INTEGRALS
54. MAXIMUM MODULUS PRINCIPLE
sec. 54 Maximum Modulus Principle 175
we find that 2π
0
|f (z0+ρeiθ)|dθ≤ 2π
0
|f (z0)|dθ =2π|f (z0)|. Thus
|f (z0)| ≥ 1 2π
2π 0
|f (z0+ρeiθ)|dθ . (5)
It is now evident from inequalities (3) and (5) that
|f (z0)| = 1 2π
2π 0
|f (z0+ρeiθ)|dθ ,
or 2π
0
[|f (z0)| − |f (z0+ρeiθ)|]dθ=0.
The integrand in this last integral is continuous in the variable θ; and, in view of condition (4), it is greater than or equal to zero on the entire interval 0≤θ≤2π. Because the value of the integral is zero, then, the integrand must be identically equal to zero. That is,
|f (z0+ρeiθ)| = |f (z0)| (0≤θ ≤2π ).
(6)
This shows that|f (z)| = |f (z0)| for all points zon the circle |z−z0| =ρ. Finally, sincez1is any point in the deleted neighborhood 0<|z−z0|< ε, we see that the equation|f (z)| = |f (z0)|is, in fact, satisfied by all pointszlying on any circle |z−z0| =ρ, where 0< ρ < ε. Consequently, |f (z)| = |f (z0)| everywhere in the neighborhood|z−z0|< ε. But we know from Example 4. Sec. 25, that when the modulus of an analytic function is constant in a domain, the function itself is constant there. Thus f (z)=f (z0) for each point z in the neighborhood, and the proof of the lemma is complete.
This lemma can be used to prove the following theorem, which is known as themaximum modulus principle.
Theorem. If a function f is analytic and not constant in a given domainD, then|f (z)|has no maximum value inD. That is, there is no pointz0in the domain such that|f (z)| ≤ |f (z0)|for all pointszin it.
Given that f is analytic inD, we shall prove the theorem by assuming that
|f (z)| does have a maximum value at some point z0 inD and then showing that f (z) must be constant throughoutD.
The general approach here is similar to that taken in the proof of the lemma in Sec. 27. We draw a polygonal line Llying in D and extending fromz0 to any other pointP inD. Also,drepresents the shortest distance from points onLto the
sec. 54 Maximum Modulus Principle 177
boundary ofD. WhenD is the entire plane, d may have any positive value. Next, we observe that there is a finite sequence of points
z0, z1, z2, . . . , zn−1, zn
alongLsuch thatzn coincides with the pointP and
|zk−zk−1|< d (k=1,2, . . . , n).
In forming a finite sequence of neighborhoods (Fig. 71) N0, N1, N2, . . . , Nn−1, Nn
where each Nk has center zk and radius d, we see that f is analytic in each of these neighborhoods, which are all contained in D, and that the center of each neighborhoodNk(k=1,2, . . . , n)lies in the neighborhoodNk−1.
z0
N0 N1 N2 L Nn
P
z2 zn – 1 zn
z1
Nn – 1
FIGURE 71
Since |f (z)| was assumed to have a maximum value in D at z0, it also has a maximum value in N0 at that point. Hence, according to the preceding lemma, f (z)has the constant valuef (z0)throughoutN0. In particular,f (z1)=f (z0). This means that|f (z)| ≤ |f (z1)|for each pointzinN1; and the lemma can be applied again, this time telling us that
f (z)=f (z1)=f (z0)
when z is in N1. Since z2 is in N1, then, f (z2)=f (z0). Hence |f (z)| ≤ |f (z2)| whenzis inN2; and the lemma is once again applicable, showing that
f (z)=f (z2)=f (z0)
whenzis inN2. Continuing in this manner, we eventually reach the neighborhood Nn and arrive at the fact thatf (zn)=f (z0).
Recalling thatzn coincides with the pointP, which is any point other thanz0
inD, we may conclude thatf (z)=f (z0)forevery pointzinD. Inasmuch asf (z) has now been shown to be constant throughoutD, the theorem is proved.
If a functionf that is analytic at each point in the interior of a closed bounded regionRis also continuous throughoutR, then the modulus|f (z)|has a maximum value somewhere inR(Sec. 18). That is, there exists a nonnegative constantMsuch that|f (z)| ≤M for all pointszinR, and equality holds for at least one such point.
Iff is a constant function, then|f (z)| =Mfor allzinR. If, however,f (z)is not constant, then, according to the theorem just proved,|f (z)| =M for any pointzin the interior ofR. We thus arrive at an important corollary.
Corollary. Suppose that a functionf is continuous on a closed bounded region R and that it is analytic and not constant in the interior ofR. Then the maximum value of |f (z)| inR, which is always reached, occurs somewhere on the boundary ofRand never in the interior.
