Ambiguity in delta equal sign

Question

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 $$x^ndelta(x) = 0$$
Does the equality in this case mean it is right for every $x$ value? For $xneq0$ it is trivial, but we can't plug in $x=0$.
What I see they do in textbooks is taking the integral of $x^ndelta(x)$ 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.

md2perpe · Answer

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
$$
langle fu, varphi rangle = langle u, fvarphi rangle.
$$
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 $x^n,delta(x)=0.$

Ambiguity in delta equal sign

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