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Math 382-0: Kursobjekt

Aaron Greicius

Section 1.26 Open mapping theorem

Definition 1.26.1. Open mapping.

Let \(f\) be a complex function with open domain \(D\text{.}\) We say \(f\) is open if for all open sets \(U\subseteq D\text{,}\) \(f(U)\) is open: i.e., \(f\) maps open sets to open sets.

Proof.

We pick \(r>0\) such
  1. \(\displaystyle \overline{B}_r(z_0)\subseteq U\)
  2. \(f(z)\ne f(z_0)\) for all \(z\ne z_0\text{,}\)
  3. \(f'(z)\ne 0\) for all \(z\ne z_0\text{.}\)
Why can we do this? We will show that there is such an \(r\) for each condition (a)-(c) separately; taking the minimum of these gives us an \(r\) that satisfies all conditions at once. It is clear from the openness of \(U\) that we can find an \(r\) satisfying (a). If there were no such \(r\) for which (b) is satisfied, then for each \(r_n=1/n\text{,}\) we could find a \(z_n\in B_{1/n}(z_0)\) such that \(f(z_n)=f(z_0)=w_0\text{.}\) Since and \(D\) is connected and \(z_0\) is a limit point of the set of \(z_n\text{,}\) Theorem 1.19.8 would imply that \(f(z)=f(z_0)=w_0\) for all \(z\in D\text{,}\) contradicting the fact the \(f\) is nonconstant. Similarly, if there were no such \(r\) for which (c) is satisfied we could find a sequence of points \(z_n\to z_0\) satisfying \(f'(z_n)=0\text{,}\) and this would imply \(f'(z)=0\) for all \(z\in D\text{,}\) contradicting again that \(f\) is nonconstant.
Having chosen an \(r> 0\) satisfying (a)-(c), we let \(\gamma(t)=z_0+e^{it}\text{,}\) \(t\in [0,2\pi]\text{.}\) Since \(f(z)\ne w_0\) for all \(z\in B_r(z_0)\text{,}\) we have \(w_0\notin f\circ \gamma\text{.}\) It follows by the extreme value theorem that we can find an \(s> 0\) such that \(B_s(w_0)\cap \im f\circ \gamma=\emptyset\text{.}\) Since \(w\notin \im f\circ \gamma\text{,}\) the winding number \(\chi_{f\circ\gamma}(w)\) defines a continuous, integer-valued function on \(B_s(w_0)\text{.}\) Since \(B_s(w_0)\) is connected, this winding number \(\chi_{f\circ\gamma}(w)\) is constant for all \(w\in B_s(w_0)\text{.}\) Since \(\chi_{f\circ\gamma}(w_0)=\ord_{f-w_0}(z_0)=m\text{,}\) we see that \(\chi_{f\circ \gamma}(w)=m\) for all \(w\in B_s(w_0)\text{.}\) On the other hand, we have
\begin{align*} m=\chi_{f\circ\gamma}(w)\amp =\frac{1}{2\pi i}\int_{f\circ \gamma}\frac{1}{z-w}\, dz\\ \amp = \int_{0}^{2\pi}\frac{f'(\gamma(t))\gamma'(t)}{f(\gamma(t))-w}\, dt\\ \amp =\frac{1}{2\pi i}\int_\gamma \frac{(f(z)-w)'}{f(z)-w}\, dz\text{,} \end{align*}
and by the Theorem 1.24.10, the last integral is equal to the number of zeros (counting multiplicity) of the function \(f(z)-w\) in \(\Int \gamma=B_r(z_0)\text{.}\) Since \(f'(z)\ne 0\) for all \(z\ne z_0\text{,}\) it follows easily that for each \(w\ne w_0\in B_s(w_0)\) and \(z\in B_r(z_0)\) satisfying \(f(z)=w\text{,}\) this \(z\) is a simple root of the function \(f-w\text{:}\) i.e., \(\ord_{f-w}(z)=1\text{.}\) It follows that for each \(w\ne w_0\in B_s(w)\) there are exactly \(m\) distinct elements \(z\in B_r(z_0)\) satisfying \(f(z)=w\text{.}\) In particular, note that for each \(w\in B_s(z_0)\) there is some element \(z\in B_r(z_0)\) such that \(f(z)=w\text{.}\) Thus
\begin{equation*} B_s(w_0)\subseteq f(B_r(z_0))\text{,} \end{equation*}
as claimed.

Proof.

Given any open \(U\subseteq D\) and any \(w_0=f(z_0)\in f(U)\text{,}\) according to Theorem 1.26.2 there are open balls \(B_r(z_0)\subseteq U\) and \(B_s(w_0)\) such that
\begin{equation*} B_s(f(z_0))\subseteq f(B_r(z_0))\subseteq f(U)\text{.} \end{equation*}
Thus for all \(w_0\in f(U)\) we can find an open ball \(B_s(w_0)\subseteq f(U)\text{.}\) We conclude that \(f(U)\) is open.
The next result, the famous inverse function theorem is largely the result of applying Theorem 1.26.2 and Theorem 1.26.3 to the case where \(f'(z_0)\ne 0\) (i.e., \(m=1\)).

