Vanishing Distributional Derivative of Distribution implies Distribution is Constant

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:

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:

We have that Test Function Space with Pointwise Addition and 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:


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

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


 * $c T_{\mathbf 1} = T_c$