Defining Dirac's Delta in one dimensional with two variable I have a problem in one dimension (x-axis) with two spatial variables $x_{1}$ and $x_{2}$ and I have an Dirac's Delta $\delta(x_{1}-x_{2})$. My question is: how can I properly define this $\delta$?
I thought about
$$\delta(x_{1}-x_{2}) \varphi(x) = \varphi(x_{2})$$ for all $\varphi \in C^{\infty}_{c}(\mathbb{R})$.
Is this definition mathematically precise? I don't know if this is correct (I thought maybe $\varphi(x_{1},x_{2})$ and $\delta(x_1 - x_2) \varphi(x_1, x_2) = \varphi(x_{2})$ but I'm not sure either).
Thanks.
 A: Let $L=\{(x,y)\in\Bbb{R}^2\,|\,\,x=y\}$, i.e this is the "diagonal line" in the plane. We can now define a (tempered) distribution $\delta_L$ on $\Bbb{R}^2$ by setting for each Schwartz function $\phi\in\mathcal{S}(\Bbb{R}^2)$,
\begin{align}
\langle\delta_L,\phi\rangle&:=\int_L\phi(x,y)\,dl(x,y)=\int_{\Bbb{R}}\phi(t,t)\,dt\,\tag{$*$}
\end{align}
Here, $dl$ denotes the "volume element" on the submanifold $L$, or in this case simply the line element, which as the final equality shows is a very simple one-dimensional integral.
This should intuitively be a reasonably definition because the quantity $\delta(x-y)$, where $(x,y)\in\Bbb{R}^2$ should be $0$ if $x\neq y$, so if we "integrate" a function $\phi$ against $\delta(x-y)$, then
\begin{align}
\int_{\Bbb{R}^2}\delta(x-y)\phi(x,y)\,dxdy&\,\,\,\, ``="\int_{L}\phi(x,y),
\end{align}
and we should be integrating with respect to the line element on $L$, rather than the Lebesgue measure on $\Bbb{R}^2$, because otherwise we get a trivial $0$ (since a line $L$ in the plane has zero Lebesgue measure). Hopefully this heuristic reasoning motivates the precise definition $(*)$ above.
