0

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?

Bernard
  • 175,478
NotSoSN
  • 213
  • 1
  • 7
  • That's exactly the right answer. If $a\ne b$ both sides are $0$. If $a=b$ both sides are $a\delta(0)$ – Andrei Mar 24 '20 at 19:23
  • 1
    It is not true. Informally, $\int a \delta(a-b) da = a $, $\int b \delta(a-b) da = b$. – copper.hat Mar 24 '20 at 19:28
  • @copper.hat ... For any continuous function $f$, we have $\int f(a) \delta(a-b);da = f(b)$, right? So taking $f(a)=a$ we get the first one, $\int a\delta(a-b);da=b$. And taking $f(a) = b$, a constant function, we get the second one $\int b\delta(a-b);da = b$. – GEdgar Mar 24 '20 at 19:38
  • 1
    @GEdgar: I guess I should have written $(a \delta_{a-b})(\phi) = a \phi(a-b)$ and $(b \delta_{a-b})(\phi) = b \phi(a-b)$, so they are equal iff $a=b$. – copper.hat Mar 24 '20 at 19:43

1 Answers1

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 $x \delta(x-y) = y \delta(x-y).$

md2perpe
  • 26,770