# Convergence to Delta Dirac Distribution

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

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

## 1 Answer

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.

• 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? – IamWill Dec 19 '19 at 14:02
• 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? – IamWill Dec 19 '19 at 15:35