The diagonal delta is a distribution supported on the diagonal

Define $d_X:X\to X\times X$ by $d_X(x)=(x,x)$. For $\mathrm{\Psi }\in \mathcal{D}(X\times X)$, the diagonal restriction $x\mapsto \mathrm{\Psi }(d_X(x))$ has first derivatives ${\mathrm{\partial }}_{x_j}\mathrm{\Psi }(x,x)+{\mathrm{\partial }}_{y_j}\mathrm{\Psi }(x,x)$ by [F2]. Iterating this formula on the smooth partial derivatives of $\mathrm{\Psi }$ shows the restriction has continuous derivatives of every order, so it is smooth. Its support is contained in the first-coordinate projection of $\mathrm{supp}\mathrm{\Psi }$, which is compact as a continuous image of a compact set. The componentwise convention in the Statement and [F3] make the displayed integral well defined.

The assignment is complex-linear. Fix a nonempty compact $K\subseteq X\times X$ and a compact rectangle $R\subseteq X$ containing its first-coordinate projection. For every $\mathrm{\Psi }$ supported in $K$, $\mathrm{\mid }⟨{\delta }_{{\mathrm{\Delta }}_X},\mathrm{\Psi }⟩\mathrm{\mid }\le \mathrm{vol}(R){\sup }_{(x,y)\in K}\mathrm{\mid }\mathrm{\Psi }(x,y)\mathrm{\mid }\mathrm{.}$ Thus the seminorm $q(\mathrm{\Psi })=\mathrm{\mid }⟨{\delta }_{{\mathrm{\Delta }}_X},\mathrm{\Psi }⟩\mathrm{\mid }$ is continuous on every fixed-support test space. By the LF topology definition in [F1], such seminorms are continuous on $\mathcal{D}(X\times X)$. Therefore ${\delta }_{{\mathrm{\Delta }}_X}\in {\mathcal{D}}^{\mathrm{\prime }}(X\times X)$.

If an open set $U\subseteq X\times X$ is disjoint from ${\mathrm{\Delta }}_X$, then every test supported in $U$ vanishes at every $(x,x)$. Its pairing with ${\delta }_{{\mathrm{\Delta }}_X}$ is zero, so [F5] gives $\mathrm{supp}{\delta }_{{\mathrm{\Delta }}_X}\subseteq {\mathrm{\Delta }}_X$.

Fix $(x_0,x_0)\in {\mathrm{\Delta }}_X$ and any open neighborhood $U$ of this point. There is a ball $B$ about $x_0$ with $\bar{B}\times \bar{B}\subseteq U$. By [F4] choose $h\in \mathcal{D}(X)$ with $h\ge 0$, $h(x_0)=1$, and support in $B$. By continuity choose a closed nondegenerate cube $Q$ inside $B$ on which $h\ge 1\mathrm{/}2$, and choose a sufficiently large cube $R$ containing $\mathrm{supp}h$ and $Q$ with the faces of $Q$ among its grid cuts. The lower Darboux sum of the nonnegative function $h^2$ on this grid is at least $\mathrm{vol}(Q)\mathrm{/}4>0$, so its real Riemann integral is positive. The product test $\mathrm{\Psi }(x,y)=h(x)h(y)$ is supported in $U$, and $⟨{\delta }_{{\mathrm{\Delta }}_X},\mathrm{\Psi }⟩={\int }_{X}h(x{)}^2{d}^{4}x>0$. Thus ${\delta }_{{\mathrm{\Delta }}_X}$ does not vanish on any neighborhood of $(x_0,x_0)$, so $(x_0,x_0)\in \mathrm{supp}{\delta }_{{\mathrm{\Delta }}_X}$ by [F5]. Since the point was arbitrary, ${\mathrm{\Delta }}_X\subseteq \mathrm{supp}{\delta }_{{\mathrm{\Delta }}_X}$. Together with step 3.1 this gives equality.

For every $\mathrm{\Psi }$, the diagonal restrictions of $\mathrm{\Psi }(x,y)$ and $\mathrm{\Psi }(y,x)$ coincide, so the distribution is symmetric. The diagonal has empty interior: every product neighborhood of a point $(x,x)$ contains $(x,y)$ with $y\mathrm{\ne }x$. If a smooth function supported on the diagonal were nonzero at a point, continuity would make it nonzero on an open neighborhood, contradicting empty interior. Therefore every such smooth function is zero.