Proving a function is entire, bounded, or constant — which theorem closes the argument?
Liouville, the maximum modulus principle, and the identity theorem answer different questions — pointed at the wrong one, none of them close the proof.
Reach for this when a proof's last step is a claim about a whole function's global behavior — that it must be constant, that it can't peak in the interior, or that it must equal some other named function everywhere. Check which hypothesis you actually have: boundedness, nonconstancy, or agreement on a set of points.
Hypothesis: entire AND bounded on the whole plane
Term
Entire
Term
Bounded set
If f is entire (analytic on all of \mathbb{C}) and bounded — there is some B with |f(z)|\le B for every z\in\mathbb{C} — then f is constant.
Example
Entire + bounded together force f to be constant — neither alone is enough.
Show proof
Fix any point z_0\in\mathbb{C}. For every radius R>0, the derivative formula from the Cauchy Integral Formula (part (b), with n=1) applied to the circle C_R of radius R centered at z_0 gives f'(z_0) = \frac{1}{2\pi i}\oint_{C_R} \frac{f(z)}{(z-z_0)^2}\,dz. On C_R, |z-z_0|=R, so |f(z)/(z-z_0)^2| \le B/R^2 using the global bound |f(z)|\le B; the ML inequality (length of C_R is 2\pi R) then gives Cauchy's inequality for the first derivative, |f'(z_0)| \le \frac{1}{2\pi}\cdot\frac{B}{R^2}\cdot 2\pi R = \frac{B}{R}. This holds for every R>0, and f being entire means the formula above is valid no matter how large R is taken — there is no boundary to run into. Letting R\to\infty forces B/R\to0, so |f'(z_0)|=0, i.e. f'(z_0)=0. Since z_0\in\mathbb{C} was arbitrary, f'\equiv0 on all of \mathbb{C}, and \mathbb{C} is connected, so by the Zero Derivative Implies Constant theorem, f is constant.
Hypothesis: analytic and nonconstant on a domain
Let f be analytic and nonconstant on a domain D. Then |f| has no interior maximum on D. That is, no point z_0 \in D has |f(z_0)| \ge |f(z)| for every z in a neighborhood of z_0.
Consequently, suppose f is also continuous on a closed, bounded region R whose interior is D. Then the maximum value of |f| on R is attained on the boundary \partial R. It is never attained only at an interior point.
|f| can never peak at an interior point — only on the boundary. Proves the opposite direction from Liouville: this rules out a max, it doesn't force f to be constant.
Show proof
Suppose toward a contradiction that f is analytic and nonconstant on D, yet some z_0 \in D has a neighborhood N \subseteq D with |f(z_0)| \ge |f(z)| for every z \in N; write M = |f(z_0)|.
The mean value property forces constant modulus nearby. Fix r>0 small enough that the closed disk \{|z-z_0|\le r\} lies in N. For each \rho \in (0,r], the mean value property — f(z_0) equals the average of f over the circle of radius \rho about z_0, itself a short consequence of the Cauchy Integral Formula — f(z_0) = \frac{1}{2\pi}\int_0^{2\pi} f(z_0+\rho e^{i\theta})\,d\theta gives M = |f(z_0)| = \left|\frac{1}{2\pi}\int_0^{2\pi} f(z_0+\rho e^{i\theta})\,d\theta\right| \le \frac{1}{2\pi}\int_0^{2\pi}|f(z_0+\rho e^{i\theta})|\,d\theta \le \frac{1}{2\pi}\int_0^{2\pi} M\,d\theta = M, the last inequality because every point z_0+\rho e^{i\theta} lies in N. Equality throughout forces the continuous, nonnegative function \theta \mapsto M - |f(z_0+\rho e^{i\theta})| to average to zero, hence to vanish identically. As \rho ranges over (0,r], this makes |f| \equiv M on the entire disk \Delta = \{|z-z_0|\le r\}.
Constant modulus forces constant value. Write f=u+iv, so u^2+v^2 \equiv M^2 on \Delta. Differentiating in x and y, then substituting the Cauchy–Riemann equations u_y=-v_x, v_y=u_x, gives the linear system uu_x + vv_x = 0, \qquad vu_x - uv_x = 0 for (u_x,v_x), with coefficient determinant -(u^2+v^2) = -M^2. If M=0, then f\equiv0 on \Delta outright. If M\neq0, the determinant is nonzero, forcing u_x=v_x=0 — and then u_y=v_y=0 too, by Cauchy–Riemann again — so f'\equiv0 throughout \Delta. Either way f is constant on \Delta, equal to f(z_0), by the Zero Derivative Implies Constant theorem (\Delta is connected).
Open-and-closed connectedness finishes the job. Let S = \{z\in D : f(z) = f(z_0)\}. S is closed in D, being the preimage of a single point under the continuous map f, and S\neq\varnothing since \Delta\subseteq S. S is also open in D: near any point of S, the same mean-value computation just run at z_0 — with that point standing in for z_0, using that f already equals the constant f(z_0) throughout a whole disk touching it — forces f to equal f(z_0) throughout a further whole disk around it (the identical mechanism the Identity Theorem's own proof uses to spread a zero outward, applied here to the zero set of f-f(z_0) instead of to f itself). This is exactly the open-and-closed connectedness argument behind the harmonic maximum principle: an equality pinned down on one neighborhood spreads outward, one overlapping disk at a time, to the entire connected domain. Since D is connected and S is a nonempty subset of D that is both open and closed in D, S=D. Hence f \equiv f(z_0) on all of D — contradicting the standing assumption that f is nonconstant, and completing the proof.
Hypothesis: two analytic functions agree on infinitely many points with a limit point in the domain
Let f be analytic on a connected domain D, and suppose the zero set Z = \{z \in D : f(z) = 0\} has a limit point z_0 \in D (a point of D every neighborhood of which contains infinitely many points of Z). Then f \equiv 0 on all of D.
Consequently, if f and g are both analytic on D and f = g on some subset of D that has a limit point in D — in particular, on any open subdisk of D or along any arc in D — then f \equiv g on all of D.
Example
Agreement on a set with an accumulation point forces the two functions to be IDENTICAL everywhere on the domain — not just close, equal.
Show proof
Apply the theorem to h = f-g, which is analytic on D and vanishes on the given subset, so the consequence follows directly from the main statement; it remains to justify that one.
Every zero of an analytic function is either isolated (some punctured neighborhood of it contains no other zero) or a limit point of other zeros. The Taylor expansion of f centered at the limit point z_0 — valid since f is analytic there — decides which case occurs: if the expansion is not identically zero, its lowest-order nonzero term of some degree n dominates near z_0, forcing f(z) \neq 0 on a punctured neighborhood of z_0, which would make z_0 an isolated zero and contradicts the hypothesis that z_0 is a limit point of zeros. So the Taylor expansion at z_0 must be identically zero, meaning f \equiv 0 on some disk around z_0.
From there, the set of points in D where f and all its derivatives vanish is shown to be both open (by the same Taylor-expansion argument applied at each of its points) and closed in D (as an intersection of zero sets of continuous functions); since D is connected and this set contains z_0, it must be all of D. This "isolated zeros" machinery is developed rigorously in the power series chapter, where the Taylor expansion argument sketched above is made precise.