Komplexe Mannigfaltigkeit/Differenzierbare Funktion/Ableitungsform/Zerlegung/Lokale Beschreibung/Fakt/Beweis

Beweis

Wir können direkt ${}U\subseteq {\mathbb {C} }^{n}$ annehmen. Sei

{}z_{j}=x_{j}+{\mathrm {i} }y_{j}\,.

Es ist

{}{\begin{aligned}df&=\sum _{j=1}^{n}{\frac {\partial f}{\partial {x_{j}}}}dx_{j}+\sum _{j=1}^{n}{\frac {\partial f}{\partial y_{j}}}dy_{j}\\&=\sum _{j=1}^{n}{\left({\frac {\partial f}{\partial {x_{j}}}}dx_{j}+{\frac {\partial f}{\partial y_{j}}}dy_{j}\right)}\\&={\frac {1}{2}}\sum _{j=1}^{n}{\left({\left({\frac {\partial f}{\partial x_{j}}}-{\mathrm {i} }{\frac {\partial f}{\partial y_{j}}}\right)}{\left(dx_{j}+{\mathrm {i} }dy_{j}\right)}+{\left({\frac {\partial f}{\partial x_{j}}}+{\mathrm {i} }{\frac {\partial f}{\partial y_{j}}}\right)}{\left(dx_{j}-{\mathrm {i} }dy_{j}\right)}\right)}\\&={\frac {1}{2}}\sum _{j=1}^{n}{\left({\left({\frac {\partial f}{\partial x_{j}}}-{\mathrm {i} }{\frac {\partial f}{\partial y_{j}}}\right)}dz_{j}+{\left({\frac {\partial f}{\partial x_{j}}}+{\mathrm {i} }{\frac {\partial f}{\partial y_{j}}}\right)}d{\overline {z}}_{j}\right)}\\&=\sum _{j=1}^{n}{\left({\frac {\partial f}{\partial z_{j}}}dz_{j}+{\frac {\partial f}{\partial {\overline {z}}_{j}}}d{\overline {z}}_{j}\right)}\\&=\sum _{j=1}^{n}{\frac {\partial f}{\partial z_{j}}}dz_{j}+\sum _{j=1}^{n}{\frac {\partial f}{\partial {\overline {z}}_{j}}}d{\overline {z}}_{j},\end{aligned}}

und die Summe links gehört zu ${}{\mathcal {E}}^{(1,0)}$ .