entire functions
Liouville's Theorem
You should know: cauchy integral formula
Overview
Liouville's theorem states that every bounded entire function is constant: if f is holomorphic on all of ℂ and there is a constant M with |f(z)| ≤ M for every z, then f must be constant. This is a striking rigidity statement with no real-analysis counterpart — plenty of bounded, smooth, non-constant real functions exist (like sin x), but complex differentiability everywhere is so restrictive that boundedness alone forces triviality. The theorem follows quickly from the generalized Cauchy integral formula applied to f' on circles of growing radius: the derivative estimate shows |f'(z)| shrinks to 0 everywhere, so f' ≡ 0 and f is constant. Liouville's theorem is best known as the key step in the fundamental theorem of algebra: assuming a non-constant polynomial had no root leads to 1/p(z) being a bounded entire function, which Liouville's theorem rules out unless it's constant — a contradiction.
Intuition
The generalized Cauchy integral formula gives f'(z₀) as a contour integral over any circle around z₀, and if |f| ≤ M everywhere, the size of that integral is bounded by M divided by the radius of the circle — because the circle's circumference grows only linearly while the 1/(z−z₀)² factor shrinks quadratically. Since f is entire, this estimate is valid for arbitrarily large circles, and letting the radius go to infinity crushes the bound on |f'(z₀)| all the way down to 0. A function with zero derivative everywhere on a connected domain (all of ℂ) is constant. The deep reason this has no real analogue is that a real bounded smooth function's derivative doesn't have to be controlled by any such global 'derivative estimate on arbitrarily large circles' — real differentiability is a local, one-directional notion, while complex differentiability everywhere ties every point's derivative to the function's global (in this case, global-boundedness) behavior via Cauchy's formula.
Formal Definition
Let f : ℂ → ℂ be entire (holomorphic on all of ℂ):
Worked Examples
sin(z) is entire (it is defined by a power series convergent for all z, e.g. sin z = (e^{iz}-e^{-iz})/2i).
sin(z) is clearly non-constant (e.g. sin(0)=0, sin(π/2)=1).
By Liouville's theorem (contrapositive), a non-constant entire function cannot be bounded on all of ℂ, so sin(z) must be unbounded.
Answer: sin(z) is unbounded on ℂ; in fact along the imaginary axis, sin(iy) = i·sinh(y) → ∞ as y → ∞, confirming this directly.
Practice Problems
Is e^z bounded on ℂ? What does Liouville's theorem tell you about e^z?
If f is entire and satisfies |f(z)| ≤ 7 for all z ∈ ℂ, and f(0) = 3, what is f(100 + 200i)?
Suppose f is entire and |f(z)| ≤ |z| + 1 for all z (not bounded, but growing at most linearly). Explain, in the spirit of the Liouville-theorem derivative estimate, why f must be a polynomial of degree at most 1 (this is the 'generalized Liouville theorem').
Quiz
Summary
- Liouville's theorem: every bounded entire function (holomorphic on all of ℂ) is constant.
- The proof uses the Cauchy derivative estimate |f'(z₀)| ≤ M/R on circles of radius R, letting R → ∞ to force f' ≡ 0.
- Non-constant entire functions like e^z and sin(z) must therefore be unbounded on ℂ, unlike their real-line counterparts.
- Liouville's theorem proves the fundamental theorem of algebra: assuming a non-constant polynomial has no roots makes 1/p(z) a bounded entire function, forcing a contradiction.
Mathematics