Vanishing Distributional Derivative of Distribution implies Distribution is Constant

From ProofWiki
Jump to navigation Jump to search

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 $\ell = T_{\mathbf 1}$.

By Kernel of Linear Transformation contained in Kernel of different Linear Transformation implies Transformations are Proportional‎:

$\exists c \in \C : T = c T_{\mathbf 1}$

By definition of multiplication of a distribution by a smooth function:

$c T_{\mathbf 1} = T_c$

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Vanishing_Distributional_Derivative_of_Distribution_implies_Distribution_is_Constant&oldid=544883"