Chapter 3: The maximum modulus principle

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1 Chapter 3: The maximum modulus principle Course 44, 23 4 December 3, 23 Theorem 3. (Identity theorem for analytic functions) Let G C be open and connected (and nonempty). Let f : G C be analytic. Then the following are equivalent for f: (i) f (ii) there is an infinite sequence (z n ) n= of distinct points of G with lim n z n = a G and f(z n ) = n (iii) there is a point a G with f (n) (a) = for n =,, 2,.... Proof. We show (i) (ii) (iii) (i). (i) (ii): is really obvious. If f, take any a G (here we need G ), choose δ > with D(a, δ) G and put z n = a + δ/(n + ). (ii) (iii): Assuming lim n z n = a G, z n distinct and f(z n ) = n, consider the power series for f centered at a. That is a n (z a) n for z a < δ n= (for some δ > with D(a, δ) G). Here a n = f (n) (a)/n! and so our aim of showing f (n) (a) = for all n is equivalent to showing a n = for all n. If that is not the case, there must be a smallest m with a m. Now, for z a < δ we can write a n (z a) n n=m = (z a) m a n (z a) n m n=m = (z a) m g(z)

2 R. Timoney and g(z) = n=m a n(z a) n m is analytic for z a < δ. Moreover g(a) = a m, g(z) is continuous at z = a and so we can find δ >, δ δ, with g(z) g(a) 2 g(a) for z a < δ g(z) 2 g(a) for z a < δ g(z) for z a < δ. But = f(z n ) and for n large enough (say n > N) we have z n a < δ so that f(z n ) = (z n a) m g(z n ) =. Thus (for n > N) z n = or g(z n ) =. However, we know g(z n ) and the z n are distinct so that at most one n can have z n = a. Hence we are faced with a contradiction. The contradiction arose from assuming that there was any a n. We must therefore have a n = n. (iii) (i): Assume now that there is a G with f (n) (a) = for all n. Then the power series expansion for f about a (which is valid in a disc D(a, δ) G with δ > ) is n= f (n) (a) (z a) n = for z a < δ. n! Thus f(z) for z a < δ and differentiating we get f (n) (z) = for n =,, 2,.... This shows that U = {a G : f (n) (a) = for all n =,, 2,...} is open (and nonempty). U is also closed relative to G. To see that take b G \ U. Then there is some n with f (n) (b). Now that f (n) is continuous at b and so there is a δ > so that f (n) (z) for all z with z b < δ. This means none of these z can be in U or in other words D(b, δ) G \ U. This means G \ U is open. As G is connected, U G nonempty and both open and closed relative to G implies U = G. This means f (n) (z) = for all n and all z G. Specifically with n = we have f. Corollary 3.2 (version with two functions) Let G C be open and connected (and nonempty). Let f, g : G C be two analytic functions. Then the following are equivalent for f and g: (i) f g (ii) there is an infinite sequence (z n ) n= of distinct points of G with lim n z n = a G and f(z n ) = g(z n ) n (iii) there is a point a G with f (n) (a) = g (n) (a) for n =,, 2,.... Proof. apply the Identity Theorem 3. to the difference f g. Remark 3.3 The significance of the Identity Theorem is that an analytic function on a connected open G C is determined on all of G by its behaviour near a single point. Thus if an analytic function is given on one part of G by a formula like and that z formula makes sense and gives an analytic function on a larger connected subset of G then it has to be that also holds in the larger set. z

3 Chapter 3 maximum modulus principle 3 by This is quite different from what happens with continuous functions like f : C C defined { z z < z z. z Even for C functions we can have different formulae holding in different places. Consider g : C C where ( z ( ) ) 2 exp z >. z The original meaning of the word analytic related to this property of analytic functions (one formula). Corollary 3.4 If G C is a connected open set and f : G C is analytic and not identically constant, then the zero set of f has no accumulation points in G. Z f = {z G : } Proof. First we should define accumulation point in case you forget it. If S C is any set and a C, then a is called an accumulation point of S if for each δ > (S \ {a}) D(a, δ). If we a is an accumulation point of S we can choose z (S \ {a}) D(a, ) ( ( )) z 2 (S \ {a}) D a, min 2, z a ( ( )) z 3 (S \ {a}) D a, min 3, z 2 a and (inductively) z n+ (S \{a}) D ( a, min ( n, z n a )). This produces a sequence (z n ) n= of distinct points z n S with lim n z n = a. (It is not hard to see that the existence of such a sequence is equivalent to a being an accumulation point of S.) Applying this to S = Z f and using Theorem 3. we get f. Corollary 3.5 Let G C be open and connected and let K G be compact. Let f, g : G C be analytic. If the equation g(z) has infinitely many solutions z K, then f g. Proof. Choose an infinite sequence (z n ) n= of distinct points z n K where f(z n ) = g(z n ). Since K is compact, the sequence has a convergent subsequence (z nj ) j= with a limit a = lim j z nj K G. By Corollary 3.2, f g.

