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

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By way of contradiction, suppose ${B}_{s/2}\left(f\left(c\right)$ 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\left(c\right)| which implies that $|w-f\left(z\right)|>s/2$ for all $z\in {C}_{r}.$ Consider the function $g$ defined on the closed disk ${\overline{B}}_{r}\left(c\right)$ by $g\left(z\right)=1/\left(w-f\left(z\right)\right).$ 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 $|g|$ only occurs on the boundary ${C}_{r}$ of this disk. That is, there exists a point ${z}^{\text{'}}\in {C}_{r}$ such that $|g\left(z\right)|<|g\left({z}^{\text{'}}\right)|$ for all $z\in {B}_{r}\left(c\right).$ But then

$\frac{2}{s}=\frac{1}{s/2}<\frac{1}{|w-f\left(c\right)|}<\frac{1}{|w-f\left({z}^{\text{'}}\right)|}\le \frac{1}{s},$

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 ${{g}^{-1}}^{\text{'}}\left(f\left(c\right)=1/{f}^{\text{'}}\left(c\right).$

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 $u\left({x}_{n},{y}_{n}\right)-u\left({x}_{n}^{\text{'}},{y}_{n}^{\text{'}}\right)=0$ for all $n.$ So, by part (c) of [link] , there must exist for each $n$ a point $\left({\stackrel{^}{x}}_{n},{\stackrel{^}{y}}_{n}\right),$ such that $\left({\stackrel{^}{x}}_{n},{\stackrel{^}{y}}_{n}\right)$ is on the line segment joining ${z}_{n}$ and ${z}_{n}^{\text{'}},$ and for which

$0=u\left({x}_{n},{y}_{n}\right)-u\left({x}_{n}^{\text{'}},{y}_{n}^{\text{'}}\right)=\frac{tialu}{tialx}\left({\stackrel{^}{x}}_{n},{\stackrel{^}{y}}_{n}\right)\left({x}_{n}-{x}_{n}^{\text{'}}\right)+\frac{tialu}{tialy}\left({\stackrel{^}{x}}_{n},{\stackrel{^}{y}}_{n}\right)\left({y}_{n}-{y}_{n}^{\text{'}}\right).$

Similarly, applying the same kind of reasoning to $v,$ there must exist points $\left({\stackrel{˜}{x}}_{n},{\stackrel{˜}{y}}_{n}\right)$ on the segment joining ${z}_{n}$ to ${z}_{n}^{\text{'}}$ such that

$0=\frac{tialv}{tialx}\left({\stackrel{˜}{x}}_{n},{\stackrel{˜}{y}}_{n}\right)\left({x}_{n}-{x}_{n}^{\text{'}}\right)+\frac{tialv}{tialy}\left({\stackrel{˜}{x}}_{n},{\stackrel{˜}{y}}_{n}\right)\left({y}_{n}-{y}_{n}^{\text{'}}\right).$

If we define vectors ${\stackrel{\to }{U}}_{n}$ and ${\stackrel{\to }{V}}_{n}$ by

${\stackrel{\to }{U}}_{n}=\left(\frac{tialu}{tialx}\left({\stackrel{^}{x}}_{n},{\stackrel{^}{y}}_{n}\right),\frac{tialu}{tialy}\left({\stackrel{^}{x}}_{n},{\stackrel{^}{y}}_{n}\right)\right)$

and

${\stackrel{\to }{V}}_{n}=\left(\frac{tialv}{tialx}\left({\stackrel{˜}{x}}_{n},{\stackrel{˜}{y}}_{n}\right),\frac{tialv}{tialy}\left({\stackrel{˜}{x}}_{n},{\stackrel{˜}{y}}_{n}\right)\right),$

then we have that both ${\stackrel{\to }{U}}_{n}$ and ${\stackrel{\to }{V}}_{n}$ are perpendicular to the nonzero vector $\left(\left({x}_{n}-{x}_{n}^{\text{'}}\right),\left({y}_{n}-{y}_{n}^{\text{'}}\right)\right).$ Therefore, ${\stackrel{\to }{U}}_{n}$ and ${\stackrel{\to }{V}}_{n}$ are linearly dependent, whence

