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The Open Mapping Theorem and the Inverse Function Theorem

Module by: Lawrence Baggett. E-mail the author

Summary: We turn next to a question about functions of a complex variable that is related to the Inverse Function Theorem. That result asserts, subject to a couple of hypotheses, that the inverse of a one-to-one differentiable function of a real variable is also differentiable. Since a function is only differentiable at points in the interior of its domain, it is necessary to verify that the point f(c)f(c) is in the interior of the domain f(S)f(S) of the inverse function f-1f-1 before the question of differentiability at that point can be addressed. And, the peculiar thing is that it is this point about f(c)f(c) being in the interior of f(S)f(S) that is the subtle part. The fact that the inverse function is differentiable there, and has the prescribed form, is then only a careful ϵ-δϵ-δ argument. For continuous real-valued functions of real variables, the fact that f(c)f(c) belongs to the interior of f(S)f(S) boils down to the fact that intervals get mapped onto intervals by continuous functions, which is basically a consequence of the Intermediate Value Theorem. However, for complex-valued functions of complex variables, the situation is much deeper. For instance, the continuous image of a disk is just not always another disk, and it may not even be an open set. Well, all is not lost; we just have to work a little harder.

We turn next to a question about functions of a complex variable that is related to (Reference), the Inverse Function Theorem. That result asserts, subject to a couple of hypotheses, that the inverse of a one-to-one differentiable function of a real variable is also differentiable. Since a function is only differentiable at points in the interior of its domain, it is necessary to verify that the point f(c)f(c) is in the interior of the domain f(S)f(S) of the inverse function f-1f-1 before the question of differentiability at that point can be addressed. And, the peculiar thing is that it is this point about f(c)f(c) being in the interior of f(S)f(S) that is the subtle part. The fact that the inverse function is differentiable there, and has the prescribed form, is then only a careful ϵ-δϵ-δ argument. For continuous real-valued functions of real variables, the fact that f(c)f(c) belongs to the interior of f(S)f(S) boils down to the fact that intervals get mapped onto intervals by continuous functions, which is basically a consequence of the Intermediate Value Theorem. However, for complex-valued functions of complex variables, the situation is much deeper. For instance, the continuous image of a disk is just not always another disk, and it may not even be an open set. Well, all is not lost; we just have to work a little harder.

Theorem 1: Open Mapping Theorem

Let SS be a piecewise smooth geometric set, and write UU for the (open) interior S0S0 of S.S. Suppose ff is a nonconstant differentiable, complex-valued function on the set U.U. Then the range f(U)f(U)of ff is an open subset of C.C.

Proof

Let cc be in U.U. Because ff is not a constant function, there must exist an r>0r>0 such that f(c)f(z)f(c)f(z) for all zz on the boundary CrCr of the disk Br(c).Br(c). See part (b) of (Reference). Let z0z0 be a point in the compact set CrCr at which the continuous real-valued function |f(z)-f(c)||f(z)-f(c)| attains its minimum value s.s. Since f(z)f(c)f(z)f(c) for any zCr,zCr, we must have that s>0.s>0. We claim that the disk Bs/2(f(c))Bs/2(f(c)) belongs to the range f(U)f(U) of f.f. This will show that the point f(c)f(c) belongs to the ihnterior of the set f(U),f(U), and that will finish the proof.

By way of contradiction, suppose Bs/2(f(c)Bs/2(f(c) is not contained in f(U),f(U),, and let wBs/2(f(c))wBs/2(f(c)) be a complex number that is not in f(U).f(U). We have that |w-f(c)|<s/2,|w-f(c)|<s/2, which implies that |w-f(z)|>s/2|w-f(z)|>s/2 for all zCr.zCr. Consider the function gg defined on the closed disk B¯r(c)B¯r(c) by g(z)=1/(w-f(z)).g(z)=1/(w-f(z)). Then gg is continuous on the closed disk B¯r(c)B¯r(c) and differentiable on Br(c).Br(c). Moreover, gg is not a constant function, for if it were, ff would also be a constant function on Br(c)Br(c) and therefore, by the Identity Theorem, constant on all of U,U, whichg is not the case by hypothesis. Hence, by the Maximum Modulus Principle, the maximum value of |g||g| only occurs on the boundary CrCr of this disk. That is, there exists a point z'Crz'Cr such that |g(z)|<|g(z')||g(z)|<|g(z')| for all zBr(c).zBr(c). But then