4 R. Timoney Theorem 3.6 (Maximum modulus theorem, basic version) Let G C be a connected open set and f : G C analytic. If there is any a G with f(a) f(z) for all z G, then f is constant. Proof. (Another way to state this is that f(z) cannot have a maximum in G, unless f is constant.) Choose δ > so that D(a, δ) G. Fix < r < δ and then we have (by the Cauchy integral formula) f(a) = i z a =r f(z) z a dz. Write this out in terms of a parametrisation z = a + re iθ with θ, dz = ire iθ dθ. f(a) = i f(a + re iθ ) ire iθ dθ f(a + re iθ ) dθ. Hence f(a) f(a + re iθ ) dθ f(a) dθ = f(a), using f(a + re iθ ) f(a) θ. We must therefore have equality in the inequalities. Since the integrand f(a + re iθ ) is a continuous function of θ, this implies f(a + re iθ ) = f(a) for all θ. Put α = Arg(f(a)). Now f(a) = e iα f(a) = e iα = R f(a) = f(a) = R = f(a + re iθ ) dθ e iα f(a + re iθ ) dθ e iα f(a + re iθ ) dθ R(e iα f(a + re iθ )) dθ e iα f(a + re iθ ) dθ using Rw w for w C f(a + re iθ ) dθ f(a) dθ = f(a) Thus we must again have equality in all the inequalities and so R(e iα f(a + re iθ )) = e iα f(a + re iθ ) = f(a)

5 Chapter 3 maximum modulus principle 5 for all θ. Thus I(e iα f(a + re iθ )) = and e iα f(a + re iθ ) = f(a) or f(a + re iθ ) = e iα f(a). Thus f(z) is constant for z in the infinite compact subset {z : z a = r} of G. By Corollary 3.5, it follows that f(z) is constant (on G). Theorem 3.7 (Maximum modulus theorem, usual version) The absolute value of a nonconstant analytic function on a connected open set G C cannot have a local maximum point in G. Proof. Let f : G C be analytic. By a local maximum point for f we mean a point a G where f(a) f(z) holds for all z D(a, δ) G, some δ >. As G is open, by making δ > smaller if necessary we can assume D(a, δ) G. By Theorem 3.6, f(a) f(z) z D(a, δ) implies f constant on D(a, δ) (since f must be analytic on D(a, δ) G and D(a, δ) is connected open). Then, by the Identity Theorem (Corollary 3.2), f must be constant. Corollary 3.8 (Maximum modulus theorem, another usual version) Let G C be a bounded and connected open set. Let f : Ḡ C be continuous on the closure Ḡ of G and analytic on G. Then sup f(z) = sup f(z). z Ḡ z G (That is the maximum modulus of the analytic function f(z) is attained on the boundary G.) Proof. Since G is bounded, its closure Ḡ is closed and bounded, hence compact. f(z) is a continuous real-valued function on the compact set and so sup z Ḡ f(z) < and the supremum is attained at some point b Ḡ. That is f(b) f(z) z Ḡ. If b is on the boundary G then we have f(b) = sup f(z) = sup f(z) z G z Ḡ but if b G, then f must be constant by the Identity Theorem 3.. So in that case sup z G f(z) = sup z Ḡ f(z) is also true. Theorem 3.9 (Fundamental theorem of algebra) Let p(z) be a nonconstant polynomial (p(z) = n k= a kz k with a n and n ). Then the equation p(z) = has a solution z C. Proof. Dividing the equation by the coefficient of the highest power of z with a nonzero coefficient (a n in the notation above) we can assume without loss of generality that the polynomial is monic (meaning that the coefficient of the highest power of z is ) and that the polynomial is n p(z) = z n + a k z k (n ). k=

6 R. Timoney Now for z = R R = max (, 2 n k= a k ), we have n p(z) z n a k z k k= n = R n a k R k k= n R n a k R n k= using R n = R n R n a k k= R n R n (R /2) R n (R R /2) R n R/2 = R n /2 = z n /2 Now if p(z) is never zero for z C, then /p(z) is an entire function. For z R we have f(z) = p(z) z n /2 = 2 z 2 n R n By the maximum modulus theorem (Corollary 3.8) sup f(z) = sup f(z) 2 z R z =R R n and so we have f(z) 2/R n for all z C. By Liouville s theorem (.26) f must be constant. But then p(z) = /f(z) is also constant, contradicting our hypotheses. The contradiction arose by assuming p(z) was never. So p(z) = for some z C. Remark 3. There are many ways to prove the fundamental theorem, but all (or at least most) rely on the same first step as above for z large, p(z) behaves like its term of highest degree a n z n. To make things a little simpler, we took a n = but that is not really essential. The idea is that, when z is large the highest term outweighs the combination of the lower degree terms. Also, although we used the maximum modulus theorem to get the same bound for f(z) for z R as for z R, we could have just used compactness to get sup z R f(z) <. (We used an argument like that once before in the proof of the winding number version of Cauchy s theorem (.3 claim 2 in the proof). Corollary 3. If p(z) = n k= a kz k is a polynomial of degree n (meaning a n ) then p(z) can be factored for some α, α 2,..., α n C. p(z) = a n n j= (z α j )

7 Chapter 3 maximum modulus principle 7 Proof. For n = the result is true (provided we interpret the empty product as ). By the Fundamental Theorem of algebra 3.9, there is some α C with p(α ) =. Then it follows from the remainder theorem that z α divides p(z). In other words p(z) = (z α )q(z) where q(z) has degree n and leading coefficient a n (same as for p(z)). If we arrange the proof more formally as induction on the degree, we have the starting n = and the induction step. Applying the result from degree n to q(z) = a n n j=2 (z α j) we are done. Richard M. Timoney December 3, 23