# proof of maximal modulus principle

$f:U\to\mathbb{C}$ is holomorphic and therefore continuous, so $|f|$ will also be continuous on $U$. $K\subset U$ is compact and since $|f|$ is continuous on $K$ it must attain a maximum and a minimum value there.

Suppose the maximum of $|f|$ is attained at $z_{0}$ in the interior of $K$.

By definition there will exist $r>0$ such that the set $S_{r}=\left\{z\in\mathbb{C}:|z-z_{0}|^{2}\leq r^{2}\right\}\subset K$.

Consider $C_{r}$ the boundary of the previous set parameterized counter-clockwise. Since $f$ is holomorphic by hypothesis, Cauchy integral formula says that

 $f(z_{0})=\frac{1}{2\pi i}\oint_{C}\frac{f(z)}{z-z_{0}}dz$ (1)

A canonical parameterization of $C_{r}$ is $z=z_{0}+re^{i\frac{\theta}{r}}$, for $\theta\in[0,2\pi r]$.

 $f(z_{0})=\frac{1}{2\pi r}\int_{0}^{2\pi r}f(z_{0}+re^{i\frac{\theta}{r}})d\theta$ (2)

Taking modulus on both sides and using the estimating theorem of contour integral

 $|f(z_{0})|\leq\operatorname{max}_{z\in C_{r}}|f(z)|$

Since $|f(z_{0})|$ is a maximum, the last inequality must be verified by having the equality in the $\leq$ verified.

The proof of the estimating theorem of contour integral (http://planetmath.org/ProofOfEstimatingTheoremOfContourIntegral) implies that equality is only verified when

 $\frac{f(z_{o}+re^{i\frac{\theta}{r}})}{re^{i\frac{\theta}{r}}}=\lambda% \overline{ie^{i\frac{\theta}{r}}}$

where $\lambda\in\mathbb{C}$ is a constant. Therefore, $f(z_{o}+re^{i\frac{\theta}{r}})$ is constant and to verify equation 2 its value must be $f(z_{0})$.

So $f$ is holomorphic and constant on a circumference. It’s a well known result that if 2 holomorphic functions are equal on a curve, then they are equal on their entire domain, so $f$ is constant.

to see this in this particular circumstance is using equation 1 to calculate the value of $f$ on a point $\xi\in$ interior $S_{r}$ different than $z_{0}$. Bearing in mind that $f(z)=f(z_{0})$ is constant in $C_{r}$ the formula reads $f(\xi)=\frac{f(z_{0})}{2\pi i}\oint_{C_{r}}\frac{1}{z-\xi}dz=f(z_{0})$. So $f$ is really constant in the interior of $S_{r}$ and the only holomorphic function defined in $K$ that is constant in the interior of $S_{r}$ is the constant function on all $K$.

Thus if the maximum of $|f|$ is attained in the interior of $K$, then $f$ is constant. If $f$ isn’t constant, the maximum must be attained somewhere in $K$, but not in its interior. Since $K$ is compact, by definition it must be attained at $\partial K$.

Title proof of maximal modulus principle ProofOfMaximalModulusPrinciple 2013-03-22 15:46:15 2013-03-22 15:46:15 cvalente (11260) cvalente (11260) 19 cvalente (11260) Proof msc 30F15 msc 31B05 msc 31A05 msc 30C80