# proof of maximal modulus principle

$f:U\to \u2102$ 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 ${S}_{r}=\{z\in \u2102:{|z-{z}_{0}|}^{2}\le {r}^{2}\}\subset K$.

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

$$f({z}_{0})=\frac{1}{2\pi i}{\oint}_{C}\frac{f(z)}{z-{z}_{0}}\mathit{d}z$$ | (1) |

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

$$f({z}_{0})=\frac{1}{2\pi r}{\int}_{0}^{2\pi r}f({z}_{0}+r{e}^{i\frac{\theta}{r}})\mathit{d}\theta $$ | (2) |

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

$$|f({z}_{0})|\le {\mathrm{max}}_{z\in {C}_{r}}|f(z)|$$ |

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

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

$$\frac{f({z}_{o}+r{e}^{i\frac{\theta}{r}})}{r{e}^{i\frac{\theta}{r}}}=\lambda \overline{i{e}^{i\frac{\theta}{r}}}$$ |

where $\lambda \in \u2102$ is a constant. Therefore, $f({z}_{o}+r{e}^{i\frac{\theta}{r}})$ 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 $f(\xi )=\frac{f({z}_{0})}{2\pi i}{\oint}_{{C}_{r}}\frac{1}{z-\xi}\mathit{d}z=f({z}_{0})$. 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 |
---|---|

Canonical name | ProofOfMaximalModulusPrinciple |

Date of creation | 2013-03-22 15:46:15 |

Last modified on | 2013-03-22 15:46:15 |

Owner | cvalente (11260) |

Last modified by | cvalente (11260) |

Numerical id | 19 |

Author | cvalente (11260) |

Entry type | Proof |

Classification | msc 30F15 |

Classification | msc 31B05 |

Classification | msc 31A05 |

Classification | msc 30C80 |