Dirac delta function as a product of Heaviside function and Exponent for large argument

Question

Is is valid to say that

$\delta( x -t) = \lim_{A\to \infty} [A e^{-A (x-t)} \theta(x-t)]$

Here $A$, $x$ and $t$ are all positive.

Answer 1

You are correct. I will show it for $t=0$. The general case is just a translation of this.

Take $\phi \in C_c^\infty(\mathbb R)$. Then we have $$ \langle A e^{-Ax} \theta(x), \phi(x) \rangle = \int A e^{-Ax} \theta(x) \, \phi(x) \, dx = \{ y = Ax \} = \int_0^\infty e^{-y} \, \phi(y/A) \, dy \\ \to \int_0^\infty e^{-y} \, \phi(0) \, dy = \int_0^\infty e^{-y} \, dy \, \phi(0) = \phi(0) = \langle \delta(x), \phi(x) \rangle $$ as $A \to \infty$.

This shows that $$\lim_{A \to \infty} A e^{-Ax} \theta(x) = \delta(x).$$