Why do you say that? The function that takes a function [itex]f(x)[/itex] and returns [itex]f(0)[/itex] is a well-defined distribution. Maybe what you mean is that if we identify two functions [itex]f \sim g[/itex] if [itex]\int |f(x) - g(x)|^2 dx = 0[/itex], then the delta-function distribution is ill-defined on the equivalence class. But it's perfectly well-defined on square-integrable functions, if we don't take equivalence classes.

[itex]\delta(x)[/itex] is not really a function, so [itex]\delta(0)[/itex] is not defined. The only thing that is defined about the delta function is integrals involving it:

[itex]\int_{a}^b \delta(x) f(x) = f(0)[/itex] if the interval contains [itex]x=0[/itex], and is zero, otherwise.

Statements involving [itex]\delta(x)[/itex] that don't explicitly involve integrals can only be made sense of by understanding them in terms of integration.

It's not continuous, since ##\lVert f - g\rVert = 0##, but ##\left| F(f)-F(g)\right| \neq 0##. So it would be an element of the algebraic dual, but not the topological dual space.

It's a little confusing. When people talk about [itex]L^2[/itex], they do mean functions modulo equivalence. But it seems to me that we can certainly talk about the set of square-integrable functions without taking equivalence relations.

Viewed as functions (as opposed to equivalence classes of functions), [itex]\int |f(x) - g(x)|^2 dx = 0[/itex] does not imply that [itex]f=g[/itex], so I don't see what's the problem with having [itex]F(f) \neq F(g)[/itex].

The space of square-integrable functions is usually denoted by ##\mathcal L^2## rather than ##L^2##, but it is not a Hilbert space, so it's not suitable for QM.

Your ##F## is well-defined on ##\mathcal L^2##. It's just not continuous and distributions are usually defined to be continuous. It's okay to use discontinuous distributions, but you must be aware that many things might not work anymore as expected.

That's the thing isn't it. If ##f = g## a.e., why would we expect the integrals ##\int f(x)\delta(x)dx ## and ##\int g(x)\delta(x)dx## to be different.
Furthermore, what will be the derivative of ##\delta## in your formalism?
And why did you not put any continuity requirements on the definition of your distribution?

I mean: ok, sure, your definition is consistent, but I don't see it as very useful.

That's the whole point--[itex]\int f(x) \delta(x) dx[/itex] is not actually an integral. It's just notation. That notation is defined to be [itex]f(0)[/itex].

I don't understand the question. [itex]\delta(x)[/itex] is not a function. It doesn't have a derivative. It only makes sense in the expression [itex]\int \delta(x) f(x) dx[/itex].

A distribution is not in general an integral. Certain distributions can be defined via an integral, but the delta function is one that can't be.

There are no continuity requirements: [itex]F(f) \equiv f(0)[/itex]. That's well-defined for any function that has a value at [itex]x=0[/itex]

Well, the context of this discussion is the delta function, and there is no way to make sense of a delta function in terms of continuous distributions.

You really don't see the difference between what you wrote and what's in the wiki link? For one, the space of test functions in the wiki link is not ##\mathcal{H}##, and second they require continuity.

The delta distrubution is continuous on the usual test function spaces like ##C^\infty_c## or ##\mathcal S##. You can embedd them into ##L^2## by sending each ##f## to its equivalence class. Of course, delta isn't continuous in the Hilbert space norm, but it's continuous in the topologies of the usual test function spaces.

(Edit: Since micromass is active, I'm going to let him continue the argument.)

Look, the functional [itex]F[/itex] defined by [itex]F(f) = f(0)[/itex] is defined on the set of all functions from [itex]R[/itex] to [itex]C[/itex], regardless of whether they are continuous, or square-integrable, or whatever. You might be right that this function is only considered a "distribution" when it is restricted to a suitable collection of functions.