Trivial or not: Dirac delta function is the unit of convolution.

Question

My task is to prove that the Dirac delta function is the unit of convolution and all I find always is this formula but no further explanation:

$$[f*\delta](t)=\int_{-\infty}^{\infty}f(t-\sigma)\delta(\sigma)d\sigma=f(t)$$

Should I see from this that it's true, or should I use it as the beginning of the proof. If it is trivial then why and if not then what should I do?

Answer 1

A useful property of Dirac-delta function $\delta(t-a)$ is

• $\int_{-\infty}^{\infty}f(t).\delta(t-a)dt=f(a)$

Answer 2

I think you have provided the answer in the statement of your question: $(f*\delta)(t)=f(t)$ , ie. $f$ convolved with delta yields $f$. So for the operation of convolution the Dirac distribution is the unit. This is analogous to $f\cdot1=f$ for the operation of multiplication where $1$ is the unit.

Answer 3

I guess, it is easy here to take the mathematical definitions and not the physicist's definitions. The delta distribution is defined as $$\delta(\varphi) = \varphi(0)$$ for each test-function $\varphi.$ The convolution of two distributions is defined by $$(T \star S)(\varphi) = T_x S_y(\varphi(x+y)).$$ Hence, for each distribution $T$ we have $$(T\star \delta)(\varphi)=T_x \delta_y (\varphi(x+y))=T_x (\varphi(x)) = T(\varphi),$$ for each test-function $\varphi.$ Hence $$T\star \delta = T.$$