7.5 The open mapping theorem and the inverse function theorem  (Page 2/2)

By way of contradiction, suppose $B_{s/2}(f\left(c\right)$ is not contained in $f\left(U\right),$ , and let $w\in B_{s/2}\left(f\left(c\right)\right)$ be a complex number that is not in $f\left(U\right).$ We have that $|w-f(c\left)\right|<s/2,$ which implies that $|w-f(z\left)\right|>s/2$ for all $z\in C_r.$ Consider the function $g$ defined on the closed disk ${\overline{B}}_r\left(c\right)$ by $g\left(z\right)=1/(w-f(z\left)\right).$ Then $g$ is continuous on the closed disk ${\overline{B}}_r\left(c\right)$ and differentiable on $B_r\left(c\right).$ Moreover, $g$ is not a constant function, for if it were, $f$ would also be a constant function on $B_r\left(c\right)$ and therefore, by the Identity Theorem, constant on all of $U,$ whichg is not the case by hypothesis. Hence, by the Maximum Modulus Principle,the maximum value of $\left|g\right|$ only occurs on the boundary $C_r$ of this disk. That is, there exists a point $z^{\text{'}}\in C_r$ such that $\left|g\left(z\right)\right|<|g\left(z^{\text{'}}\right)|$ for all $z\in B_r\left(c\right).$ But then

which gives the desired contradiction. Therefore, the entire disk $B_{s/2}\left(f\left(c\right)\right)$ belongs to $f\left(U\right),$ and hence the point $f\left(c\right)$ belongs to the interior of the set $f\left(U\right).$ Since this holds for any point $c\in U,$ it follows that $f\left(U\right)$ is open, as desired.

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

Let $S$ be a piecewise smooth geometric set, and suppose $f:S\to C$ is continuously differentiable at a point $c=a+bi,$ and assume that $f^{\text{'}}\left(c\right)\ne 0.$ Then:

1. There exists an $r>0,$ such that ${\overline{B}}_r\left(c\right)\subseteq S,$ for which $f$ is one-to-one on ${\overline{B}}_r\left(c\right).$
2.   $f\left(c\right)$ belongs to the interior of $f\left(S\right).$
3. If $g$ denotes the restriction of the function $f$ to $B_r\left(c\right),$ then $g$ is one-to-one, $g^{-1}$ is differentiable at the point $f\left(c\right),$ and ${g^{-1}}^{\text{'}}(f\left(c\right)=1/f^{\text{'}}\left(c\right).$

Arguing by contradiction, suppose that $f$ is not one-to-one on any disk ${\overline{B}}_r\left(c\right).$ Then, for each natural number $n,$ there must exist two points $z_n=x_n+i{y}_n$ and $z_n^{\text{'}}=x_n^{\text{'}}+i{y}_n^{\text{'}}$ such that $|z_n-c|<1/n,$ $|z_n^{\text{'}}-c|<1/n,$ and $f\left(z_n\right)=f\left(z_n^{\text{'}}\right).$ If we write $f=u+iv,$ then we would have that $u(x_n,y_n)-u(x_n^{\text{'}},y_n^{\text{'}})=0$ for all $n.$ So, by part (c) of [link] , there must exist for each $n$ a point $({\widehat{x}}_n,{\widehat{y}}_n),$ such that $({\widehat{x}}_n,{\widehat{y}}_n)$ is on the line segment joining $z_n$ and $z_n^{\text{'}},$ and for which

Similarly, applying the same kind of reasoning to $v,$ there must exist points $({\tilde{x}}_n,{\tilde{y}}_n)$ on the segment joining $z_n$ to $z_n^{\text{'}}$ such that

If we define vectors ${\overrightarrow{U}}_n$ and ${\overrightarrow{V}}_n$ by

and

then we have that both ${\overrightarrow{U}}_n$ and ${\overrightarrow{V}}_n$ are perpendicular to the nonzero vector $((x_n-x_n^{\text{'}}),(y_n-y_n^{\text{'}})).$ Therefore, ${\overrightarrow{U}}_n$ and ${\overrightarrow{V}}_n$ are linearly dependent, whence

Now, since both $\{{\widehat{x}}_n+i{\widehat{y}}_n\}$ and $\{{\tilde{x}}_n+i{\tilde{y}}_n\}$ converge to the point $c=a+ib,$ and the partial derivatives of $u$ and $v$ are continuous at $c,$ we deduce that

Now, from [link] , this would imply that $f^{\text{'}}\left(c\right)=0,$ and this is a contradiction. Hence, there must exist an $r>0$ for which $f$ is one-to-one on ${\overline{B}}_r\left(c\right),$ and this proves part (1).

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

Now write $g$ for the restriction of $f$ to the disk $B_r\left(c\right).$ Then $g$ is one-to-one. According to part (2) of [link] , we can prove that $g^{-1}$ is differentiable at $f\left(c\right)$ by showing that

That is, we need to show that, given an $ϵ>0,$ there exists a $\delta >0$ such that if $0<|z-f(c\left)\right|<\delta$ then

First of all, because the function $1/w$ is continuous at the point $f^{\text{'}}\left(c\right),$ there exists an ${ϵ}^{\text{'}}>0$ such that if $|w-f^{\text{'}}\left(c\right)|<{ϵ}^{\text{'}},$ then

Next, because $f$ is differentiable at $c,$ there exists a ${\delta }^{\text{'}}>0$ such that if $0<|y-c|<{\delta }^{\text{'}}$ then

Now, by [link] , $g^{-1}$ is continuous at the point $f\left(c\right),$ and therefore there exists a $\delta >0$ such that if $|z-f(c\left)\right|<\delta$ then

So, if $|z-f(c\left)\right|<\delta ,$ then

But then,

from which it follows that

as desired.