Distributional Partial Derivatives Commute
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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
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis: Chapter $\S 6.2$: A glimpse of distribution theory. Derivatives in the distributional sense