# Smoothness of family of distributions

Let $$X$$ be a compact manifold. Denote by $$\mathscr{D}^\prime(X \times X)$$ the space of tempered distributions on the cartesian product $$X \times X$$. Given two test functions $$\varphi, \psi \in \mathscr{D}(X)$$, an element $$T \in \mathscr{D}^\prime(X \times X)$$ can be evaluated at the function $$\varphi \otimes \psi$$ on $$X \times X$$ defined by $$(\varphi \otimes \psi)(x, y) := \varphi(x) \psi(y)$$.

Suppose now that $$T_\lambda \in \mathscr{D}^\prime(X \times X)$$, $$\lambda \in\mathbb{R}$$, is a family of distributions such that $$\lambda \longmapsto T_\lambda[\varphi \otimes \psi]$$ is a smooth function from $$\mathbb{R}$$ to $$\mathbb{R}$$ for any two $$\varphi, \psi \in \mathscr{D}(X)$$.

Q: Does it follow that also the function $$\lambda \longmapsto T_\lambda[\Phi]$$ is smooth for every $$\Phi \in \mathscr{D}(X \times X)$$?

• My rapid guess is the answer is yes. I think the compact hypothesis is a distraction and should be removed. I would try first $X=\mathbb{R}$ and if it works, it should adapt to reasonable smooth manifolds (second countable). For $X=\mathbb{R}$, I would also first consider the analogue for $\mathscr{S},\mathscr{S}'$ instead of $\mathscr{D},\mathscr{D}'$. The condition on the functions of $\lambda$ being smooth, i.e., being in $\mathscr{E}$, should be replaced in the analogue by being in $\mathscr{O}_{\rm M}$, then the multiplier space characterization of the latter and the Kernel Theorem... – Abdelmalek Abdesselam Jan 23 '20 at 15:14
• ..might be enough. – Abdelmalek Abdesselam Jan 23 '20 at 15:15
• If the answer is yes, then the family $T_\lambda$ for $\lambda$ in a compact set should be pointwise bounded on $\mathscr D(X\times X)$ and hence equicontinuous, which roughly means that the continuity estimates for $T_\lambda$ only depend on a fixed number of derivatives with uniform constants. On the other hand, if you have this kind of equicontinuity a priori for $T_\lambda$ and all of its derivatives w.r.t. $\lambda$, an approximation argument might yield the desired result. – Jochen Wengenroth Jan 23 '20 at 15:44
• Another guess: for good behavior one needs the kernel $T_{\lambda}(x,y)$ to be a distribution in $(\lambda,x,y)$. I don't yet see how this follows from the hypotheses. Perhaps using the uniform boundedness principle? – Abdelmalek Abdesselam Jan 23 '20 at 16:19
• In other words: the hypotheses define a bilinear map $\mathscr{D}(X)^2\rightarrow \mathscr{E}(\mathbb{R})$, but do they imply this bilinear form is continuous? – Abdelmalek Abdesselam Jan 23 '20 at 16:31

Here is a proof using convenient analysis: By the kernel theorem $$\mathscr{D}^\prime(X \times X) = L(\mathscr{D}(X),\mathscr{D}^\prime(X))$$ and by The Convenient setting of Global Analysis, 5.18 (which is just the uniform boundedness principle) and 2.14 we have \begin{align*} &\lambda\mapsto T_\lambda(\Phi) \in \mathbb R \text{ is }C^\infty \quad\forall \Phi \in \mathscr{D}(X \times X) \\ \iff &\lambda\mapsto T_\lambda \in \mathscr{D}^\prime(X \times X) \text{ is }C^\infty \quad &\text{by 2.14} \\ \iff &\lambda\mapsto T_\lambda \in L(\mathscr{D}(X),\mathscr{D}^\prime(X)) \text{ is }C^\infty \\ \iff & \lambda\mapsto T_\lambda(\varphi) \in \mathscr{D}^\prime(X) \text{ is }C^\infty\quad \forall \varphi\in \mathscr{D}(X) &\text{by 5.18} \\ \iff & \lambda\mapsto T_\lambda(\varphi)(\psi) \in \mathbb R \text{ is }C^\infty \quad\forall \varphi,\psi\in \mathscr{D}(X) \quad &\text{by 2.14} \end{align*} Up to Frechet spaces convenient smoothness equals each other reasonable notion, but beyond it differs. A short description of convenient analysis can be found in Wikipedia.
• Interesting. I thought you would contribute an answer because, if I understood correctly, convenient analysis builds on the notion of smooth curves in LCTVS's. Do you know if $T_{\lambda}(x,y)$ belongs to $\mathscr{D}'(\mathbb{R}\times X\times X)$? – Abdelmalek Abdesselam Jan 23 '20 at 23:31
• @ Abdelmalek Abdesselam $T_\lambda(x,y)$ belongs to $C^\infty(\mathbb R, \mathscr{D}^\prime(X \times X) = C^\infty(\mathbb R)\bar\otimes \mathscr{D}^\prime(X \times X)\subset \mathscr{D}^\prime(\mathbb R\times X \times X)$. – Peter Michor Jan 24 '20 at 10:05