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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 $\mathbb{R}^{n}$.

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  • $\begingroup$ You can consider the dirac delta on any kind of vector space of functions. $\endgroup$ – Botond Dec 19 '19 at 13:53
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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.

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  • $\begingroup$ thanks for your answer. The case I'm dealing concerns convergence of functions belonging to $\mathcal{S}(\mathbb{R}^{n})$, so my embedding is $\mathcal{S}(\mathbb{R}^{n}) \to C(\mathbb{R}^{n})^{*}$. This satisfies the fast decay property you mentioned. Does this imply the existence of such a sequence converging for a given $\delta_{x_{0}}$? Is it easy to see that? $\endgroup$ – IamWill Dec 19 '19 at 14:02
  • $\begingroup$ Yes, my mistake. I meant an embedding $\mathcal{S}(\mathbb{R}^{n}) \to [C(\mathbb{R}^{n})\cap L^{\infty}(\mathbb{R}^{n})]^{*}$. Now, this seems well-defined. What about the existence of such a sequence converging to $\delta$? Is it easy to construct such a sequence? $\endgroup$ – IamWill Dec 19 '19 at 15:35

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