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What is the proper definition of the distribution underneath this notation?

Since "$x\mapsto \delta(x)$" is define with respect to the distribution $\delta$ such that $$\langle \delta, f \rangle = f(0) $$ is "$(x,y)\mapsto \delta(x+y)$" define with respect to the distribution $T$ such that (with $g\in \mathcal{S}(\mathbb{R}^2)$) $$\langle T, g \rangle = \int_\mathbb{R} g(x,-x) d x \ ?$$

Attempt :

If I try to use "$(x,y)\mapsto \delta(x+y)$" to define my distribution as if it were a true function (and assuming Fubini is working for the sake of the guess), I would have $$\hspace{-50pt} \begin{align*} \langle T, g \rangle & =\iint g(x,y)\delta(x+y)dx dy \\ & = \int \bigg(\int g(x,y)\delta(x+y) dx\bigg) dy \\ & = \int g(-y,y) dy \tag{$= \int g(x,-x) dx$, with $x=-y$} \end{align*} $$ (since $\int g(x,y)\delta(x+y) dx=\langle \delta_{-y}, g(.,y) \rangle=g(-y,y)\ $).

Generalisation

If $h:\mathbb{R}^n\rightarrow \mathbb{R}$ the distribution $\delta_h$ denoted by $\delta(h(x))$ is define by $$ \langle \delta_h, g \rangle = \int_{\{h(y)=0\}} g(x) d x, \quad g\in \mathcal{S}(\mathbb{R}^n) $$

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  • $\begingroup$ @XanderHenderson see math.stackexchange.com/questions/2657870/an-identity-on-deltaxy ; people use $\delta(x+y)$ as a "function" of 2 variables, so it implies a distribution over test-functions define on $\mathbb{R}^2$, yes! $\endgroup$ – Netchaiev Feb 26 '18 at 14:39
  • $\begingroup$ @XanderHenderson see also math.stackexchange.com/questions/1705991/… $\endgroup$ – Netchaiev Feb 26 '18 at 14:41
  • $\begingroup$ Your hunch is pretty much correct. The delta only kills one of the two integrals and enforces the relationship between x and y. $\endgroup$ – Spencer Feb 26 '18 at 14:56
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    $\begingroup$ With regards to your generalization it seems to be missing an important property $\delta(g(x)) =\sum \delta(x-x_i) / |g'(x_i)|$ $\endgroup$ – Spencer Feb 26 '18 at 15:26
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    $\begingroup$ I originally voted to close the question because it consisted of about four lines, none of which were clear. The edits are a vast improvement, and I have retracted my close vote. $\endgroup$ – Xander Henderson Feb 26 '18 at 16:26
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In order to define $T(x,y)=\delta(x+y)$ as an element of $\mathcal{D}'(\mathbb{R}^2)$ one can use at least three different methods.

Method 1:

As in the OP, after the formal manipulations one can use the last line as a definition. Namely, for any smooth compactly supported function $g(x,y)$ on $\mathbb{R}^2$, one sets $$ \langle T, g\rangle:=\int g(x,-x)\ dx\ . $$

Method 2:

Take the tensor product of the two distribution in one variable given by $\delta(x)$ and the constant function (of $y$) equal to one. This gives $S=\delta\otimes 1\in \mathcal{D}'(\mathbb{R}^2)$ which one can write formally as $$ S(x,y)=\delta(x)\ . $$ Then use the definition of pull-backs of distributions by diffeomorphisms in order to define $T(x,y):=S(x+y,y)$.

Method 3:

For $n\ge 1$ let $$ \phi_n(x)=\frac{n}{\sqrt{2\pi}} e^{-\frac{n^2 x^2}{2}} $$ and $$ T_n(x,y)=\phi_n(x+y)\ . $$ The function $T_n$ is locally integrable and thus defines a distribution $T_n$. Then let $T:=\lim_{n\rightarrow \infty} T_n$ in the (strong) topology of $\mathcal{D}'(\mathbb{R}^2)$.

All these methods are equivalent, i.e., produce the same distribution $T$. For the generalization with $h$, I recommend trying Method 3 first (in order to realize that this requires some "niceness" hypotheses on $h$).

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