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A Dirac distribution or Dirac $\delta$-distribution $\delta(p)$ is the distribution that is given by evaluating a function at a point $p$.

That is, the Dirac $\delta(p)$ function is the distribution defined by $$\langle\delta(p),\phi \rangle=\phi(p)$$

This allows us to write $$\phi(p)=\int_0^1\delta(p)\phi(x)dx$$

Suppose we want to write

$$\int\phi(x)\phi(x)dx=\int\int_0^1\delta( x-y)\phi(x)\phi(y)dxdy$$

How should we define $\delta( x-y)$?

In this page Free quantum fields on example 14.4 they have
$\delta( x-y) \in \Gamma'(E \boxtimes E)$ where $\Gamma'(E \boxtimes E)$ is the dual of the space of section of the bundle $\Gamma(E \boxtimes E)$ and And $E$ is the real line bundle

amilton moreira
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Your notation is inconsistent here: If you define $\delta(p)$ via $\langle \delta(p),f \rangle = f(p)$, then you have to write something like $f(p) = \int \delta(p)(x)f(x)\mathrm{d}x$ for the integral notation.

A more common notation would be to write the point at which the $\delta$ evaluates as a subscript, i.e. $\langle \delta_p,f\rangle = f(p)$ and then the integral notation is $f(p) = \int \delta_p(x)f(x)\mathrm{d}x$. In that notation, integrating against $\delta(x-y)$ is simply another way to write $\delta_x(y)$ or $\delta_y(x)$, there is no need to define the action of a delta function on multiple functions: Your double integral in terms of the normal $\delta$ is $$ \int \langle \delta_y, \phi(y)\cdot \phi \rangle\mathrm{d}y.$$

ACuriousMind
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    why would $δ(p)$ depend on $x$. According to the defenition it is alinear map that eats a tests function $f$ and splits a number $f(p)$? – amilton moreira Jan 14 '23 at 14:02
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    @amiltonmoreira It's just how the notation works. The inner product $\langle f,g\rangle$ of two functions is $\int f(x)g(x)\mathrm{d}x$, so when you want to write $\langle \delta_p,f\rangle$ as an integral you write $\int \delta_p(x)f(x)\mathrm{d}x$ even though $\delta_p$ technically is not a function and $\delta_p(x)$ outside of an integral doesn't have any meaning. – ACuriousMind Jan 14 '23 at 14:36
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    I edited my question – amilton moreira Jan 14 '23 at 18:21
  • @amiltonmoreira That doesn't change my answer. You can interpret what I wrote as a distribution on pairs of functions simply as $(f,g)\mapsto \int f(y)\langle \delta_y, g\rangle\mathrm{d}y$, which for $(f,g)$ from some space $V$ makes this a member of the dual space $V'$. There is no need to define the $\delta$ any differently. – ACuriousMind Jan 14 '23 at 18:30
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    in the integral $\int \langle \delta_y, \phi(y)\cdot \phi \rangle\mathrm{d}y.$ means for each $y$ we have a distribution $\delta_y$? – amilton moreira Jan 14 '23 at 18:52