Is the Dirac delta a function?

Question

Is Dirac delta a function? What is its contribution to analysis?

What I know about it: It is infinite at 0 and 0 everywhere else. Its integration is 1 and I know how does it come.

Answer 1

The delta distribution is not a function from $\mathbb{R} \to \mathbb{R}$. For any such function $f$ that is zero except at finitly many points, as the delta distribution is, $\int f= 0$ (this holds for both the riemann and the lebesgue integral). Yet $\int \delta = 1$...

You can distributions as a function that maps functions (well, rather some functions) to real values though, i.e as a function $\mathbb{R}^\mathbb{R} \to \mathbb{R}$ (since it doesn't may every function to a real value, the domain is not really $\mathbb{R}^\mathbb{R}$ but a subset thereof).

The delta distribution is particularly simply, it just evaluates the function at $0$, i.e. $$ \delta(f) = f(0) \text{.} $$ Instead of $\delta(f)$, people often write $\int f \delta$, and that way you get $$ \int f \delta = f(0) \text{.} $$ But this is just notation - just as the $df$ and $dx$ in $\frac{df}{dx}$ aren't real numbers, yet sometimes you may treat them as if they were, $\delta$ isn't a function - yet sometimes it behaves like one. For example, you may use partial integration to compute $\delta'$ by doing $$ d'(f) = \int f \delta' = - \int f' \delta \text{.} $$ Thus, $\delta'$ is the distribution which, given a function $f$, evaluates the derivative of $f$ at $0$ (and multiplities by $-1$), i.e. returns $-f'(0)$. This is, in fact, the very definition the derivative of a distribution btw.

This syntax also allows you to treat every function (well, actually every interable function) $h$ as a distribution. Following the path laid out by the integral notation for distributions, you get $$ h(f) = \int h f = \int h(x) f(x) \, dx \text{.} $$

Note that this allows you to assign a derivative to a lot of functions which aren't otherwise differentiable. If you convert them into a distribution first, you can then set $$ h'(f) = \int h' f = -\int h f' \text{.} $$ provided that $f$ is suitable smooth.

This leaves the question of which $f$ are actually allowed here, i.e. which functions are mapped to real values open. You'll generally want them to be differentiable arbitrarily often, but that still leaves multiple choices. You'll have to read up on the theory of distributions to understand all the details.

Answer 2

To have a better understanding of what is the Delta Dirac "function", it is good to know what is a distribution (but this is not necessary). Let $\Omega\subset\mathbb{R}^N$ be a open set and $\mathcal{D}(\Omega)=C_0^\infty(\Omega)$. We define in $\mathcal{D}(\Omega)$ the following notion of convergence: we say that $\phi_n\to 0$ in $\mathcal{D}(\Omega)$ if

I - $\operatorname{spt}(\phi_n)\subset K\subset\Omega$, where $K$ is a fixed compact ($\operatorname{spt}(\phi_n)$ is the supoort of $\phi_n$),

II - For all $j$, $D_j\phi_n\to 0$ uniformly in $\Omega$.

We denote by $\mathcal{D}'(\Omega)$ the set of linear transformations $F:\mathcal{D}(\Omega)\to\mathbb{R}$ that are continuous with respect to the pseudo topology (the notion of convergence) that we defined, i.e. if $\phi_n\to 0$ in $\mathcal{D}(\Omega)$, then $\langle T,\phi_n\rangle=T(\phi_n)\to 0$. We call the elements of $\mathcal{D}'(\Omega)$ distributions.

Now we are able to define the famous delta Dirac "distribution". It is a distribution $\delta_{x}:\mathcal{D}(\Omega)\to\mathbb{R}$ defined by the formula ($x\in\Omega$) $$\langle\delta_{x},\phi\rangle=\phi(x)$$

You can easily check by using the definition that $\delta_x\in\mathcal{D}'(\Omega)$. Also, any funcion $u\in L^1_{loc}(\Omega)$ de fined a distribution by the formula $$\langle T_u,\phi\rangle=\int_\Omega u\phi$$

On the other hand, we can define a notion of convergence in $\mathcal{D}'(\Omega)$, to wit, $T_n\to T$ in $\mathcal{D}'(\Omega)$ if $\langle T_n,\phi\rangle\to\langle T,\phi\rangle $. You can check by using this definition that if $h_\epsilon$ is a mollifier sequence, then $T_{h_\epsilon}\to \delta _x$ in the sense of distributions.

From the last paragraph you can see, why usullay is said that the Delta Dirac "function" is one that satisfies $\delta(x)=1$, $\delta(y)=0$ if $y\neq x$ and $\int \delta(x)=1$. It is obvius that in the standard sense, this is impossible, but if we were to take the limit of the sequence $h_\epsilon$ pointwise, we would have something like this.

Also there is a lot of places in mathematics where the distribution of Dirac is used, for example, it is know that the Fundamental solution $\Gamma$ of the Laplace equation satisfies $-\Delta\Gamma=\delta$ in the sense of distributions, hence, if you wanna solve the problem $-\Delta u=f$, you can write (at least formally and with some assumptions) $u=\Gamma\ast f$, where $\ast$ stands for convolution.

There are a bunch of other applications of this distribution, but this place here is to small to write them all.

Answer 3

$ \delta(x) = \begin{cases} +\infty, & x = 0 \\ 0, & x \ne 0 > \end{cases}$

and which is also constrained to satisfy the identity $ > \int_{-\infty}^\infty \delta(x) \, dx = 1$

For example, the objects f(x) = δ(x) and g(x) = 0 are equal everywhere except at x = 0 yet have integrals that are different. According to Lebesgue integration theory, if f and g are functions such that f = g almost everywhere, then f is integrable if and only if g is integrable and the integrals of f and g are identical.

This is merely a heuristic characterization. The Dirac delta is not a function in the traditional sense as no function defined on the real numbers has these properties. The Dirac delta function can be rigorously defined either as a distribution or as a measure.

Source: Wikipedia