Vanishing Distributional Derivative of Distribution implies Distribution is Constant
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Theorem
Let $T \in \map {\DD'} \R$ be a distribution.
Let $\mathbf 0$ be the zero distribution.
Suppose the distributional derivative of $T$ vanishes:
- $\ds \dfrac \d {\d x} T = \mathbf 0$
Then $T$ is a constant distribution.
Proof
Let $\phi \in \map \DD \R$ be a test function.
Then:
\(\ds 0\) | \(=\) | \(\ds \map {\mathbf 0} \phi\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {T'} \phi\) | Assumption of the Theorem | |||||||||||
\(\ds \) | \(=\) | \(\ds - \map T {\phi'}\) | Definition of Distributional Derivative |
Hence:
- $\set {\phi' : \phi \in \map \DD \R} \subseteq \ker T$
where $\ker$ denotes the kernel.
Let $\mathbf 1$ be a constant mapping such that $\mathbf 1 : \R \to 1$.
Then the associated distribution reads:
- $\ds \map {T_{\mathbf 1}} \phi = \int_{-\infty}^\infty \map \phi x \rd x$
Furthermore:
\(\ds \ker T_{\mathbf 1}\) | \(=\) | \(\ds \set {\psi \in \map \DD \R : \int_{-\infty}^\infty \map \psi x \rd x = 0}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \set {\phi' : \phi \in \map \DD \R}\) | Characterization of Derivative of Test Function | |||||||||||
\(\ds \) | \(\subseteq\) | \(\ds \ker T\) |
We have that Test Function Space with Pointwise Addition and Pointwise Scalar Multiplication forms Vector Space.
Let $V = \map \DD \R$, $L = T$ and $\ell = T_{\mathbf 1}$.
- $\exists c \in \C : T = c T_{\mathbf 1}$
By definition of multiplication of a distribution by a smooth function:
- $c T_{\mathbf 1} = T_c$
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 6.2$: Derivatives in the distributional sense