Offene Menge/R^n/Vektorfelder/Hintereinanderschaltung/Explizit/Fakt/Beweis

Beweis

Es ist

${}{\begin{aligned}{\left(\sum _{i=1}^{n}f_{i}\partial _{i}\right)}{\left({\left(\sum _{j=1}^{n}g_{j}\partial _{j}\right)}(h)\right)}&={\left(\sum _{i=1}^{n}f_{i}\partial _{i}\right)}{\left(\sum _{j=1}^{n}g_{j}\partial _{j}(h)\right)}\\&=\sum _{i=1}^{n}f_{i}\partial _{i}{\left(\sum _{j=1}^{n}g_{j}\partial _{j}(h)\right)}\\&=\sum _{i=1}^{n}f_{i}{\left(\sum _{j=1}^{n}{\left(\partial _{i}(g_{j})\partial _{j}(h)+g_{j}\partial _{i}\partial _{j}(h)\right)}\right)}\\&=\sum _{j=1}^{n}{\left(\sum _{i=1}^{n}f_{i}\partial _{i}g_{j}\right)}\partial _{j}h+\sum _{j=1}^{n}\sum _{i=1}^{n}f_{i}g_{j}\partial _{i}\partial _{j}h.\end{aligned}}$

Zur bewiesenen Aussage