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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.

- Specific problem on distribution theory.
- Inequality about disbtribution in Functional Analysis by Rudin
- Origin of the name 'test functions'
- Tensor products of functions generate dense subspace?
- Is the Dirac Delta “Function” really a function?
- distribution with point support

- Difficulty in understanding a part in a proof from Stein and Shakarchi Fourier Analysis book.
- On the derivative of a Heaviside step function being proportional to the Dirac delta function
- Integral of Dirac delta function/distribution $\delta(x)$ with upper boundary equal to zero
- What do physicists mean with this bra-ket notation?
- Integration and differentiation of Fourier series
- When is a Fourier series analytic?
- How to prove that inverse Fourier transform of “1” is delta funstion?
- What happens to the frequency-spectrum if this sine-signal gets reset periodically?
- a question about Fourier transforms
- $\nabla^2(\|\boldsymbol{x}-\boldsymbol{x}_0\|^{-1})=-4\pi\delta(\boldsymbol{x}-\boldsymbol{x}_0)$ with distributions defined on Schwartz space

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.

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 = 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.

$ \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

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