Is it correct to set $\delta(t) f(t) = \delta(t) f(0)$ within a distribution?

Question

I have some complicated probability distributions which come out as $$ P(x,t) = \delta(t)G(x,t) + K(x,t),$$ where $G(x,t)$ and $K(x,t)$ are continuous in time. Is it permissible to simplify such distributions as $$ P(x,t) = \delta(t) G(x,0) + K(x,t)?$$ Are any properties of $P$ lost in making such a simplification? At a glance it seems normalization and time derivatives are unchanged. The latter since \begin{align} \frac{\partial}{\partial t} P(x,t) &= \delta(t) \frac{\partial}{\partial t} G(x,t) + \delta'(t) G(x,t) + \frac{\partial}{\partial t} K(x,t) \\&= \delta(t) \frac{\partial}{\partial t} G(x,t) - \delta(t) \frac{\partial}{\partial t} G(x,t) + \frac{\partial}{\partial t} K(x,t)\\ &= \frac{\partial}{\partial t}K(x,t).\end{align} Is there anything I could be missing through?

Answer 1

Yes, the identity $f(t)\delta(t) = f(0)\delta(t),$ where $f\in C^\infty,$ is valid.

Proof:
For any test function $\phi$ we have $$ \langle f\delta, \phi \rangle = \langle \delta, f\phi \rangle = (f\phi)(0) = f(0)\phi(0) = f(0) \langle \delta, \phi \rangle = \langle f(0)\delta, \phi \rangle . $$ Thus, $f\delta=f(0)\delta.$