Complex Analysis/Extremum principles, open mapping theorem, Schwarz' lemma

We continue our quest of proving general properties of holomorphic functions, this time even better equipped, since we have the theorems from last chapter.

Extremum principles

Under certain circumstances, holomorphic functions assume their maximal resp. minimal absolute value on the boundary. Before making this precise, we need a preparatory lemma.

Lemma 8.1:

Let $f:B_r(z_0)\to ℂ$ be holomorphic, where $r>0$ and $z_0\in ℂ$ are arbitrary, and assume that it even satisfies $|f(z)|=\alpha$ for a constant $\alpha >0$. Then $f$ itself is constant.

Proof:

In case $|f(z)|=0$ in $B_r(z_0)$, we may conclude $f(z)=0$ in $B_r(z_0)$ and are done. Otherwise, we proceed as follows:

If $|f(z)|$ is constant, so is $|f(z){|}^2$. We write $f=f_1+i{f}_2$. Then ${\alpha }^2=|f(z){|}^2=f_1(z{)}^2+f_2(z{)}^2$ for all $z\in B_r(z_0)$. Thus, taking partial derivatives, we get

${\partial }_x|f(z){|}^2=f_1(z){\partial }_x{f}_1(z)+f_2(z){\partial }_x{f}_2(z)=0$ and ${\partial }_y|f(z){|}^2=f_1(z){\partial }_y{f}_1(z)+f_2(z){\partial }_y{f}_2(z)=0$.

From the Cauchy–Riemann equations we may further infer

$f_1(z){\partial }_x{f}_1(z)-f_2(z){\partial }_y{f}_1(z)=0$ and $f_1(z){\partial }_y{f}_1(z)+f_2(z){\partial }_x{f}_1(z)=0$,

from which follow (after some algebra) that

$(f_1(z{)}^2+f_2(z{)}^2){\partial }_x{f}_1(z)=0$, $(f_1(z{)}^2+f_2(z{)}^2){\partial }_y{f}_1(z)=0$, $(f_1(z{)}^2+f_2(z{)}^2){\partial }_x{f}_2(z)=0$ and $(f_1(z{)}^2+f_2(z{)}^2){\partial }_y{f}_2(z)=0$,

that is ${\partial }_x{f}_1\equiv {\partial }_y{f}_1\equiv {\partial }_x{f}_2\equiv {\partial }_y{f}_2\equiv 0$.$◻$

Now we are ready to explicate the extremum principles in the form of the following two theorems.

Proof:

Assume $z_M\notin \partial S$, that is, $z_M\in \stackrel{\circ }{S}$. Let $ϵ>0$ be arbitrary such that $\overline{B_{ϵ}(z_m)}\subseteq \stackrel{\circ }{S}$. Then Cauchy's integral formula implies

$f(z_M)=\frac{1}{2\pi i}\underset{\partial B_{ϵ}(z_M)}{\int }\frac{f(z)}{z-z_M}dz=\frac{1}{2\pi }\underset{0}{\overset{2\pi }{\int }}f(z_M+ϵe^{i\phi })d\phi$.

If now $|f(z)|<|f(z_M)|$ for some $z\in \partial B_{ϵ}(z_M)$, then by the continuity of $f$

$|\frac{1}{2\pi }\underset{0}{\overset{2\pi }{\int }}f(z_M+ϵe^{i\phi })d\phi |\le \frac{1}{2\pi }\underset{0}{\overset{2\pi }{\int }}|f(z_M+ϵe^{i\phi })|d\phi <|f(z_M)|$,

a contradiction. Hence, $|f(z)|=|f(z_M)|$ on all of $|z|=ϵ$, and since $ϵ$ was arbitrary (provided that $\overline{B_{ϵ}(z_m)}\subseteq \stackrel{\circ }{S}$), $|f(z)|=|f(z_M)|$ in a small ball around $z_M$. From lemma 8.1, it follows that $f$ is constant there, and hence the identity theorem implies that $f$ is constant on the whole connected component containing $z_M$.$◻$

Similarly, we have:

Proof:

If $f$ does not have a zero inside $\stackrel{\circ }{S}$, the chain rule implies that the function

$z\mapsto \frac{1}{f(z)}$

is holomorphic in $\stackrel{\circ }{S}$. Hence, the maximum principle applies and either $z\mapsto \frac{1}{f(z)}$ has no maximum in the interior (and thus $f$ has no minimum in the interior) or $z\mapsto \frac{1}{f(z)}$ is constant (and hence $f$ is constant as well).$◻$

The open mapping theorem

That is, as topologists would say, $f$ is an open map.

Proof:

Let $z_0\in U$. We prove that there exists a ball around $f(z_0)$ which is contained within $f(U)$. To this end, we pick (due to the openness of $U$) a $\delta >0$ such that $\overline{B_{\delta }(z_0)}\subseteq U$ and furthermore $f(w)\ne f(z_0)$ on $\overline{B_{\delta }(z_0)}\setminus \{z_0\}$ (by the identity theorem) and set

$ϵ:=\frac{1}{3}\underset{z\in \partial B_{\delta }(z_0)}{\min }|f(z)-f(z_0)|$;

since $\partial B_{\delta }(z_0)$ is compact, $f$ assumes a minimum there and it's not equal to zero by choice of $\delta$, which is why $ϵ>0$. Now for every $w\in B_{ϵ}(f(z_0))$ we define the function

$\overline{B_{\delta }(z_0)}\to ℂ,z\mapsto f(z)-w$.

In $z_0$, the absolute value of this function is less than $ϵ$ by choice of $w$. However, for $z\in \partial B_{\delta }(z_0)$, we have

$|f(z)-w|\ge |f(z)|-|w|>ϵ$.

Hence, the minimum principle implies that the function $\overline{B_{\delta }(z_0)}\to ℂ,z\mapsto f(z)-w$ has a zero in $B_{\delta }(z_0)$, and this proves (since $w\in B_{ϵ}(f(z_0))$ was arbitrary) that $f$ assumes every value in $B_{ϵ}(f(z_0))$.$◻$

Schwarz' lemma

Proof:

First, we consider the following function:

$g(z):=\{\begin{array}{cc}\frac{f(z)}{z}& z\ne 0\\ f^{\prime }(0)& z=0\end{array}$.

Since this map is bounded, continuous and holomorphic everywhere except in $0$, it is even holomorphic in $0$ due to Riemann's theorem (the extension in $0$ must be uniquely chosen s.t. continuity is satisfied). Furthermore, we have

$|f(z)|\le 1\Rightarrow |g(z)|\le \frac{1}{|z|}$

for all $z\in B_1(0)$ by assumption; in particular, if $|z|=r$, then $|g(z)|\le 1/r$, and thus, by the maximum principle, $|g(z)|\le 1/r$ also for $|z|<r$. Taking $r\to 1$ gives $|g(z)|<1$ in $B_1(0)$, and hence $|f(z)|\le |z|$ for $z\in B_1(0)$.

For the second part, if either $|f^{\prime }(0)|=1$ or $|f(z)|=|z|$ for a $z\in B_1(0)\setminus 0$, then $|g(z)|=1$ somewhere inside $B_1(0)$, and hence, again by the maximum principle, $g$ must be constant, from which follows $f(z)=f^{\prime }(0)z$, that is, we may pick $\lambda =f^{\prime }(0)$.$◻$