FURMAN MTH 350 - Lecture 32: Liouville’s Theorem

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Lecture 32:Liouville’s TheoremDan SloughterFurman UniversityMathematics 39May 3, 200432.1 Liouville’s TheoremThe following remarkable result is known as Liouville’s theorem.Theorem 32.1. If f : C → C is entire and bounded, then f (z) is constantthroughout the plane.The proof of Liouville’s theorem follows easily from the following lemma.Lemma 32.1. Let CRbe the circle |z − z0| = R, R > 0, and suppose f isanalytic on the region consisting of CRand the points in its interior. If MRis the maximum value of |f (z)| on CR, then, for n = 1, 2, 3, . . .,f(n)(z0)≤n!MRRn.Proof. Sincef(n)(z0) =n!2πiZCRf(z)(z − z0)n+1dzandf(z)(z − z0)n+1=|f(z)||z − z0|n+1≤MRRn+1,we havef(n)(z0)≤n!2π·MRRn+1· 2πR =n!MRRn.1We may now return to the proof of Liouville’s theorem.Proof. Suppose f is entire and f(z) ≤ M for all z ∈ C. From the lemma, wehave, for any z0∈ C and any R > 0,|f0(z0)| ≤MRR≤MR.Letting R → ∞, we have |f0(z0)| = 0, and hence f0(z0) = 0 for every z0∈ C.Thus f(z) = c for some constant c and all z ∈ C.32.2 PolynomialsNow consider a polynomialP (z) = a0+ a1z + a2z2+ · · · + anzn,with an6= 0. Suppose there does not exists a z0∈ C for which P (z0) = 0.Letf(z) =1P (z).Then f is entire. Moreover, if n ≥ 1,limz→∞f(z) = 0sincelimz→0f1z= limz→01P1z= limz→01a0+a1z+a2z2+ · · · +anzn= limz→0zna0zn+ a1zn−1+ a2zn−2+ · · · + an=0an= 0.Hence there exists R > 0 such that |f(z)| < 1 whenever |z| > R. Sincef is continuous on the closed disk |z| ≤ R, there exists M > 0 such that|f(z)| ≤ M whenever |z| ≤ R. It follows that f is bounded on C. But then,by Liouville’s theorem, f is a constant function, which is true only if n = 0.Hence we have proven the following fundamental theorem of algebra.2Theorem 32.2. If P is a polynomial of degree n ≥ 1, then there exists atleast one point z0∈ C such that P (z0) = 0.Given a polynomial P of degree n ≥ 1 and a point z1for which P (z1) = 0,one may show that there exists a polynomial Q of degree n − 1 such thatP (z) = (z − z1)Q(z).Proceeding in this way, it now follows that there exist constants c and zk,k = 1, 2, 3, . . . , n, such thatP (z) = c(z − z1)(z − z2) · · · (z − zn).This is the Fundamental Theorem of Algebra.Corollary 32.1. Every algebraist needs an


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