Complex Variables - Max Modulus Inequality

In summary, we have shown that if $f$ is analytic on the disc $\vert{z}\vert<1$ and satisfies $\vert{f(z)}\vert\le{M}$ for all $\vert{z}\vert<1$, and $f(\alpha)=0$ for some $\alpha, \vert{\alpha}\vert<1$, then $\vert{f(z)}\vert\le{M\vert{\frac{z-\alpha}{1-\overline{\alpha}z}}\vert}$ for all $z, \vert{z}\vert<1$. We have also shown that $f$ satisfies the inequality $M\vert{\frac{f(z)-f(z_0)}{M
  • #1
joypav
151
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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?
 
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  • #2
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 $z\in \Bbb D$.
 
  • #3
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?
 
  • #4
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 $z\in \Bbb D$.
 
  • #5
Euge said:
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!
 

Related to Complex Variables - Max Modulus Inequality

1. What is the Max Modulus Inequality in complex variables?

The Max Modulus Inequality is a fundamental theorem in complex analysis that states that the maximum modulus of a complex-valued function on a closed and bounded domain is always less than or equal to the maximum modulus on the boundary of that domain. This inequality is important in understanding the behavior of complex functions and their limits.

2. How is the Max Modulus Inequality applied in complex analysis?

The Max Modulus Inequality is used to prove many other theorems and results in complex analysis. It is often used to find the maximum or minimum values of a complex function on a given domain, or to show that a function has no zeros within a certain region. It also helps in understanding the behavior of complex functions near singularities or poles.

3. What is the geometric interpretation of the Max Modulus Inequality?

The Max Modulus Inequality can be interpreted geometrically as stating that the maximum modulus of a complex function on a closed and bounded domain is always located on the boundary of that domain. This means that the function cannot attain a greater magnitude within the interior of the domain, and this can be visualized as a circle being inscribed within a larger circle.

4. Can the Max Modulus Inequality be extended to higher dimensions?

Yes, the Max Modulus Inequality can be extended to higher dimensions through the use of multivariable complex analysis. In this case, the maximum modulus of a complex-valued function on a bounded domain is related to the maximum modulus on the boundary of that domain, but the boundary is now a higher-dimensional surface instead of a one-dimensional curve.

5. How does the Max Modulus Inequality relate to the Cauchy Integral Formula?

The Max Modulus Inequality is closely related to the Cauchy Integral Formula, which states that the value of a complex function at a point inside a closed and bounded domain is equal to the average of the function's values on the boundary of that domain. The Max Modulus Inequality is often used in conjunction with the Cauchy Integral Formula to prove other important results in complex analysis.

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