Kurs:Funktionentheorie/Übungen/2. Zettel/Aufgabe 4

Aufgabe (Kettenregel, 5 Punkte)

Seien $f,g\colon \mathbf {C} \to \mathbf {C}$ stetig differenzierbar. Beweise, dass

${\frac {\partial }{\partial z}}(f\circ g)={\frac {\partial f}{\partial z}}\circ g\cdot {\frac {\partial g}{\partial z}}+{\frac {\partial f}{\partial {\bar {z}}}}\circ g\cdot {\frac {\partial {\bar {g}}}{\partial z}}$

und

${\frac {\partial }{\partial {\bar {z}}}}(f\circ g)={\frac {\partial f}{\partial z}}\circ g\cdot {\frac {\partial g}{\partial {\bar {z}}}}+{\frac {\partial f}{\partial {\bar {z}}}}\circ g\cdot {\frac {\partial {\bar {g}}}{\partial {\bar {z}}}}$

gelten.

Lösung

Wir erinnern uns (Fischer/Lieb, Seite 21 unten): Für eine differenzierbare Funktion $f\colon U\to \mathbf {C}$ sind die partiellen Ableitungen nach $z$ und ${\bar {z}}$ wie folgt charakterisiert: Sind $A_{1},A_{2}\colon U\to \mathbf {C}$ stetige Funktionen mit

$f(z)=f(z_{0})+A_{1}(z)(z-z_{0})+A_{2}(z)({\bar {z}}-{\bar {z}}_{0})$

so ist $\partial _{z}f(z_{0})=A_{1}(z_{0})$ und $\partial _{\bar {z}}f(z_{0})=A_{2}(z_{0})$.

Diese Beschreibung der Wirtinger-Ableitungen wollen wir hier benutzen. Sei $z_{0}\in \mathbf {C}$. Da $g$ in $z_{0}$ differenzierbar ist, haben wir stetige Funktionen $B_{1},B_{2}$ so dass

$g(z)=g(z_{0})+B_{1}(z)(z-z_{0})+B_{2}(z)({\bar {z}}-{\bar {z}}_{0})$

gilt. Damit ist

${\begin{array}{rl}(f\circ g)(z)&=f{\bigl (}g(z){\bigr )}\\&=f{\bigl (}g(z_{0})+B_{1}(z)(z-z_{0})+B_{2}(z)({\bar {z}}-{\bar {z}}_{0}){\bigr )}\end{array}}$

Setze nun

${\begin{array}{rl}w&:=g(z_{0})+B_{1}(z)(z-z_{0})+B_{2}(z)({\bar {z}}-{\bar {z}}_{0})=g(z)\\w_{0}&:=g(z_{0})\end{array}}$

Da $f$ in $w_{0}$ differenzierbar ist, existieren stetige Funktionen $A_{1}$, $A_{2}$ so, dass

$f(w)=f(w_{0})+A_{1}(w)(w-w_{0})+A_{2}(w)({\bar {w}}-{\bar {w}}_{0})$

Setzen wir ein, ergibt das

${\begin{array}{rl}(f\circ g)(z)&=f{\bigl (}g(z){\bigr )}\\&=f{\bigl (}g(z_{0})+B_{1}(z)(z-z_{0})+B_{2}(z)({\bar {z}}-{\bar {z}}_{0}){\bigr )}\\&=f(w_{0})+A_{1}(w)(w-w_{0})+A_{2}(w)({\bar {w}}-{\bar {w}}_{0})\\&=f{\bigl (}g(z_{0}){\bigr )}+A_{1}{\bigl (}g(z){\bigr )}{\bigl (}B_{1}(z)(z-z_{0})+B_{2}(z)({\bar {z}}-{\bar {z}}_{0}){\bigr )}+A_{2}{\bigl (}g(z){\bigr )}{\bigl (}{\overline {B_{1}(z)(z-z_{0})+B_{2}(z)({\bar {z}}-{\bar {z}}_{0})}}{\bigr )}\\&=(f\circ g)(z_{0})+\underbrace {{\Bigl (}A_{1}{\bigl (}g(z){\bigr )}B_{1}(z)+A_{2}{\bigl (}g(z){\bigr )}{\overline {B_{2}(z)}}{\Bigl )}} _{C_{1}(z):=}(z-z_{0})+\underbrace {{\Bigl (}A_{1}{\bigl (}g(z){\bigr )}B_{2}(z)+A_{2}{\bigl (}g(z){\bigr )}{\overline {B_{1}(z)}}{\Bigl )}} _{C_{2}(z):=}({\bar {z}}-{\bar {z}}_{0})\end{array}}$

