I'm unsure about $(\delta(t-a)*f)(t)=f(t-a)$, where $\delta$ is Dirac delta function.

Question

I'm reading some notes from MIT OCW on function convolution defined by:

$$(f*g)(t) = \int_{0^{-}}^{t^{+}} f(s)g(t-s) \ ds$$

It says that $(\delta(t-a)*f)(t)=f(t-a)$ and the proof is that:

$(\delta(t-a)*f)(t)=\int_{0^{-}}^{t^{+}}\delta(s-a)f(t-s)ds = f(t-a)$.

I don't see why this proof holds for values of $t$ less than $a$. Because then, the $s$ in the integration never reaches $a$ and $\delta$ is $0$ on the whole integration interval.

Am I right that this proof only holds for $t\ge a$?

Thanks!

Answer 1

Yes $t<a \implies \int_0^t \delta(s-a)f(t-s)ds=0$

Answer 2

Perhaps the simplest way to see this is to regard $\delta $ as atomic measure. Then, the result is immediate:

If $E\subseteq \mathbb R$, then $\delta_a (E)=1$ if $a\in E$, and $0$ otherwise.

Then

$\int_{0^{-}}^{t^{+}}f(t-s)d\delta _a(s) =\int_{[0,t]}f(t-s)d\delta _a(s)=f(t-a)\ $ if $0\leq a\leq t\ $ and $0$ otherwise.