Proof.

  1. Since \(f'(z_0)\ne 0\text{,}\) \(z_0\) is a simple zero of the function \(h(z)=f(z)-f(z_0)\text{.}\) We are thus in the \(m=1\) situation of Theorem 1.26.2. Let \(B_r(z_0)\) and \(B_s(f(z_0))\) be open balls satisfying the conditions of that theorem. The set
    \begin{equation*} V=f^{-1}(B_s(f(z_0)))\cap B_r(z_0) \end{equation*}
    is an open set containing \(z_0\) on which \(f\) is injective (1-to-1) and \(f'\) is nonzero. The restricted function \(f\colon V\rightarrow f(V)\) is thus bijective.
  2. We first show that \(f^{-1}\colon f(V)\rightarrow V\) is continuous. This will follow from the fact \(f\) is an open map. For notational clarity, we let \(g=f^{-1}\text{.}\) Using the defining property of the inverse function, \(g\) satisfies
    \begin{align*} g(f(z)) \amp = z, \text{ for all } z\in V \amp f(g(w))\amp =w \text{ for all } w\in f(V)\text{.} \end{align*}
    To show \(g\) is continuous we will prove the following equivalent property (shown in one of your homework assignments): if \(W\subseteq V\) is open, then \(g^{-1}(W)\subseteq f(V)\) is open. But this is easy: given any open \(W\subseteq V\) it is easy to see that
    \begin{equation*} g^{-1}(W)=(f^{-1})^{-1}(W)=f(W)\text{,} \end{equation*}
    which is open since \(f\) is an open map. This proves \(g=f^{-1}\colon f(V)\rightarrow V\) is continuous.
    It remains only to prove that \(f^{-1}\) is differentiable at all \(w\in f(V)\text{:}\) it turns out this is not so difficult to prove once we know \(f^{-1}\) is continuous. Although it is straightforward to prove the formula (1.97) using the limit definition of the derivative, we will instead use Caratheory’s equivalent notion of differentiability (mentioned in homework) as an illustration of this technique’s pleasing algebraic nature. Given \(w\in f(V)\text{,}\) we have \(w=f(z)\) for some \(z\in V\text{.}\) Since \(f\) is differentiable at \(z\) there is a function \(\phi\) satisfying the following conditions:
    1. \(f(z')=\phi(z')(z'-z)+f(z)\) for all \(z'\in V\text{,}\)
    2. \(\phi\) is continuous at \(z\) and satisfies \(\phi(z)=f'(z)\text{.}\)
    Note that since \(f'\) is nonzero on \(V\text{,}\) we have \(\phi(z)\ne 0\text{.}\) Since \(\phi\) is continuous at \(z\text{,}\) we can find an open ball \(B\) around \(z\) such that \(\phi(z')\ne 0\) for all \(z'\in B\text{.}\) After a little algebra we see that
    \begin{equation*} z'=\frac{1}{\phi(z')}(f(z')-f(z))+z \end{equation*}
    for all \(z'\in B\text{,}\) or equivalently,
    \begin{equation*} f^{-1}(w')=\frac{1}{\phi(f^{-1}(w'))}(w'-w)+f^{-1}(w) \end{equation*}
    for all \(w'\) in the open set \(f(B)\text{.}\) Since \(f^{-1}\) is continuous, the function \(\psi=\frac{1}{\phi\circ f^{-1}}\) is continuous at \(w\) and satisfies
    \begin{equation*} f^{-1}(w')=\psi(w')(w'-w)+f^{-1}(w) \end{equation*}
    for all \(w'\in f(B)\text{.}\) Caratheodory’s theorem now implies that \(f^{-1}\) is differentiable at \(w\) and we have
    \begin{equation*} (f^{-1})'(w)=\psi(w)=\frac{1}{\phi(f^{-1}(w))}=\frac{1}{f'(f^{-1}(w))}\text{,} \end{equation*}
    as claimed.

Definition 1.26.5. Biholomorphic function.

Let \(U\) be an open subset of \(\C\text{.}\) A complex function \(f\) is biholomorphic on \(U\) if \(f\) is analytic and injective on \(U\text{,}\) and its inverse function \(f^{-1}\colon f(U)\rightarrow U\) is holomorphic.

Remark 1.26.7. Integral formula.

With the notation of Theorem 1.26.4, given any closed ball \(\overline{B}\subseteq V\) and \(w_0\in f(B)\text{,}\) we have the integral formula
\begin{equation} f^{-1}(w_0) =\frac{1}{2\pi i}\int_{\gamma}\frac{zf'(z)}{f(z)-w_0}\, dz\text{,}\tag{1.98} \end{equation}
where \(\gamma\) is a simple counterclockwise parametrization of \(\partial B\text{.}\) We do not include this as an official result (with proof) as we will not make use of it in the course and not wish to overburden the reader with formulae.

