(z)}}\vert\leq{\vert{\frac{z-z_0}{1-\overline{z_0}z}}\vert}}$$

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Additionally, there is a problem...

Suppose that $f^{(k)}(0)=0$ for k=0,...,N. Show that
$\vert{f(z)}\vert\leq{M\vert{z}\vert^{N+1}}$ for all $z, \vert{z}\vert<1$.

Are those inequalities used to prove this?

 

[Euge]

Your proofs look good. (Yes)

As for your latest question, you don't need those inequalities. A proof similar to the standard proof of the Schwarz lemma will suffice. The conditions on $f$ imply that that there is an analytic function $g$ on $\Bbb D$ such that $f(z) = z^{N+1} g(z)$ for all $z\in \Bbb D$. Since $\lvert f(z)\rvert \le M$ for all $\lvert z\rvert < 1$, then on every circle $\lvert z\rvert = \rho < 1$, $\lvert g(z)\rvert \le M\rho^{-N-1}$. Use the maximum principle, then take the limit as $\rho \to 1^{-}$ to obtain $\lvert g(z)\vert \le M$ for every $z\in \Bbb D$, i.e., $\lvert f(z)\rvert \le M \lvert z\rvert^{N+1}$ for all $z\in \Bbb D$.

 

[joypav]

Your proofs look good. (Yes)

As for your latest question, you don't need those inequalities. A proof similar to the standard proof of the Schwarz lemma will suffice. The conditions on $f$ imply that that there is an analytic function $g$ on $\Bbb D$ such that $f(z) = z^{N+1} g(z)$ for all $z\in \Bbb D$. Since $\lvert f(z)\rvert \le M$ for all $\lvert z\rvert < 1$, then on every circle $\lvert z\rvert = \rho < 1$, $\lvert g(z)\rvert \le M\rho^{-N-1}$. Use the maximum principle, then take the limit as $\rho \to 1^{-}$ to obtain $\lvert g(z)\vert \le M$ for every $z\in \Bbb D$, i.e., $\lvert f(z)\rvert \le M \lvert z\rvert^{N+1}$ for all $z\in \Bbb D$.

Oh okay, it is almost exactly like that proof. Thanks!