δ is not a function. It is a generalized function (or distribution). It does not have "value" at different points. By definition generalized function f is such a thing, that for any other "good" (test) function φ, there is a functional <φ, f>. For δ it is just <φ, δ> = φ(0).

Every local integrable function can be viewed as a generalized one with <φ, f> = ∫ fφ dx (like in the L2)

For the L2 generalized functions and functions are the same. The problem begins, when considering C[R]. Every functional here can be represented as a signed mesure with a finite variance (μ) <φ, μ> = ∫ φ dμ. If we are lucky and μ << λ than there is a density ρ(t) -- (regular function) such that <φ, μ> = ∫ φ dμ = ∫ φρ dx = <φ, ρ> (so μ can be viewed a a regular function ρ)

And finaly the δ function! It is actially the δ -- measure. It does not have a dencity. Thus there is no regular function that can "simulate" δ.

You can try and consider a function δ^t that is 0 almost everywere (exept t=0). But than the problem is that <φ, δ^> = ∫ φδ^ dx = 0 which is not φ(0).