Thm. 3. Maximum Modulus Principal

Supp. A is a domain and f : A -> C is analytic

(a) if |f(a)| attains its maximum in A, then f is constant

(b) if A is bounded and f is continuous on closure A, then |f(z)| attains its maximum on bd(A)


pf.) For (a), Suppose |f(z)| attains a maximum at z* in A,

Let h(z) = |f(z)|,

U = {z in A : |f(z)| = |f(z*)|}

V = {z in A : |f(z)| < |f(z*)|} = {z in A : |f(z)| =/= |f(z*)|}

Then U ∪ V = A and U ∩ V = empty

Moreover U is open by the local version of MMP, i.e. suppose w in U, then there exists R>0 s.t. D(w,R) and |f(z)| = |f(w)| for all z in D(w,R) in U => U is open

and V is open since h : A -> R is continuous and

V = h^(-1)((-infinity, |f(z*)|))

여기서, V=h^(-1)([0, |f(z*)|)) 아닌가요...? 이게 질문입니다.

Then because A is connected, we must have either U = empty or V = empty.

But z* in U, so we conclude that V = empty and U = A; i.e., |f(z)|=|f(z*)| for all z in A.

Then f is analytic with constant absolute values on A, so f must be constant.


To prove (b), suppose A is bounded. Then closure A is compact. so the continuous function h attains its maximim on closure A.

If h(z) = |f(z)| attains its maximum on int(A) = A, then f is constant on A.

Thus |f(z)| attains its maximum on bd(A), since it is continuous on closure A.

If |f(z)| does not attain its maximum on A, then its maxmum must occur on bd(A) = closureA - A, as desired.