Why the Minimum Modulus Principle Follows from the Maximum Modulus Principle

The minimum modulus principle can feel like a separate theorem, but its usual proof is a direct application of the maximum modulus principle. The key observation is that taking a reciprocal reverses the size comparison: a small nonzero value of $f$ corresponds to a large value of $1/f$.

The precise statement is that if $f$ is holomorphic and nonconstant on a connected domain, then $|f|$ cannot have a positive local minimum at an interior point. Thus, if $|f|$ does have an interior local minimum, that minimum must be $0$, meaning $f$ vanishes there.

The reciprocal proof, step by step

Suppose $z_0$ is an interior point where $|f|$ has a local minimum. There are two cases.

If $f(z_0)=0$, then $|f(z_0)|=0$. Since a modulus cannot be negative, this is already a minimum. This case does not imply that $f$ is constant; for example, $f(z)=z$ has a local minimum of its modulus at $0$.

Now suppose $f(z_0)\ne 0$. By continuity, $f(z)$ stays nonzero in some small neighborhood of $z_0$. So the reciprocal

$$g(z)=\frac{1}{f(z)}$$

is holomorphic in that neighborhood. Because $|f|$ has a local minimum there, taking reciprocals reverses the inequality:

$$|g(z)|=\frac{1}{|f(z)|}\leq \frac{1}{|f(z_0)|}=|g(z_0)|.$$

In other words, $|g|$ has a local maximum at the interior point $z_0$. The maximum modulus principle says that a holomorphic function with an interior local maximum of its modulus must be constant (on the relevant connected neighborhood). Therefore $g$ is constant nearby, and hence $f=1/g$ is constant nearby.

If the original domain is connected, the identity theorem then implies that $f$ is constant throughout the domain. This contradicts the assumption that $f$ is nonconstant. So the proposed local minimum cannot have a nonzero value.

Why a minimum becomes a maximum

For positive numbers, the function $x\mapsto 1/x$ is decreasing. If $|f(z)|\geq |f(z_0)|>0$ near $z_0$, then

$$\frac{1}{|f(z)|}\leq \frac{1}{|f(z_0)|}.$$

That inequality is exactly the local-maximum condition for $|1/f|$. The assumption $f(z_0)\ne0$ matters: without it, the reciprocal may not exist near the point, and the argument cannot be used.

A concrete example

Consider $f(z)=2+z$ near $z_0=0$. Here $f(0)=2\ne0$, so if $|f|$ had a local minimum at $0$, then $g(z)=1/(2+z)$ would have a local maximum there. But moving a small distance in the negative real direction gives $z=-t$ for $0<t<1$, and

$$|f(-t)|=|2-t|=2-t<2=|f(0)|.$$

Thus $|f|$ actually decreases immediately in that direction; $0$ is not a local minimum. The reciprocal view gives the matching increase:

$$|g(-t)|=\frac{1}{2-t}>\frac{1}{2}=|g(0)|.$$

This illustrates the reversal, while the maximum modulus principle supplies the general proof for every holomorphic function.

Common mistakes

• Forgetting the zero case: A nonconstant holomorphic function can have an interior minimum of its modulus when that minimum is zero. The principle rules out positive interior minima, not zero ones.
• Taking the reciprocal without checking for zeros: The reciprocal $1/f$ is holomorphic only where $f$ does not vanish. Continuity and $f(z_0)\ne0$ guarantee a sufficiently small zero-free neighborhood.
• Confusing local and global statements: The proof first uses the maximum modulus principle locally. To conclude that $f$ is constant on the whole domain, use the identity theorem and assume the domain is connected.
• Assuming the modulus itself is holomorphic: In general, $|f|$ is not holomorphic. The maximum modulus principle applies to the holomorphic function $g=1/f$, through its modulus $|g|$.