I know that the Dirac delta function is not really a function but a distribution.

It means that when we write $\int \delta(x-x_0) f(x) dx$ we want to say in fact $<\delta_{x_0},f>$.

So the dirac "eats" a function and not a number.

But sometimes in physics we don't have expression like :

$$\int \delta(x-x_0) f(x) dx $$ but just something like : $$\delta(x-x_0) f(x)$$

without any integral around it.

I know how to interpret the last line physically (it will be $f(x_0)$ if $x=x_0$ and $0$ elsewhere). But how to give it a mathematical sense ? Indeed it is not a dirac that "eats" f here, but it is something else that I don't know how to give a mathematical sense.

I insist on the fact that it's not a Kronecker symbol that we have (in my example we obtained this term by derivating $\Theta(x-x_0)f(x)$ where $\Theta$ is the Heaviside function).

(I just know basics distribution theory)


If you have some text that deals with $$ g(x) = f(x)\delta(x-x_0), $$ with no integration, then $g(x)$ is also a distribution, to be used only in the form $$ \langle g, h\rangle = \int g(x) h(x) \mathrm dx= \int f(x)\delta(x-x_0) h(x) \mathrm dx = f(x_0)h(x_0). $$ In this particular example, it's easy to see that you can simply replace the $f(x)$ by $f(x_0)$, i.e. in the form $$ g(x) = f(x_0)\delta(x-x_0), $$ so $g(x)$ is just a delta function multiplied by a constant $f(x_0)$, but that's a bit secondary. The point, really, is that if you start handling distributions, then everything is a distribution (instead of a function) unless proven otherwise.


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