2 s = 1 s / 2 < 1 | w - f ( c ) | < 1 | w - f ( z ' ) | 1 s , 2 s = 1 s / 2 < 1 | w - f ( c ) | < 1 | w - f ( z ' ) | 1 s ,
(1)

which gives the desired contradiction. Therefore, the entire disk Bs/2(f(c))Bs/2(f(c)) belongs to f(U),f(U), and hence the point f(c)f(c) belongs to the interior of the set f(U).f(U). Since this holds for any point cU,cU, it follows that f(U)f(U) is open, as desired.

Now we can give the version of the Inverse Function Theorem for complex variables.

Theorem 2

Let SS be a piecewise smooth geometric set, and suppose f:SCf:SC is continuously differentiable at a point c=a+bi,c=a+bi, and assume that f'(c)0.f'(c)0. Then:

  1. There exists an r>0,r>0, such that B¯r(c)S,B¯r(c)S, for which ff is one-to-one on B¯r(c).B¯r(c).
  2.  f(c)f(c) belongs to the interior of f(S).f(S).
  3. If gg denotes the restriction of the function ff to Br(c),Br(c), then gg is one-to-one, g-1g-1 is differentiable at the point f(c),f(c), and g-1'(f(c)=1/f'(c).g-1'(f(c)=1/f'(c).

Proof

Arguing by contradiction, suppose that ff is not one-to-one on any disk B¯r(c).B¯r(c). Then, for each natural number n,n, there must exist two points zn=xn+iynzn=xn+iyn and zn'=xn'+iyn'zn'=xn'+iyn' such that |zn-c|<1/n,|zn-c|<1/n,|zn'-c|<1/n,|zn'-c|<1/n, and f(zn)=f(zn').f(zn)=f(zn'). If we write f=u+iv,f=u+iv, then we would have that u(xn,yn)-u(xn',yn')=0u(xn,yn)-u(xn',yn')=0 for all n.n. So, by part (c) of (Reference), there must exist for each nn a point (x^n,y^n),(x^n,y^n), such that (x^n,y^n)(x^n,y^n) is on the line segment joining znzn and zn',zn', and for which

0 = u ( x n , y n ) - u ( x n ' , y n ' ) = t i a l u t i a l x ( x ^ n , y ^ n ) ( x n - x n ' ) + t i a l u t i a l y ( x ^ n , y ^ n ) ( y n - y n ' ) . 0 = u ( x n , y n ) - u ( x n ' , y n ' ) = t i a l u t i a l x ( x ^ n , y ^ n ) ( x n - x n ' ) + t i a l u t i a l y ( x ^ n , y ^ n ) ( y n - y n ' ) .
(2)

Similarly, applying the same kind of reasoning to v,v, there must exist points (x˜n,y˜n)(x˜n,y˜n) on the segment joining znzn to zn'zn' such that

0 = t i a l v t i a l x ( x ˜ n , y ˜ n ) ( x n - x n ' ) + t i a l v t i a l y ( x ˜ n , y ˜ n ) ( y n - y n ' ) . 0 = t i a l v t i a l x ( x ˜ n , y ˜ n ) ( x n - x n ' ) + t i a l v t i a l y ( x ˜ n , y ˜ n ) ( y n - y n ' ) .
(3)

If we define vectors UnUn and VnVn by

U n = ( t i a l u t i a l x ( x ^ n , y ^ n ) , t i a l u t i a l y ( x ^ n , y ^ n ) ) U n = ( t i a l u t i a l x ( x ^ n , y ^ n ) , t i a l u t i a l y ( x ^ n , y ^ n ) )
(4)

and

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

then we have that both UnUn and VnVn are perpendicular to the nonzero vector ((xn-xn'),(yn-yn')).((xn-xn'),(yn-yn')). Therefore, UnUn and VnVn are linearly dependent, whence

det ( ( tial u tial x ( ( x ^ n , y ^ n ) tial u tial y ( x ^ n , y ^ n ) tial v tial x ( x ˜ n , y ˜ n ) tial v tial y ( x ˜ n , y ˜ n ) ) ) = 0 . det ( ( tial u tial x ( ( x ^ n , y ^ n ) tial u tial y ( x ^ n , y ^ n ) tial v tial x ( x ˜ n , y ˜ n ) tial v tial y ( x ˜ n , y ˜ n ) ) ) = 0 .
(6)

