Equality of Integrands Involving Dirac Delta Functions

Question

Say, I have the following integral: $$\int_{-\infty}^{\infty} f(x) \delta(x-y) \, dx = f(y) \int_{-\infty}^{\infty} \delta(x-y) \, dx = f(y).$$

Again consider the following integral: $$\int_{-\infty}^{\infty} f(y) \delta(x-y) \, dx = f(y) \int_{-\infty}^{\infty} \delta(x-y) \, dx = f(y).$$

My Question

In this case, is $$f(x) \delta(x-y) = f(y) \delta(x-y) \, ?$$

The reason I am hesitating to conclude that (using Fundamental Theorem of Calculus) is $f(x) \delta(x-y)$ not being a function (strictly speaking) due to the presence of the Dirac delta 'function'. For more details, see here.

Although, it is quite common in Physics, for instance, here in the normalization of one-particle states.

Answer 1

Yes. As you say, both of them are equal to $f(x)\delta(0)$ when $x=y,$ while they are equal to $0$ when $x\neq y.$

I think maybe the point you're confused about is how to define "equal" for non-ordinary functions, and how to understand "multiply" for non-ordinary functions like $f(x)\delta(x-y)$. You can see generalized function for help.

In brief, $\delta(x)$ can be seen as a functional on $C^\infty(\mathbb{R}^n),$ which acts as $$\langle \delta(x), h(x)\rangle=\int_{\mathbb{R}^n}\delta(x) h(x)dx=h(0),\quad \forall h\in C^\infty(\mathbb{R}^n),$$ so for given $y$, $h\in C^\infty(\mathbb{R}^n)$, $$ \begin{aligned} \langle f(x)\delta(x-y),h(x)\rangle &=\int_{\mathbb{R}^n}f(x)\delta(x-y) h(x)dx=f(y)h(y)\\ &=\int_{\mathbb{R}^n}f(y)\delta(x-y)h(x)dx=\langle f(y)\delta(x-y),h(x)\rangle \end{aligned}$$ Thus regarded as functional on $C^\infty(\mathbb{R^n_x})$, for any give $y,$ $f(x)\delta(x-y)=f(y)\delta(x-y).$