# Proof of the construction of Dirac Delta

The Dirac Delta function pops up in a wide variety of applications, especially in applications that require Laplace and Fourier transforms.

But my question is: what's the proof that the distribution can be constructed in the first place? How can we determine that there exists a function that is zero at all points but one, this exception is at infinity, and that the integration of this distribution across its entire domain is one?

## 4 Answers

The Dirac delta is the distribution defined by $$\langle \delta_{x_0},\varphi \rangle = \varphi(x_0)$$ for all $\varphi \in \mathcal{D}(\mathbb{R}^n)$. It is not a function, so your question cannot be answered.

There is no such "function". But in mathematics we use a construction in functional analysis. The "functional" $F$ defined by $F(g) = g(0)$ for all continuous $g$ is the one called "Dirac delta". In physics, they "pretend" there is an actual function, and do computations with it. Since it often produces useful results, there is little reason to stop doing that.

• Yep, the same voodoo like the Leibniz Kalkül for differentials. Nice for separation of variables, chain rule etc. – mvw May 29 '15 at 14:28

But my question is: what's the proof that the distribution can be constructed in the first place? How can we determine that there exists a function that is zero at all points but one, this exception is at infinity, and that the integration of this distribution across its entire domain is one?

Strictly only the last claim holds. $\langle f, \delta\rangle = \int f(x) \delta(x) \, dx = f(x_0)$.

You can find an elementary treatment in Lighthill's book on Generalised Functions.

http://www.amazon.com/gp/aw/d/0521091284/ref=mp_s_a_1_1?qid=1432930439&sr=8-1&pi=SL75_QL70&keywords=Generalised+functions