Komplexe Zahlen/Reell-differenzierbare Funktionen/Antiholomorphe Ableitung/Produktregel/Fakt/Beweis

Beweis

Wir verwenden die Produktregel für die beiden partielle Ableitungen, also

${}{\frac {\partial (fg)}{\partial x}}=f{\frac {\partial g}{\partial x}}+g{\frac {\partial f}{\partial x}}\,$

und

${}{\frac {\partial (fg)}{\partial y}}=f{\frac {\partial g}{\partial y}}+g{\frac {\partial f}{\partial y}}\,,$

die in dieser Form auf Aufgabe beruht. Somit folgt die Aussage mit

${}{\begin{aligned}{\frac {\partial {\left(fg\right)}}{\partial {\overline {z}}}}&={\frac {1}{2}}{\left({\frac {\partial {\left(fg\right)}}{\partial x}}+{\mathrm {i} }{\frac {\partial {\left(fg\right)}}{\partial y}}\right)}\\&={\frac {1}{2}}{\left(f{\frac {\partial g}{\partial x}}+g{\frac {\partial f}{\partial x}}+{\mathrm {i} }{\left(f{\frac {\partial g}{\partial y}}+g{\frac {\partial f}{\partial y}}\right)}\right)}\\&=f{\left({\frac {1}{2}}{\left({\frac {\partial g}{\partial x}}+{\mathrm {i} }{\frac {\partial g}{\partial y}}\right)}\right)}+g{\left({\frac {1}{2}}{\left({\frac {\partial f}{\partial x}}+{\mathrm {i} }{\frac {\partial f}{\partial y}}\right)}\right)}\\&=f{\frac {\partial g}{\partial {\overline {z}}}}+g{\frac {\partial f}{\partial {\overline {z}}}}.\end{aligned}}$

Zur bewiesenen Aussage