EXAMPLE. LetRdenote the rectangular region 0≤x≤π,0≤y≤1. The corollary tells us that the modulus of the entire functionf (z)=sinzhas a maximum value inRthat occurs somewhere on the boundary ofRand not in its interior. This can be verified directly by writing (see Sec. 34)
|f (z)| =
sin2x+sinh2y
and noting that the term sin2x is greatest whenx=π/2 and that the increasing function sinh2y is greatest wheny=1. Thus the maximum value of|f (z)| in R occurs at the boundary pointz=(π/2,1)and at no other point inR(Fig. 72).
O x 1 y
FIGURE 72
When the functionf in the corollary is writtenf (z)=u(x, y)+iv(x, y),the component function u(x, y)also has a maximum value in R which is assumed on the boundary ofRand never in the interior, where it is harmonic (Sec. 26). This is because the composite function g(z)=exp[f (z)] is continuous in R and analytic and not constant in the interior. Hence its modulus|g(z)| =exp[u(x, y)], which is continuous in R, must assume its maximum value in R on the boundary. In view of the increasing nature of the exponential function, it follows that the maximum value ofu(x, y)also occurs on the boundary.
Properties ofminimum values of|f (z)|andu(x, y)are treated in the exercises.
EXERCISES
1. Suppose thatf (z)is entire and that the harmonic functionu(x, y)=Re[f (z)] has an upper boundu0; that is,u(x, y)≤u0for all points(x, y) in thexy plane. Show that u(x, y)must be constant throughout the plane.
Suggestion:Apply Liouville’s theorem (Sec. 53) to the functiong(z)=exp[f (z)].
sec. 54 Exercises 179
2. Show that forRsufficiently large, the polynomialP (z)in Theorem 2, Sec. 53, satisfies the inequality
|P (z)|<2|an||z|n whenever |z| ≥R.
[Compare with the first of inequalities (5), Sec. 53.]
Suggestion:Observe that there is a positive numberR such that the modulus of each quotient in expression (3), Sec. 53, is less than|an|/nwhen|z|> R.
3. Let a functionf be continuous on a closed bounded regionR, and let it be analytic and not constant throughout the interior of R. Assuming that f (z)=0 anywhere in R, prove that|f (z)|has aminimum value minRwhich occurs on the boundary ofR and never in the interior. Do this by applying the corresponding result for maximum values (Sec. 54) to the functiong(z)=1/f (z).
4. Use the functionf (z)=zto show that in Exercise 3 the conditionf (z)=0 anywhere in R is necessary in order to obtain the result of that exercise. That is, show that
|f (z)|can reach its minimum value at an interior point when the minimum value is zero.
5. Consider the functionf (z)=(z+1)2and the closed triangular regionRwith vertices at the pointsz=0, z=2, andz=i. Find points inRwhere|f (z)|has its maximum and minimum values, thus illustrating results in Sec. 54 and Exercise 3.
Suggestion: Interpret|f (z)|as the square of the distance betweenz and−1.
Ans.z=2, z=0.
6. Letf (z)=u(x, y)+iv(x, y) be a function that is continuous on a closed bounded region R and analytic and not constant throughout the interior ofR. Prove that the component functionu(x, y)has a minimum value inRwhich occurs on the boundary ofRand never in the interior. (See Exercise 3.)
7. Letf be the functionf (z)=ezandR the rectangular region 0≤x≤1,0≤y≤π.
Illustrate results in Sec. 54 and Exercise 6 by finding points inRwhere the component functionu(x, y)=Re[f (z)] reaches its maximum and minimum values.
Ans.z=1, z=1+π i.
8. Let the function f (z)=u(x, y)+iv(x, y) be continuous on a closed bounded region R, and suppose that it is analytic and not constant in the interior of R.
Show that the component function v(x, y) has maximum and minimum values in R which are reached on the boundary of R and never in the interior, where it is harmonic.
Suggestion:Apply results in Sec. 54 and Exercise 6 to the functiong(z)= −if (z).
9. Letz0 be a zero of the polynomial
P (z)=a0+a1z+a2z2+ · · · +anzn (an=0) of degreen (n≥1). Show in the following way that
P (z)=(z−z0)Q(z) whereQ(z)is a polynomial of degreen−1.
(a) Verify that
zk−zk0=(z−z0)(zk−1+zk−2z0+ · · · +z zk−0 2+zk−0 1) (k=2,3, . . .).
(b) Use the factorization in part(a)to show that
P (z)−P (z0)=(z−z0)Q(z)
whereQ(z)is a polynomial of degreen−1, and deduce the desired result from this.