Maximum Modulus Principle

Why Holomorphic Functions Have No Nonconstant Interior Modulus Maximum

For a real-valued function, an interior maximum is often an ordinary feature of its graph. Holomorphic functions behave differently: unless the function is constant, its modulus cannot attain a local maximum at an interior point. This restriction is a consequence of the Cauchy integral formula, which relates the value at the center of a circle to the average of the values around it.

The Fundamental Theorem of Algebra used polynomial growth and Liouville’s Theorem to establish root existence. Here, Cauchy’s integral formula gives a different kind of global information: it constrains where the largest modulus can occur. We first make the averaging argument precise, then use it to prove the Maximum Modulus Principle.

The Cauchy Mean-Value Formula

If \(f\) is holomorphic on an open set containing the closed disk \(\overline{D(a,r)}\), Cauchy’s integral formula can be applied on the circle \(|z-a|=r\). At the center, it says that \(f(a)\) is the average of its values on that circle:

The triangle inequality immediately gives \(|f(a)|\leq \max_{|z-a|=r}|f(z)|\). The important point for the maximum principle is what happens when equality holds in the corresponding modulus bound.

Proof. If \(M=0\), the bound says \(f=0\) on the circle and \(f(a)=0\). Cauchy’s integral formula then gives \(f(w)=0\) for every \(|w-a|<r\), and continuity gives the same conclusion on the circle.

Now suppose \(M>0\). Choose a complex number \(\lambda\) with \(|\lambda|=1\) and \(\lambda f(a)=M\); for example, \(\lambda=M/f(a)\). Set \(g=\lambda f\). On the circle, \(|g|\leq M\), and \(g(a)=M\). The mean-value formula gives

Taking real parts shows that the average of \(\operatorname{Re}g(a+re^{it})\) is \(M\). But at every point of the circle, \(\operatorname{Re}g\leq |g|\leq M\). Thus the continuous function \(t\mapsto M-\operatorname{Re}g(a+re^{it})\) is nonnegative and has integral zero. It must be zero everywhere: if it were positive at one point, continuity would make it positive on an interval, giving a positive integral. Therefore \(\operatorname{Re}g=M\) on the circle. Since \(|g|\leq M\), this forces \(\operatorname{Im}g=0\) there, so \(g=M\), and hence \(f=f(a)\), on the circle. Applying Cauchy’s integral formula at each point inside the circle gives \(f(w)=f(a)\) for \(|w-a|<r\) as well. \(\square\)

The equality argument matters: an average can equal its upper bound only if every value being averaged reaches that bound. Here the complex values must also point in the same direction, which is why the rotation by \(\lambda\) lets us use real parts.

The Maximum Modulus Principle

Proof. Suppose \(|f|\) has a local maximum at \(a\). Choose \(r>0\) small enough that \(\overline{D(a,r)}\subseteq U\) and \(|f(z)|\leq |f(a)|\) throughout that disk. The Equality in the Cauchy Mean-Value Formula Lemma shows that \(f(z)=f(a)\) for \(|z-a|\leq r\). Thus \(f\) is constant on a neighborhood of \(a\).

It remains to show that this local constancy extends throughout \(U\). Write \(c=f(a)\), and define \(A\) to be the set of points of \(U\) at which \(f\) is identically \(c\) on some neighborhood. The disk just found shows \(A\) is nonempty, and its definition shows it is open in \(U\).

To see that \(A\) is closed in \(U\), take a sequence \((z_n)\) in \(A\) converging to \(z\in U\). The function \(h=f-c\) is holomorphic. Since it vanishes on a neighborhood of each \(z_n\), every derivative \(h^{(k)}(z_n)\) is zero for every integer \(k\geq0\). Holomorphic functions are analytic, so their derivatives are continuous; consequently \(h^{(k)}(z)=0\) for every \(k\geq0\). The Taylor series of \(h\) at \(z\) therefore has every coefficient zero, and analyticity implies that \(h\) vanishes on a neighborhood of \(z\). Hence \(z\in A\), proving relative closedness. As \(U\) is connected and \(A\) is nonempty, open, and closed in \(U\), \(A=U\). Thus \(f=c\) throughout \(U\). \(\square\)

The connectedness hypothesis ensures that constancy near the maximum propagates across the whole domain. On a disconnected open set, the theorem’s conclusion applies only to the connected component containing the point of local maximum; the function may take different values on other components.

Maximum Modulus on a Bounded Domain

Proof. Since \(\Omega\) is bounded, its closure \(\overline{\Omega}\) is closed and bounded in \(\mathbb{C}\), hence compact. The function \(|f|\) is continuous on \(\overline{\Omega}\), so the Extreme Value Theorem gives a point \(z_0\in\overline{\Omega}\) at which \(|f|\) attains its maximum. If \(z_0\) were in \(\Omega\), this would be a local maximum of \(|f|\). The Maximum Modulus Principle would then make \(f\) constant on \(\Omega\), contrary to the hypothesis. Thus, when \(f\) is nonconstant, \(z_0\in\partial\Omega\). \(\square\)

The theorem says more than that a maximum can be found somewhere in the closure: for a nonconstant function, no interior point can be where the global maximum occurs. The boundary need not be a circle, and the domain need not be a disk. Compactness supplies existence of a maximum; holomorphicity rules out an interior location.

A Minimum Modulus Consequence

There is a useful minimum principle when a holomorphic function has no zeros. If \(f\) never vanishes on an open set, the Reciprocal of a Holomorphic Function theorem says that \(1/f\) is holomorphic there. Applying the maximum principle to \(1/f\) turns a maximum of its modulus into a minimum of \(|f|\).

More generally, if \(f\) is continuous on the closure of a bounded connected domain, holomorphic and nonvanishing in the domain, and has no zeros on the boundary either, then \(1/f\) is continuous on the closure and holomorphic in the domain. The boundary maximum principle applied to \(1/f\) shows that the minimum of \(|f|\) on the closure occurs on the boundary, unless \(f\) is constant. The nonvanishing condition is essential: a zero would make the reciprocal undefined there.

Common Pitfalls

The Maximum Modulus Principle does not claim that the modulus has no maximum. On a compact closure it must have one, but for a nonconstant holomorphic function that maximum must be on the boundary. Nor does the theorem say that \(|f|\) is largest at every boundary point; it only rules out an interior location for the maximum.

The hypotheses also matter. A function with a pole inside the domain is not holomorphic there, so the principle does not apply. For example, \(f(z)=1/z\) is unbounded near \(0\) and is not holomorphic on any open set containing \(0\). Finally, connectedness is needed for the conclusion that \(f\) is constant on the whole domain, rather than merely on the component containing an interior maximum.