Proof of Dirac delta property

Question

I am trying to prove $$a\delta(a-b)=b\delta(a-b).$$ I tried using the Dirac delta identities, specifically $\delta(ax)=\frac{1}{a}\delta(x)$ However, I keep coming up with a=b as the only solution which is not true. Is there another identity I should use as well? Or am I starting completely off base?

Answer 1

Multiply $x \delta(x-y)$ with a test function $\varphi(x,y)$ and integrate over $x$ and $y$, and use the property $\int \delta(x-a) \varphi(x) \, dx = \varphi(a)$ twice: $$ \iint x \delta(x-y) \varphi(x,y) \, dx \, dy = \int \left( \int \delta(x-y) \left( x \varphi(x,y) \right) \, dx \right) \, dy \\ = \int y \varphi(y,y) \, dy = \int y \left( \int \delta(x-y) \varphi(x,y) \, dx \right) \, dy \\ = \iint y \delta(x-y) \varphi(x,y) \, dx \, dy $$ Since we get that $\iint x \delta(x-y) \varphi(x,y) \, dx \, dy = \iint y \delta(x-y) \varphi(x,y) \, dx \, dy$ for every test function $\varphi$ we get that $x \delta(x-y) = y \delta(x-y).$