📂Distribution Theory

The Dirac Delta Function Rigorously Defined through Distributions

Definition¹

Let’s define the functional of the space of test functions $\mathcal{D}(\mathbb{R}^{n})$ as follows and call it the Dirac delta function.

$$\delta_{a}(\phi):=\phi (a)$$

Then, the Dirac delta function becomes a distribution. It is briefly represented as follows if $a=0$.

$$\delta=\delta_{0}$$

Explanation

The Dirac delta function, which was not strictly defined due to having divergent values and was roughly termed as a function, is rigorously defined by the above definition.

$$\delta_{a} (\phi) = \int \delta (x-a)\phi (x)dx=\phi (a)$$

However, since it cannot be defined as a locally integrable function, it is not a regular distribution. To avoid confusion with the conventional Dirac delta function, the delta function as a distribution will be referred to as the delta distribution.

Proof

• Part 1. Linearity

  For $\alpha, \beta \in \mathbb{C}$ and $\phi, \psi \in \mathcal{D}$,

  $$\begin{align*} \delta_{a}(\alpha \phi + \beta \psi) &= (\alpha \phi+\beta\psi)(a) \\ &=\alpha \psi (a) +\beta \psi (a) \\ &= \alpha\delta_{a}+\beta \delta (a) \end{align*}$$ hence, the delta distribution is linear.

• Part 2. Continuity

  Assume that $\phi_{j} \to \phi \text{ in } \mathcal{D}$. Then, the following holds.

  $$\begin{align*} \left| \delta_{a}(\phi _{j}) -\delta_{a}(\phi) \right| &= \left| \phi_{j}(a)-\phi (a)\right| \end{align*}$$

  If $\phi_{j} \to \phi \text{ in } \mathcal{D}$, then $\lim \limits_{j \to \infty}\left| \phi_{j}(a)-\phi (a) \right|=0$, hence $\delta_{a}(\phi_{j}) \to \delta_{a}(\phi)$.

Because $\delta_{a}$ is linear and continuous, it is a distribution.

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