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The Maximum Modulus Principle

Module by: Lawrence Baggett. E-mail the author

Summary: Our next goal is to examine so-called “max/min” problems for coplex-valued functions of complex variables. Since order makes no sense for complex numbers, we will investigate max/min problems for the absolute value of a complex-valued function. For the corresponding question for real-valued functions of real variables, we have as our basic result the First Derivative Test. Indeed, when searching for the poinhts where a differentiable real-valued function ff on an interval [a,b][a,b] attains its extreme values, we consider first the poinhts where it attains a local max or min. Of course, to find the absolute minimum and maximum, we must also check the values of the function at the endpoints.

Our next goal is to examine so-called “max/min” problems for coplex-valued functions of complex variables. Since order makes no sense for complex numbers, we will investigate max/min problems for the absolute value of a complex-valued function. For the corresponding question for real-valued functions of real variables, we have as our basic result the First Derivative Test ((Reference)). Indeed, when searching for the poinhts where a differentiable real-valued function ff on an interval [a,b][a,b] attains its extreme values, we consider first the poinhts where it attains a local max or min, to which purpose end (Reference) is useful. Of course, to find the absolute minimum and maximum, we must also check the values of the function at the endpoints.

An analog of (Reference) holds in the complex case, but in fact a much different result is really valid. Indeed, it is nearly impossible for the absolute value of a differentiable function of a complex variable to attain a local maximum or minimum.

Theorem 1

Let ff be a continuous function on a piecewise smooth geometric set S,S, and assume that ff is differentiable on the interior S0S0 of S.S. Suppose cc is a point in S0S0 at which the real-valued function |f||f| attains a local maximum. That is, there exists an ϵ>0ϵ>0 such that |f(c)||f(z)||f(c)||f(z)| for all zz satisfying |z-c|<ϵ.|z-c|<ϵ. Then ff is a constant function on S;S; i.e., f(z)=f(c)f(z)=f(c) for all zS.zS. In other words, the only differentiable functions of a complex variable, whose absolute value attains a local maximum on the interior of a geometric set, are constant functions on that set.

Proof

If f(c)=0,f(c)=0, then f(z)=0f(z)=0 for all zBϵ(c).zBϵ(c). Hence, by the Identity Theorem ((Reference)), f(z)f(z) would equal 0 for all zS.zS. so, we may as well assume that f(c)0.f(c)0. Let rr be any positive number for which the closed disk B¯r(c)B¯r(c) is contained in Bϵ(c).Bϵ(c). We claim first that there exists a point zz on the boundary CrCr of the disk B¯r(c)B¯r(c) for which |f(z)|=|f(c)|.|f(z)|=|f(c)|. Of course, |f(z||f(c)||f(z||f(c)| for all zz on this boundary by assumption. By way of contradiction, suppose that |f(ζ)|<|f(c)||f(ζ)|<|f(c)| for all ζζ on the boundary CrCr of the disk. Write MM for the maximum value of the function |f||f| on the compact set Cr.Cr. Then, by our assumption, M<|f(c)|.M<|f(c)|. Now, we use the Cauchy Integral Formula:

| f ( c ) | = | 1 2 π i C r f ( ζ ) ζ - c d ζ | = 1 2 π | 0 2 π f ( c + r e i t ) r e i t i r e i t d t | 1 2 π 0 2 π | f ( c + r e i t ) | d t 1 2 π 0 2 π M d t = M < | f ( c ) | , | f ( c ) | = | 1 2 π i C r f ( ζ ) ζ - c d ζ | = 1 2 π | 0 2 π f ( c + r e i t ) r e i t i r e i t d t | 1 2 π 0 2 π | f ( c + r e i t ) | d t 1 2 π 0 2 π M d t = M < | f ( c ) | ,
(1)

and this is a contradiction.

