delta functions $e^{x}\delta (x)=\delta (x)$

Question

How would you prove that; $$e^{x} \delta (x)= \delta (x)$$

Is it anything to do with the following relationship; $$ \int_{-\infty}^{\infty} g'(x)h(x)\,dx = \int_{-\infty}^{\infty} g(x)h'(x)\,dx.$$

Thanks in advance

Answer 1

Hint: $\delta(x)$ is a linear functional on some space of suitably nice (in particular, continuous) functions of $\mathbb{R}$. Given a suitably nice function $f$, what does $\delta(x)[f]$ equal?

When you say $e^x \delta(x) = \delta(x)$, what you're saying is that the left hand side and the right hand side act on continuous functions in the same way. Is that true?

Answer 2

Depends on how you define the Dirac delta, if it is defined as a linear functional acting on an $f\in C^{\infty}_0(\mathbb{R})$: $$ \delta[f](x) = \int^{\infty}_{-\infty} \delta(x)f(x) \,dx = f(0) $$ Then we can see for any test function $\phi$: $$ (e^{x}\delta(\cdot) - \delta(\cdot))[\phi](x) = \int^{\infty}_{-\infty} (e^x\delta(x)\phi(x) - \delta(x)\phi(x)) \,dx = e^0\phi(0) - \phi(0) = 0 $$ Hence $e^{x} \delta(x) = \delta(x)$.

Answer 3

The $\delta_0$ functional is representable as a measure. You have $$\delta_0(E) = \cases{1 & if $0 \in E$\cr 0 & otherwise }$$ Then the linear functional here is represented as $$f \mapsto f(0) = \int_{R^d} f(\xi)d\delta_0(\xi). $$ It is not hard to show this relation will hold for any continuous function. Your result will follow right away.

Answer 4

$$ \int_{-\infty}^\infty f(x)\delta(x)\,dx = f(0). $$ $$ \int_{-\infty}^\infty f(x)\Big(e^x \delta(x)\Big)\,dx = \text{what?} $$ But look at that last integral this way: $$ \int_{-\infty}^\infty \Big(f(x)e^x\Big) \delta(x)\,dx. $$ This is equal to the value of the function $x\mapsto f(x)e^x$ at $x=0$, because the delta function is defined that way. Letting $g(x)=e^x\delta(x)$, we now have $$ \int_{-\infty}^\infty f(x)\delta(x)\,dx = \int_{\infty}^\infty f(x)g(x)\,dx \text{ for all suitable functions }f. $$ The definition of generalized functions is such that that implies that $g=\delta$. ("Suitable" will mean test functions or Schwarz functions or whatever it is you're using in that role in the context in which you're working.)

So be careful to understand the definition.