Now, since both {x^n+iy^n}{x^n+iy^n} and {x˜n+iy˜n}{x˜n+iy˜n} converge to the point c=a+ib,c=a+ib, and the partial derivatives of uu and vv are continuous at c,c, we deduce that

det ( ( tial u tial x ( a , b ) tial u tial y ( a , b ) tial v tial x ( a , b ) tial v tial y ( a , b ) ) ) = 0 . det ( ( tial u tial x ( a , b ) tial u tial y ( a , b ) tial v tial x ( a , b ) tial v tial y ( a , b ) ) ) = 0 .
(7)

Now, from (Reference), this would imply that f'(c)=0,f'(c)=0, and this is a contradiction. Hence, there must exist an r>0r>0 for which ff is one-to-one on B¯r(c),B¯r(c), and this proves part (1).

Because ff is one-to-one on Br(c),Br(c),ff is obviously not a constant function. So, by the Open Mapping Theorem, the point f(c)f(c) belongs to the interior of the range of f,f, and this proves part (2).

Now write gg for the restriction of ff to the disk Br(c).Br(c). Then gg is one-to-one. According to part (2) of (Reference), we can prove that g-1g-1 is differentiable at f(c)f(c) by showing that

lim z f ( c ) g - 1 ( z ) - g - 1 ( f ( c ) ) z - f ( c ) = 1 f ' ( c ) . lim z f ( c ) g - 1 ( z ) - g - 1 ( f ( c ) ) z - f ( c ) = 1 f ' ( c ) .
(8)

That is, we need to show that, given an ϵ>0,ϵ>0, there exists a δ>0δ>0 such that if 0<|z-f(c)|<δ0<|z-f(c)|<δ then

| g - 1 ( z ) - g - 1 ( f ( c ) ) z - f ( c ) - 1 f ' ( c ) | < ϵ . | g - 1 ( z ) - g - 1 ( f ( c ) ) z - f ( c ) - 1 f ' ( c ) | < ϵ .
(9)

First of all, because the function 1/w1/w is continuous at the point f'(c),f'(c), there exists an ϵ'>0ϵ'>0 such that if |w-f'(c)|<ϵ',|w-f'(c)|<ϵ', then

| 1 w - 1 f ' ( c ) | < ϵ . | 1 w - 1 f ' ( c ) | < ϵ .
(10)

Next, because ff is differentiable at c,c, there exists a δ'>0δ'>0 such that if 0<|y-c|<δ'0<|y-c|<δ' then

| f ( y ) - f ( c ) y - c - f ' ( c ) | < ϵ ' . | f ( y ) - f ( c ) y - c - f ' ( c ) | < ϵ ' .
(11)

Now, by (Reference), g-1g-1 is continuous at the point f(c),f(c), and therefore there exists a δ>0δ>0 such that if |z-f(c)|<δ|z-f(c)|<δ then

| g - 1 ( z ) - g - 1 ( f ( c ) | < δ ' . | g - 1 ( z ) - g - 1 ( f ( c ) | < δ ' .
(12)

So, if |z-f(c)|<δ,|z-f(c)|<δ, then

| g - 1 ( z ) - c | = | g - 1 ( z ) - g - 1 ( f ( c ) ) | < δ ' . | g - 1 ( z ) - c | = | g - 1 ( z ) - g - 1 ( f ( c ) ) | < δ ' .
(13)

But then,

| f ( g - 1 ( z ) ) - f ( c ) g - 1 ( z ) - c - f ' ( c ) | < ϵ ' , | f ( g - 1 ( z ) ) - f ( c ) g - 1 ( z ) - c - f ' ( c ) | < ϵ ' ,
(14)

from which it follows that

| g - 1 ( z ) - g - 1 ( f ( c ) ) z - f ( c ) - 1 f ' ( c ) | < ϵ , | g - 1 ( z ) - g - 1 ( f ( c ) ) z - f ( c ) - 1 f ' ( c ) | < ϵ ,
(15)

as desired.

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