📂Complex Anaylsis

Proof of the Maximum Absolute Value Theorem

Theorem ¹

Let us say the function $f$ is continuous on the simple closed path $\mathscr{C}$, analytic in the interior, and is not a constant function at any point. Then, the point $z = z_{0}$ that maximizes $|f(z)|$ on $\mathscr{C}$ exists on $\mathscr{C}$.

Description

Simply put, in complex analysis, the maximum value of $|f|$ within a closed path exists on its boundary. At this point, it becomes a level of intuition that is hard to grasp; for some reason, it’s quite fascinating though inexplicable. For embodiment, it’s good to think about various functions directly and confirm their truth.

Usually, though it’s easier to accept theories as facts, understanding their proofs can be difficult. However, the maximum-minimum modulus theorem may be quite the opposite. As absolute values are used, the geometric shape in real functions keeps coming to mind, which can interfere with understanding. Even conceding a hundred times that some function exists and the maximum modulus is on the boundary of the path integral interval, it’s hard to imagine that the minimum modulus also exists on the same boundary. So, don’t worry too much about the shape, and try to accept it by applying it directly to various functions as explained.

Considering $\displaystyle {{1} \over {f}}$ next, it naturally follows that the point which maximizes $\displaystyle \left| {{1} \over {f}} \right|$ exists on $\mathscr{C}$ by the maximum modulus theorem. That point, conversely, is the one that minimizes $|f|$, hence the minimum modulus theorem can be deduced.

Minimum Modulus Theorem

Let the function $f$ be continuous on the simple closed path $\mathscr{C}$, analytic in the interior, and not a constant function at any point.

If $|f(z)| \ne 0$ in the interior of $\mathscr{C}$, then the point $z = z_{0}$ that minimizes $|f(z)|$ exists on $\mathscr{C}$.

Proof

Assume that the point $z = z_{0}$, where $|f(z)|$ is maximum, exists in the interior of $\mathscr{C}$. Then, by the density of real numbers, there always exists some $r>0$ that makes $|z - z_{0}| = r$ exist in the interior of $\mathscr{C}$.

Meanwhile, since $\left| f(z_{0}) \right|$ has the maximum value at $z = z_{0}$, it should be $|f(z_{0} + r e ^{ i \theta } )| \le \left| f(z_{0}) \right|$, but since it is not a constant function at any point, it must be $|f(z_{0} + r e ^{ i \theta } )| < \left| f(z_{0}) \right|$.

Gauss’s Mean Value Theorem: If the function $f$ is analytic on the closed circle $| z - z_{0} | \le r$, $$f(z_{0}) = {{1} \over {2 \pi}} \int_{0}^{2 \pi} f(z_{0} + r e ^{i \theta } ) d \theta$$

By Gauss’s Mean Value Theorem, $$ f(z_{0}) = {{1} \over {2 \pi}} \int_{0}^{2 \pi} f(z_{0} + r e ^{i \theta } ) d \theta $$ Taking the absolute value of both sides, $$ \begin{align*} \left| f(z_{0}) \right| =& \left| {{1} \over {2 \pi}} \int_{0}^{2 \pi} f(z_{0} + r e ^{i \theta } ) d \theta \right| \\ \le & {{1} \over {2 \pi}} \int_{0}^{2 \pi} | f(z_{0} + r e ^{i \theta } ) | d \theta \end{align*} $$ But, $$ \begin{align*} \left| f(z_{0}) \right| \le & {{1} \over {2 \pi}} \int_{0}^{2 \pi} | f(z_{0} + r e ^{i \theta } ) | d \theta \\ <& {{1} \over {2 \pi}} \int_{0}^{2 \pi} | f(z_{0}) | d \theta \\ =& \left| f(z_{0}) \right| \end{align*} $$ Therefore, $$ \left| f(z_{0}) \right| < \left| f(z_{0}) \right| $$ This is a contradiction, hence $z=z_{0}$ cannot exist in the interior of $\mathscr{C}$.

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