Which tempered distributions have the delta function as their derivative?

Question

I am looking for the general solution to the equation $T'= \delta_{x_0}$ where $T$ is a tempered distribution.

I started by using the following equality for derivatives of distributions: $\langle T', \phi \rangle = - \langle T, \phi' \rangle \quad (I)$ where $\phi \in S$ (the space of Schwartz functions).

I am told to use the fact that, if a distribution satisfies $T'=0$ a constant $\alpha$ exists so that $T = \alpha$.

I wrote out the integral form of $(I)$ and plugged in $\alpha$ for $T$ and tried to integrate by parts but I am not getting near a solution. I would appreciate any sort of help!

Answer 1

I guess that it is assumed that you know that $H'=\delta,$ and consequently that $H'(x-x_0) = \delta(x-x_0).$

Let $T \in S'$ be such that $T'=\delta.$ Then $(T-H)' = T'-H' = \delta-\delta=0$ so $T-H = C$ for some constant $C.$ Thus all solutions to $T'=\delta$ can be written as $T=H+C,$ where $C$ is a constant.

Now, extend this argument to $T' = \delta_{x_0}.$

Answer 2

The Heaviside function satisfy your requirements:

$$H(x) = \left \{ \begin{array}{ll} 1 \quad \text{if} \quad x\geq 0 \\ 0 \quad \text{if} \quad x <0 \end{array}\right.$$ It's distributional derivative satisfy $\frac{d}{dx}H=\delta_0$.