Now for each natural number nn for which 1/n<ϵ,1/n<ϵ, let znzn be a point for which |zn-c|=1/n|zn-c|=1/n and |f(zn)|=|f(c)|.|f(zn)|=|f(c)|. We claim that the derivative f'(zn)f'(zn) of ff at zn=0zn=0 for all n.n. What we know is that the real-valued function F(x,y)=|f(x+iy)|2=(u(x,y)2+(v(x,y))2F(x,y)=|f(x+iy)|2=(u(x,y)2+(v(x,y))2 attains a local maximum value at zn=(xn,yn).zn=(xn,yn). Hence, by (Reference), both partial derivatives of FF must be 0 at (xn,yn).(xn,yn). That is

2 u ( x n , y n ) t i a l u t i a l x ( x n , y n ) + 2 v ( x n , y n ) t i a l v t i a l x ( x n , y n ) = 0 2 u ( x n , y n ) t i a l u t i a l x ( x n , y n ) + 2 v ( x n , y n ) t i a l v t i a l x ( x n , y n ) = 0
(2)

and

2 u ( x n , y n ) t i a l u t i a l y ( x n , y n ) + 2 v ( x n , y n ) t i a l v t i a l y ( x n , y n ) = 0 . 2 u ( x n , y n ) t i a l u t i a l y ( x n , y n ) + 2 v ( x n , y n ) t i a l v t i a l y ( x n , y n ) = 0 .
(3)

Hence the two vectors

V 1 = ( t i a l u t i a l x ( x n , y n ) , t i a l v t i a l x ( x n , y n ) ) V 1 = ( t i a l u t i a l x ( x n , y n ) , t i a l v t i a l x ( x n , y n ) )
(4)

and

V 2 = ( t i a l u t i a l y ( x n , y n ) , t i a l v t i a l y ( x n , y n ) ) V 2 = ( t i a l u t i a l y ( x n , y n ) , t i a l v t i a l y ( x n , y n ) )
(5)

are both perpendicular to the vector V3=(u(xn,yn),v(xn,yn)).V3=(u(xn,yn),v(xn,yn)). But V30,V30, because V3=|f(zn)|=|f(c)|>0,V3=|f(zn)|=|f(c)|>0, and hence V1V1 and V2V2 are linearly dependent. But this implies that f'(zn)=0,f'(zn)=0, according to (Reference).

Since c=limzn,c=limzn, and f'f' is analytic on S0,S0, it follows from the Identity Theorem that there exists an r>0r>0 such that f'(z)=0f'(z)=0 for all zBr(c).zBr(c). But this implies that ff is a constant f(z)=f(c)f(z)=f(c) for all zBr(c).zBr(c). And thenm, again using the Identity Theorem, this implies that f(z)=f(c)f(z)=f(c) for all zS,zS, which completes the proof.

REMARK Of course, the preceding proof contains in it the verification that if |f||f| attains a maximum at a point cc where it is differentiable, then f'(c)=0.f'(c)=0. This is the analog for functions of a complex variable of (Reference). But, Theorem 1 certainly asserts a lot more than that. In fact, it says that it is impossible for the absolute value of a nonconstant differentiable function of a complex variable to attain a local maximum. Here is the coup d'grâs:

Corollary 1: Maximum Modulus Principle

Let ff be a continuous, nonconstant, complex-valued function on a piecewise smooth geometric set S,S, and suppose that ff is differentiable on the interior S0S0 of S.S. Let MM be the maximum value of the continuous, real-valued function |f||f| on S,S, and let zz be a point in SS for which |f(z)|=M.|f(z)|=M. Then, zz does not belong to the interior S0S0 of S;S; it belongs to the boundary of S.S. In other words, |f||f| attains its maximum value only on the boundary of S.S.

Exercise 1

  1. Prove the preceding corollary.
  2. Let ff be an analytic function on an open set U,U, and let cUcU be a point at which |f||f| achieves a local minimum; i.e., there exists an ϵ>0ϵ>0 such that |f(c)||f(z)||f(c)||f(z)| for all zBϵ(c).zBϵ(c). Show that, if f(c)0,f(c)0, then ff is constant on Bϵ(c).Bϵ(c). Show by example that, if f(c)=0,f(c)=0, then ff need not be a constant on Bϵ(c).Bϵ(c).
  3. Prove the “Minimum Modulus Principle:” Let ff be a nonzero, continuous, nonconstant, function on a piecewise smooth geometric set S,S, and let mm be the minimum value of the function |f||f| on S.S. If zz is a point of SS at which this minimum value is atgtained, then zz belongs to the boundary CSCS of S.S.

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