$det\left(\left(\frac{\mathrm{tial}\mathrm{u}}{\mathrm{tial}\mathrm{x}}\left(\left({\stackrel{^}{\mathrm{x}}}_{\mathrm{n}},{\stackrel{^}{\mathrm{y}}}_{\mathrm{n}}\right)\frac{\mathrm{tial}\mathrm{u}}{\mathrm{tial}\mathrm{y}}\left({\stackrel{^}{\mathrm{x}}}_{\mathrm{n}},{\stackrel{^}{\mathrm{y}}}_{\mathrm{n}}\right)\frac{\mathrm{tial}\mathrm{v}}{\mathrm{tial}\mathrm{x}}\left({\stackrel{˜}{\mathrm{x}}}_{\mathrm{n}},{\stackrel{˜}{\mathrm{y}}}_{\mathrm{n}}\right)\frac{\mathrm{tial}\mathrm{v}}{\mathrm{tial}\mathrm{y}}\left({\stackrel{˜}{\mathrm{x}}}_{\mathrm{n}},{\stackrel{˜}{\mathrm{y}}}_{\mathrm{n}}\right)\right)\right)=0.$

Now, since both $\left\{{\stackrel{^}{x}}_{n}+i{\stackrel{^}{y}}_{n}\right\}$ and $\left\{{\stackrel{˜}{x}}_{n}+i{\stackrel{˜}{y}}_{n}\right\}$ converge to the point $c=a+ib,$ and the partial derivatives of $u$ and $v$ are continuous at $c,$ we deduce that

$det\left(\left(\frac{\mathrm{tial}\mathrm{u}}{\mathrm{tial}\mathrm{x}}\left(\mathrm{a},\mathrm{b}\right)\frac{\mathrm{tial}\mathrm{u}}{\mathrm{tial}\mathrm{y}}\left(\mathrm{a},\mathrm{b}\right)\frac{\mathrm{tial}\mathrm{v}}{\mathrm{tial}\mathrm{x}}\left(\mathrm{a},\mathrm{b}\right)\frac{\mathrm{tial}\mathrm{v}}{\mathrm{tial}\mathrm{y}}\left(\mathrm{a},\mathrm{b}\right)\right)\right)=0.$

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

$\underset{z\to f\left(c\right)}{lim}\frac{{g}^{-1}\left(z\right)-{g}^{-1}\left(f\left(c\right)\right)}{z-f\left(c\right)}=\frac{1}{{f}^{\text{'}}\left(c\right)}.$

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

$|\frac{{g}^{-1}\left(z\right)-{g}^{-1}\left(f\left(c\right)\right)}{z-f\left(c\right)}-\frac{1}{{f}^{\text{'}}\left(c\right)}|<ϵ.$

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

$|\frac{1}{w}-\frac{1}{{f}^{\text{'}}\left(c\right)}|<ϵ.$

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

$|\frac{f\left(y\right)-f\left(c\right)}{y-c}-{f}^{\text{'}}\left(c\right)|<{ϵ}^{\text{'}}.$

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

$|{g}^{-1}\left(z\right)-{g}^{-1}\left(f\left(c\right)|<{\delta }^{\text{'}}.$

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

$|{g}^{-1}\left(z\right)-c|=|{g}^{-1}\left(z\right)-{g}^{-1}\left(f\left(c\right)\right)|<{\delta }^{\text{'}}.$

But then,

$|\frac{f\left({g}^{-1}\left(z\right)\right)-f\left(c\right)}{{g}^{-1}\left(z\right)-c}-{f}^{\text{'}}\left(c\right)|<{ϵ}^{\text{'}},$

from which it follows that

$|\frac{{g}^{-1}\left(z\right)-{g}^{-1}\left(f\left(c\right)\right)}{z-f\left(c\right)}-\frac{1}{{f}^{\text{'}}\left(c\right)}|<ϵ,$

as desired.

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