I tried asking this yesterday but my wording was too bad. I will try again because it really confuses me.
Let's look at an identity like $$x^n\delta(x) = 0$$ Does the equality in this case mean it is right for every $x$ value? For $x\neq0$ it is trivial, but we can't plug in $x=0$.
What I see they do in textbooks is taking the integral of $x^n\delta(x)$ around $0$, and see it is indeed zero. But why does that mean the expression itself is zero?
I seem to be missing an important notion concerning delta "functions". Are they defined only under an integral in the singularity? I see nowhere such definition.
If I'm still unclear, please let me know.
Distributions, like $\delta$, are defined as linear functionals over a space of smooth functions with compact support ($C_c^\infty$).
The action of a distribution $u$ on a test function $\varphi \in C_c^\infty$ is often denoted $\langle u, \varphi \rangle$ or with some abuse of notation, as an integral $\int_{-\infty}^{\infty} u(x) \, \varphi(x) \, dx.$
Multiplication of a distribution with a smooth function $f \in C^\infty$ is defined by $$ \langle fu, \varphi \rangle = \langle u, f\varphi \rangle. $$ In integral notation this is just the quite obvious $$ \int_{-\infty}^{\infty} (f(x)\,u(x)) \, \varphi(x) \, dx = \int_{-\infty}^{\infty} u(x) \, (f(x)\,\varphi(x)) \, dx. $$
The distribution $\delta$ is defined by $\langle \delta, \varphi \rangle = \varphi(0).$
By what I have now defined, we get $$ \langle x^n \, \delta(x), \varphi(x) \rangle = \langle \delta(x), x^n \, \varphi(x) \rangle = (x^n\,\varphi(x))[x:=0] = 0^n\,\varphi(0) = 0. $$ This is the quite exact meaning of $x^n\,\delta(x)=0.$