Da $C_{1}$ und $C_{2}$ als Komposition stetiger Funktionen stetig sind, ist $f\circ g$ partiell differenzierbar und

${\begin{array}{rl}{\frac {\partial }{\partial z}}(f\circ g)(z_{0})&=C_{1}(z_{0})\\&=A_{1}{\bigl (}g(z_{0}){\bigr )}B_{1}(z_{0})+A_{2}{\bigl (}g(z_{0}){\bigr )}{\overline {B_{2}(z_{0})}}\\&={\frac {\partial f}{\partial z}}(w_{0}){\frac {\partial g}{\partial z}}(z_{0})+{\frac {\partial f}{\partial {\bar {z}}}}(w_{0}){\overline {{\frac {\partial g}{\partial {\bar {z}}}}(z_{0})}}\end{array}}$

Den letzten Term schauen wir uns noch einmal an. Wenn wir

$g(z)=g(z_{0})+B_{1}(z)(z-z_{0})+B_{2}(z)({\bar {z}}-{\bar {z}}_{0})$

konjugieren, erhalten wir

${\bar {g}}(z)={\bar {g}}(z_{0})+{\overline {B_{2}(z)}}(z-z_{0})+{\overline {B_{1}(z)}}({\bar {z}}-{\bar {z}}_{0})$

da die Konjugation ein Körperautomorphismus ist. Wir lesen ab, dass

${\frac {\partial {\bar {g}}}{\partial z}}(z_{0})={\overline {{\frac {\partial g}{\partial {\bar {z}}}}(z_{0})}}$ und ${\frac {\partial {\bar {g}}}{\partial {\bar {z}}}}(z_{0})={\overline {{\frac {\partial g}{\partial z}}(z_{0})}}$

Setzen wir oben fort, folgt damit

${\begin{array}{rl}{\frac {\partial }{\partial z}}(f\circ g)(z_{0})&=C_{1}(z_{0})\\&=A_{1}{\bigl (}g(z_{0}){\bigr )}B_{1}(z_{0})+A_{2}{\bigl (}g(z_{0}){\bigr )}{\overline {B_{2}(z_{0})}}\\&={\frac {\partial f}{\partial z}}(w_{0}){\frac {\partial g}{\partial z}}(z_{0})+{\frac {\partial f}{\partial {\bar {z}}}}(w_{0}){\overline {{\frac {\partial g}{\partial {\bar {z}}}}(z_{0})}}\\&={\frac {\partial f}{\partial z}}(w_{0}){\frac {\partial g}{\partial z}}(z_{0})+{\frac {\partial f}{\partial {\bar {z}}}}(w_{0}){\frac {\partial {\bar {g}}}{\partial z}}(z_{0})\\&={\frac {\partial f}{\partial z}}\circ g(z_{0}){\frac {\partial g}{\partial z}}(z_{0})+{\frac {\partial f}{\partial {\bar {z}}}}\circ g(z_{0}){\frac {\partial {\bar {g}}}{\partial z}}(z_{0})\\\end{array}}$

wie behauptet. Analog folgt

${\begin{array}{rl}{\frac {\partial }{\partial {\bar {z}}}}(f\circ g)(z_{0})&=C_{2}(z_{0})\\&=A_{1}{\bigl (}g(z_{0}){\bigr )}B_{2}(z_{0})+A_{1}{\bigl (}g(z_{0}){\bigr )}{\overline {B_{1}(z_{0})}}\\&={\frac {\partial f}{\partial z}}(w_{0}){\frac {\partial g}{\partial {\bar {z}}}}(z_{0})+{\frac {\partial f}{\partial {\bar {z}}}}(w_{0}){\overline {{\frac {\partial g}{\partial z}}(z_{0})}}\\&={\frac {\partial f}{\partial z}}(w_{0}){\frac {\partial g}{\partial {\bar {z}}}}(z_{0})+{\frac {\partial f}{\partial {\bar {z}}}}(w_{0}){\frac {\partial {\bar {g}}}{\partial {\bar {z}}}}(z_{0})\\&={\frac {\partial f}{\partial z}}\circ g(z_{0}){\frac {\partial g}{\partial z}}(z_{0})+{\frac {\partial f}{\partial {\bar {z}}}}\circ g(z_{0}){\frac {\partial {\bar {g}}}{\partial {\bar {z}}}}(z_{0})\\\end{array}}$