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The Cauchy estimate with the radius allowed to grow: an entire function bounded on the whole plane is constant — and from that one sentence, the fundamental theorem of algebra, the maximum modulus principle and a description of every entire function of polynomial growth.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to state Liouville's theorem exactly, reproduce its proof from the Cauchy estimate, distinguish it from weaker or stronger statements that resemble it, use it to prove the fundamental theorem of algebra, state the maximum modulus principle, and build an entire bounded function out of a problem in order to apply the theorem to it.
The Cauchy estimate: a bound on an analytic function over a circle bounds each of its derivatives at the centre, with the radius in the denominator. Everything in this lesson is that inequality with the radius allowed to grow, and it is worth noticing how little else is needed.
Entire means analytic on the whole plane. A function is bounded when one number exceeds its modulus everywhere. The multiplicity of a root is how many times the corresponding factor divides the polynomial. The maximum modulus principle says a non-constant analytic function attains no maximum of its modulus at an interior point.
Liouville's theorem. If $f$ is entire and $|f| \le M$ everywhere, then $f$ is constant.
The proof is three lines. The Cauchy estimate gives $|f'(z_0)| \le M/R$ for every radius $R$. Being entire, $f$ is analytic on discs of every size, so $R$ may be taken as large as we like; being bounded on the whole plane, $M$ does not grow with it. So $|f'(z_0)|$ is below every positive number and is therefore zero — at every point, since $z_0$ was arbitrary. A vanishing derivative on a connected region means constant.
Why this is startling. Over the reals, $\sin x$ is infinitely differentiable, bounded by $1$ and not constant. Over the plane there is no such function. Boundedness and analyticity together leave nothing but the constants, and $\sin z$ escapes only because it is not bounded on the plane: $|\sin(iy)| = \sinh|y|$ grows without limit.
The fundamental theorem of algebra. Suppose a polynomial $p$ of degree $n \ge 1$ had no root. Then $1/p$ is entire. Far from the origin $|p|$ grows without bound, so $1/p \to 0$ out there, and on the remaining closed disc a continuous function is bounded. So $1/p$ is bounded and entire, hence constant by Liouville, hence $p$ is constant — contradicting $n \ge 1$. So $p$ has a root; divide it out and repeat to get $n$ roots with multiplicity.
An algebraic theorem, with no known proof of comparable length that stays inside algebra. It is the standard advertisement for this subject, and it is fair.
The maximum modulus principle. If $|f|$ attained a maximum at an interior point, the mean value property would force $f$ to be constant near it, and (by analytic continuation) constant throughout. So the largest values of $|f|$ on a closed bounded region sit on the boundary — which is the estimate that makes numerical work with analytic functions possible, and the reason a harmonic function has no hot spot in the middle of a plate.
Another way: picture
Think of an entire function as a landscape over the plane. Liouville says a landscape that is analytic everywhere and never rises above a fixed height must be perfectly flat. There is no way to build a bump: a bump needs curvature, curvature shows up in the derivative, and the Cauchy estimate bounds the derivative by the height divided by a radius that can be taken as large as you please.
Another way: steps
Liouville is almost never applied to a function that was handed over. It is applied to one that was built — and the building is the skill.
The standard shape: you want to prove that something is impossible, or that two things are equal. You form a function that would be entire and bounded if the thing you want to rule out were true, conclude it is constant, and derive a contradiction. The fundamental theorem of algebra takes a reciprocal; the proof that two entire functions agreeing off a bounded set are equal takes a difference; the proof that a non-constant entire function's image is dense takes a reciprocal of a shift.
Three things the theorem does not say, each of which is a common misreading.
It does not say bounded analytic functions are rare. On a disc they are everywhere: every analytic function on a closed disc is bounded on it. The hypothesis is boundedness on the whole plane, and that is what no interesting function satisfies.
It does not say an entire function that is bounded on some large region is constant. It has to be bounded on all of it. A function bounded on a half plane can be wild on the other half.
It does not say a bounded entire function is a polynomial. The conclusion is stronger than that — constant — and a learner who remembers the weaker version has remembered a true statement that is no use.
The hypothesis is global and the temptation is to use it locally. Every analytic function is bounded on every closed disc — that is just continuity on a compact set — so an argument that says $f$ is analytic and bounded on $|z| \le 5$, therefore constant proves that every analytic function is constant, which is plainly false. The bound has to hold everywhere at once.
The second misreading weakens the conclusion. Bounded entire implies polynomial is true, because constants are polynomials, and it is useless: every application of the theorem needs the full strength. If the remembered version is the weak one, the fundamental theorem of algebra cannot be proved from it.
The third is to forget that entire is also global. A function analytic on the plane minus a point is not entire, and Liouville has nothing to say about it — $1/z$ is bounded outside the unit circle and is certainly not constant.
One more, quieter: the theorem is about the modulus, not about the values. A function whose values all lie in some half plane is not obviously bounded, and yet it too must be constant — by composing with a Möbius map into the disc first. That kind of move, building a new function so that Liouville applies, is what the theorem is really for.
Suppose $f$ and $g$ are entire and $|f - g| \le 5$ everywhere. Let $h = f - g$.
Build a function out of the hypothesis.
$h$ is entire, as a difference of entire functions, and bounded by $5$.
Check both hypotheses.
So $h$ is constant: two entire functions that stay a bounded distance apart differ by a constant. Nothing like this is true for smooth real functions.
Translate back.
$\sin z$ is entire and is not constant, so by Liouville it cannot be bounded on the plane.
Run the theorem backwards.
And indeed $\sin(iy) = i\sinh y$, whose modulus grows without limit up the imaginary axis. The familiar bound of $1$ was a fact about the real axis only.
The theorem predicted where to look.
Bounded and entire, so constant by Liouville.
Check the hypotheses first.
A constant cannot take two different values, so no such function exists.
Build the proof that an entire function bounded on the whole plane is constant.
This task has no paper form; do it on a device.
Is this Liouville's theorem: an entire function that misses two values is constant?
Match each theorem to what it says.
| A bound on the function on a circle bounds each derivative at its centre | An entire function bounded on the whole plane is constant | Every polynomial of degree at least one has a complex root | A non-constant analytic function attains no maximum of its modulus inside a region | |
|---|---|---|---|---|
| The Cauchy estimate | ||||
| Liouville's theorem | ||||
| The fundamental theorem of algebra | ||||
| The maximum modulus principle |
Select every statement that is true.
This task has no paper form; do it on a device.
A polynomial of degree $3$ with complex coefficients. Counted with multiplicity, how many roots does it have in the complex numbers?
Answer:
$f$ is entire and satisfies $|f(z)| \le 7|z|^{1}$ for every $z$ with $|z|$ large. Then $f$ is a polynomial. What is the largest degree it could have?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A polynomial of degree $8$ with complex coefficients. Counted with multiplicity, how many roots does it have in the complex numbers?
Answer:
You can state and prove Liouville's theorem and use it on a problem that mentions no integral. Say in your own words why sine does not contradict it. Next: unit 4, where the point a function misbehaves at stops being an obstacle and starts being the source of the answer.
9. Your turn: can an entire function have $|f(z)| \le 3$ for all $z$ and $f(0) = 1$, $f(1) = 2$?, step 2