Convergence to Delta Dirac Distribution

Question

This question derived from my previous question. When I took a course on the theory of distributions, I was first introduced to the Dirac delta as an usual distribution, that is, as a linear functional on $C_{0}^{\infty}(\Omega)$ (the notation is explained below) satisfying some continuity property. After that, It was shown that the Dirac delta is compactly supported, so it is actually a linear functional on $C^{\infty}(\Omega)$. Now, It seems to me that one could define the Dirac delta as a functional on $C(\Omega)$ in the same way it is done for $C^{\infty}(\Omega)$ by setting $\delta_{x_{0}}f := f(x_{0})$. On the other hand, I understand the importance of defining it on $C^{\infty}(\Omega)$, once one can also define the notion of derivatives of $\delta$.

Here's what bothers me: I have encountered some papers/texts in which the Dirac delta is used in less restrictive domains than $C^{\infty}(\Omega)$. The link I posted at the begining of this question provides an example for such matter. We have a sequence of $\{f_{n}\}_{n \in \mathbb{N}}$ in $\mathcal{S}(\mathbb{R^{n}})$ that seems to converge to $\delta_{x_{0}}$ in the dual of $C(\Omega)\cap L^{\infty}(\Omega)$ (see the comments following the question on the link). Now, as I understand, this implies that the sequence of linear maping induced by $f_{n}$ converges to $\delta_{x_{0}}$, that is: $$\int f_{n}(x)\phi(x)dx \to \phi(x_{0})$$ for every bounded and continuous function $\phi$. But this seems to imply that the Dirac delta here is being considered as a functional on $C(\Omega)$ rather than $C^{\infty}(\Omega)$.

In summary, I want to know: can I (and in what extent) consider the Dirac delta as a linear functional on $C(\Omega)$ rather than $C^{\infty}(\Omega)$? Do I lose anything besides the notion of derivatives of $\delta$? Besides, can I always find a sequence $T_{n}$ of linear functionals on $C(\Omega)$ such that $T_{n}\to \delta_{x_{0}}$?

Notation: Here $C_{0}^{\infty}(\Omega)$ denotes the vector space of all $C^{\infty}$ functions with compact support and defined on some open set $\Omega \subset \mathbb{R}^{n}$, while $C^{\infty}(\Omega)$ denotes the $C^{\infty}$ functions defined on $\Omega$. In addition, $C(\Omega)$ is the vector space of continuous functions on $\Omega$. Finally, $\mathcal{S}(\mathbb{R}^{n})$ denotes the schwartz space of rapid decrease functions on $\mathbb{R}^{n}$.

Answer 1

Of course you can consider $\delta_x$ as a linear functional on $C(\Omega)$. Some crucial points are:
- Can we define $\delta_x$ as a continuous linear functional on $C(\Omega)$?
- Which sequences of which class of functions converge to $\delta_x$ in $C(\Omega)^*$?

So, you need a good topology on $C(\Omega)$, that, for $\delta_x$ to be a meaningful continuous functional should have the property "convergence in $C(\Omega)$ implies pointwise convergence". Furthermore, to make sense of the "approximate identities" you mentioned, you would want to have an embedding $C(\Omega) \to C(\Omega)^*$. This could e.g be induced by compactness of $\Omega$ or sufficiently fast decay of your continuous functions at infinity.

Edit: Also, in general you do not have the embedding $S(\mathbb{R}^n) \to C(\mathbb{R}^n)^*$ because there are functions $f \in S(\mathbb{R}^n), g \in C(\mathbb{R}^n)$ with $\int f g = \infty$ (or even undefined).
In fact, it is even worse: For any $f \in S(\mathbb{R}^n)$ you can find a $g \in C(\mathbb{R}^n)$ such that $\int f g$ is not defined.
Therefore, you would probably need to restrict yourself to a subset of $C(\mathbb{R}^n)$ with sufficient decay properties at infinity.

Answer 2

Although much more could be said (perhaps upon request?), it is certainly true that "distributions of finite order" (which would provably be the case for compactly-supported...) extend to continuous functionals on larger spaces of "nice-ish" functions.

One can keep in mind that the natural model of the (continuous) dual to (the natural, complete ... therefore, Frechet) space of smooth functions on (e.g.) the real line, is the space of compactly-supported distributions.

Yes, certainly, Dirac $\delta$ is compactly supported...