Distributional Partial Derivatives Commute

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $T \in \map {\DD'} {\R^d}$ be a distribution.


Then in the distributional sense:

$\dfrac {\partial^2 T} {\partial x_i \partial x_j} = \dfrac {\partial^2 T} {\partial x_j \partial x_i}$

where:

$i, j \in \N : 1 \le i, j \le d$


Proof

Let $\phi \in \map \DD {\R^d}$ be a test function.

\(\ds \map {\dfrac {\partial^2 T} {\partial x_i \partial x_j} } \phi\) \(=\) \(\ds -\map {\dfrac {\partial T} {\partial x_j} } {\dfrac {\partial \phi} {\partial x_i} }\) Definition of Distributional Partial Derivative
\(\ds \) \(=\) \(\ds \map T {\dfrac {\partial^2 \phi} {\partial x_j \partial x_i} }\) Definition of Distributional Partial Derivative
\(\ds \) \(=\) \(\ds \map T {\dfrac {\partial^2 \phi} {\partial x_i \partial x_j} }\) Clairaut's Theorem, Definition of Test Function
\(\ds \) \(=\) \(\ds - \map {\dfrac {\partial T} {\partial x_i} } {\dfrac {\partial \phi} {\partial x_j} }\) Definition of Distributional Partial Derivative
\(\ds \) \(=\) \(\ds \map {\dfrac {\partial^2 T} {\partial x_j \partial x_i} } \phi\) Definition of Distributional Partial Derivative

$\blacksquare$


Sources