I found this equation in thi s question How to differentiate $f(t)\theta(t)$, the product of a function with the Heaviside unit step? $$ f(t)\delta(t) = f(0) \delta(t) \tag{1} $$ Now I have a similar formula that needs to be simplified, its expression is like $ H(-t)g(-t-a) \delta(t)$, where H is the unit step function. Can it be simplified to: $$ \begin{align} H(-t)g(-t-a) \delta(t) =& H(0) g(-a) \delta(t) \\ =&g(-a) \delta(t) \tag{2} \end{align} $$
I think it's right, but I don't know how to prove eq(1) and is the usage in eq(2) correct?
As mentioned in the comments, you cannot simply multiply a Dirac delta with the Heaviside function using the standard theory of distributions, i.e., Schwartz' theory. This theory is generally linear, i.e., you can only do linear operations on distributions. Multiplication; $f.\theta$, where $f$ is a distribution, is only defined if $\theta$ is a smooth function, which is not the case for the Heaviside function. There are several scenarios in physics that require such an operation. There are some extensions of Schwartz' theory that can allow for such an operation; see, e.g., Colombeau algebra.