Sine Is Bounded on a Line, and a Line Is Almost None of the Plane

Sine is differentiable everywhere and never leaves the band from minus one to one on the real axis, which makes it the counterexample everyone reaches for, and the modulus of sine at 10i is already 11013.23. The Cauchy estimate caps every Taylor coefficient by M over r to the n on a circle of radius r, so growing r kills every coefficient above the constant one and nothing but a constant survives. The same estimate with a polynomial growth bound gives more: an entire function bounded by C times one plus the modulus of z, all to the k, is a polynomial of degree at most k, and one corollary of that is the fundamental theorem of algebra.

A function is differentiable at every point of the complex plane and its modulus never exceeds some fixed number. Show that it is constant.

The statement is startling the first time you read it, because the real analogue is false in the most obvious way possible. The sine function is smooth on the whole real line and never leaves the band between $-1$ and 1, and it is certainly not constant. So the first move most people make is to look for a counterexample by analogy, and the search fails in an instructive way.

Sine, and the axis it is bounded on

Sine is entire. Its power series converges everywhere and it is differentiable at every complex number, so it clears the first requirement without difficulty. The question is whether it clears the second one, and the answer arrives as soon as you leave the real axis. Write sine through exponentials and evaluate at a purely imaginary argument:

\sin z = \frac{e^{iz} - e^{-iz}}{2i}, \qquad \sin(iy) = \frac{e^{-y} - e^{y}}{2i} = i\sinh y

(1)

So a step of ten units in the imaginary direction gives $|\sin(10i)| = \sinh(10) = 11013.23$ , and twice that distance gives $\sinh(20) = 2.4258 \times 10^{8}$ . The growth is exponential in the distance from the real axis, and computing it with ordinary complex arithmetic, which knows nothing about the hyperbolic sine, returns the same numbers.

The bound of 1 on sine was never a statement about the plane. It was a statement about a line in the plane, and a line is a vanishingly thin slice of the surface it sits in. Boundedness on the real axis constrains a complex function far less than it looks like it does, which is exactly the gap the theorem exploits.

Fig. 1 — The same function, measured two ways. The vertical scale is logarithmic, so the straight climb is exponential growth.

Entire

A function is entire when it is complex differentiable at every point of $\mathbb{C}$ . Complex differentiability is a much heavier demand than the real kind: an entire function equals its Taylor series about any point, with the series converging on the whole plane, so the countable list of coefficients $a_0, a_1, a_2, \dots$ determines the function everywhere. Killing those coefficients is therefore the same as killing the function.

The estimate that does the work

Write $f(z) = \sum_{n \ge 0} a_n z^n$ . The Cauchy integral formula recovers each coefficient from a circle of any radius $r$ around the origin:

a_n = \frac{1}{2\pi i}\oint_{|z| = r} \frac{f(z)}{z^{\,n+1}}\,dz

(2)

Bounding the integral by the length of the contour times the largest value of the integrand on it gives the estimate the whole argument rests on. Writing $M(r) = \max_{|z| = r}|f(z)|$ , the contour has length $2\pi r$ and the integrand has modulus at most $M(r)/r^{\,n+1}$ , so:

|a_n| \le \frac{1}{2\pi}\cdot 2\pi r \cdot \frac{M(r)}{r^{\,n+1}} = \frac{M(r)}{r^{\,n}}

(3)

Now use the hypothesis. If $|f| \le M$ on the whole plane then $M(r) \le M$ for every $r$ , so the right-hand side of (3) is at most $M/r^{\,n}$ and that is free to be made as small as you like:

|a_n| \le \frac{M}{r^{\,n}} \;\;\text{for all } r > 0 \;\;\Longrightarrow\;\; a_n = 0 \;\;(n \ge 1)

(4)

A non-negative number smaller than every element of a sequence tending to 0 is 0. So every coefficient above the constant one vanishes and $f(z) = a_0$ . The result is Liouville's theorem, and the step doing all of the lifting is the freedom to grow $r$ without $M(r)$ growing with it.

Fig. 2 — Why the argument needs the whole plane. Each larger circle gives a smaller bound on the same fixed coefficient, and the bounds have to hold simultaneously.

This machinery can be run rather than admired. Discretising (2) on 4096 equally spaced angles recovers the coefficients 2, $-3$ , 0.5 and 4 of a test polynomial on circles of radius 0.5, 1, 3 and 10, agreeing to eight decimal places at every radius. Applied to a constant, the same computation returns coefficients below $10^{-12}$ for $n = 1, 2, 3$ at radii up to 1000. And the quantity in (3) behaves as advertised: $M(r)/r$ grows for the polynomial and collapses towards zero for a bounded function.

Removing one hypothesis at a time

Both conditions are load-bearing, and dropping either one produces an easy counterexample.

Drop boundedness and sine already answers, as does any non-constant polynomial. Drop entirety and the identity function on the unit disc answers: $f(z) = z$ is holomorphic there and bounded by 1 there, and is not constant. So the theorem is not really about boundedness. It is about boundedness on a domain large enough for (4) to be applied with arbitrarily large $r$ , which the disc is not.

The size of the bound is irrelevant. Nothing in (3) or (4) cares whether $M$ is 1 or $10^{100}$ , only that one such number exists. That is worth saying because the informal version of the problem often fixes the bound at 1, which makes the estimate read as $1/r^{\,n}$ and can leave the impression that the value matters.

The same estimate, pushed further

Equation (3) does more than the theorem needs, and the extra is free. Suppose $f$ is entire and grows no faster than a polynomial of degree $k$ , meaning $|f(z)| \le C(1 + |z|)^k$ everywhere. Then $M(r) \le C(1+r)^k$ and (3) gives:

|a_n| \le \frac{C(1+r)^k}{r^{\,n}} \xrightarrow[r \to \infty]{} 0 \qquad \text{whenever } n > k

(5)

So $f$ is a polynomial of degree at most $k$ . Liouville's theorem is the case $k = 0$ , and the general statement says that an entire function is pinned down by how fast it grows.

One consequence is worth carrying away, because it settles an older question. Let $p$ be a non-constant polynomial with no root anywhere in the plane. Then $1/p$ is entire, and since $|p(z)| \to \infty$ as $|z| \to \infty$ , the reciprocal tends to 0 out there and is continuous on the compact disc that remains, so it is bounded on the whole plane. By (4) it is constant, so $p$ is constant, contradicting the assumption. Every non-constant polynomial therefore has a root, which is the fundamental theorem of algebra, obtained from an estimate about circles.

Sources and further reading

The result derived in (4) — Liouville's theorem
The class of functions the theorem is about — Entire function
Where (2) comes from — Cauchy integral formula
Sine off the real axis — Hyperbolic functions
The corollary at the end of the article — Fundamental theorem of algebra
An alternative route to the same conclusion — Maximum modulus principle

The counterexample was evaluated numerically rather than trusted: $|\sin(iy)|$ matches $\sinh(y)$ at $y = 1, 2, 5, 10$ and 20, and the maximum of $|\sin|$ over 700,000 samples of the real axis is at most 1. The coefficient machinery behind (2) and (3) was exercised as a discrete contour integral at four radii, and the contrast between a growing and a collapsing bound was measured on a polynomial and on a constant.