Prove that $\int\delta(x-b)\delta(x-a)\ \mathsf dx =\delta(a-b)$.

Question

Show that the convolution of two $\delta$ functions at different points is again a Dirac $δ$ function. Convolution of a Dirac $\delta$ function with a function $f$ is defined as : $$\int\delta(x-y)f(x)\ \mathsf dx = f(y).$$

I'm not good with Dirac. May you help me? Thank you

Answer 1

Note that for all test functions $f$ we have

$$\int_{-\infty}^{\infty}\delta (b-a)f(b)\,db=f(a)$$

Now, for all test functions we have

$$\begin{align} \int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}\delta (x-a)\delta(x-b)\,dx\right)\,f(b)\,db&=\int_{-\infty}^{\infty}\delta (x-a)\left(\int_{-\infty}^{\infty}\delta(x-b)\,f(b)\,db\right)\,dx\\\\ &=\int_{-\infty}^{\infty}\delta (x-a)f(x)\,dx\\\\ &=f(a) \end{align}$$

Therefore, we observe that as distributions, $\delta(b-a)$ yields the same result as $\int_{-\infty}^{\infty}\delta (x-a)\delta(x-b)\,dx$. Therefore, we can write

$$\int_{-\infty}^{\infty}\delta (x-a)\delta(x-b)\,dx=\delta (b-a)$$

as was to be shown!