Proof.

Since \(f\) is analytic and nonconstant, it is an open map. Each proof below will make use of this fact.
  1. Given any \(z_0\in U\) and open ball \(B_r(z_0)\subseteq U\text{,}\) since \(f(B_r(z_0))\) is open, there is an open ball \(B_s(f(z_0))\subseteq f(B_r(z_0))\text{.}\) Clearly \(B_s(f(z_0))\) will contain an element \(w\) with \(\abs{w}> \abs{f(z_0)}\text{.}\) Since \(B_s(f(z_0))\subseteq f(B_r(z_0))\text{,}\) we have \(w=f(z)\) for some \(z\in B_r(z_0)\text{.}\) Thus for all open balls \(B_r(z_0)\) we can find a \(z\in B_r(z_0)\) such that \(\abs{f(z)}> f(z_0)\text{.}\) This proves \(\abs{f(z_0)}\) is not a local maximum value on \(U\text{,}\) as desired.
  2. Assume by contradiction that \(\abs{f(z_0)}=c > 0\) is a local minimum value of \(f\) on \(U\text{.}\) We could then find an open ball \(B_r(z_0)\) such that \(\abs{f(z)}\geq c> 0\) for all \(z\in B_r(z_0)\text{.}\) But then the function \(g=1/f\) is a nonconstant analytic function on \(B_r(z_0)\) and \(\abs{g(z_0)}\geq \abs{g(z)}\) for all \(z\in B_r(z_0)\text{.}\) In other words, \(\abs{g(z_0)}\) is a maximum value on the open ball \(B_r(z_0)\text{.}\) This contradicts the result in (1).
  3. By the multivariable extreme value theorem, \(\abs{f}\) attains both a maximum and minimum value on \(K\text{:}\) that is, there are points \(z_1, z_2\in K\) such that
    \begin{equation*} \abs{f(z_1)}\leq \abs{f(z)}\leq \abs{f(z_2)} \end{equation*}
    for all \(z\in K\text{.}\)
    We can write \(K=K^\circ\cup \partial K\text{,}\) where \(K^\circ\text{,}\) the interior of \(K\text{,}\) is an open set. By (1), we cannot have \(z_2\in K^\circ\text{,}\) since then \(\abs{f}\) would attain a maximum value on this open set. Thus \(z_2\in \partial K\text{,}\) showing that the maximum value of \(\abs{f}\) is attained only on the boundary.
    If the minimum value \(\abs{f(z_1)}\) is nonzero, then (2) implies that \(z_1\notin K^\circ\text{.}\) Thus \(z_1\in \partial K\text{,}\) and we see in this case that the minimum value of \(\abs{f}\) is attained only on \(\partial K\text{.}\)

Proof.

Since \(f(0)=0\text{,}\) we may write \(f(z)=zg(z)\) where \(g(z)\) is analytic on \(B_1(0)\text{.}\) Since \(\abs{f(z)}\leq 1\) for all \(z\in B_1(0)\text{,}\) we have
\begin{equation*} \abs{g(z)}\leq \frac{1}{\abs{z}} \end{equation*}
for all \(z\ne 0\text{.}\) Given any \(r\in (0,1)\text{,}\) since \(g\) is analytic, on \(K=\overline{B}_r(0)\) the function \(\abs{g}\) attains its maximum value on the boundary \(\partial K\text{.}\) Since \(\abs{g(z)}\leq \frac{1}{r}\) on \(\partial K\text{,}\) we conclude that \(\abs{g(z)}\leq \frac{1}{r}\) for all \(z\in \overline{B}_r(0)\text{.}\) Since this is true for all \(r\in (0,1)\text{,}\) we conclude that \(\abs{g(z)}\leq 1\) for all \(z\in B_1(0)\text{.}\) Since \(f(z)=zg(z)\text{,}\) it follows that \(\abs{f(z)}\leq \abs{z}\) for all \(z\in B_1(0)\text{.}\) Lastly, since \(f'(z)=g(0)\text{,}\) we see that \(\abs{f'(z)}\leq 1\text{.}\)
Lastly, if there is an element \(z\in B_1(0)\) satisfying \(\abs{f(z)}=\abs{z}\text{,}\) or if \(\abs{f'(0)}=\abs{g(0)}\text{,}\) then \(\abs{g}\) attains an absolute maximum value of 1 on \(B_1(0)\text{.}\) By Theorem 1.26.8 we must have \(g\) a constant function of modulus 1. It follows that \(g(z)=e^{i\alpha}\) for some \(\alpha\in \R\text{,}\) and thus \(f(z)=e^{i\alpha}z\text{,}\) as claimed.
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