Aufgabe (Kettenregel, 5 Punkte)

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

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

und

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

gelten.

Lösung

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

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

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

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

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

gilt. Damit ist

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

Setze nun

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

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

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

Setzen wir ein, ergibt das

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

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

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

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

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

konjugieren, erhalten wir

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

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

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

Setzen wir oben fort, folgt damit

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

wie behauptet. Analog folgt

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