Satz von Schwarz/Dreierumordnung/Aufgabe/Lösung

Unter mehrfacher Anwendung des Satzes von Schwarz ist (die Klammern verdeutlichen, wie der Satz angewendet wird. Für die vierte Gleichung wird verwendet, dass ${\frac {\partial }{\partial y}}f$ zweimal stetig differenzierbar ist)

${}{\begin{aligned}{\frac {\partial }{\partial z}}{\frac {\partial }{\partial y}}{\frac {\partial }{\partial x}}f&={\frac {\partial }{\partial z}}{\left({\frac {\partial }{\partial y}}{\frac {\partial }{\partial x}}f\right)}\\&={\frac {\partial }{\partial z}}{\left({\frac {\partial }{\partial x}}{\frac {\partial }{\partial y}}f\right)}\\&={\left({\frac {\partial }{\partial z}}{\frac {\partial }{\partial x}}\right)}{\left({\frac {\partial }{\partial y}}f\right)}\\&={\left({\frac {\partial }{\partial x}}{\frac {\partial }{\partial z}}\right)}{\left({\frac {\partial }{\partial y}}f\right)}\\&={\frac {\partial }{\partial x}}{\left({\frac {\partial }{\partial z}}{\frac {\partial }{\partial y}}f\right)}\\&={\frac {\partial }{\partial x}}{\left({\frac {\partial }{\partial y}}{\frac {\partial }{\partial z}}f\right)}\\&={\frac {\partial }{\partial x}}{\frac {\partial }{\partial y}}{\frac {\partial }{\partial z}}f.\end{aligned}}$

Zur gelösten Aufgabe