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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 $\Phi \in \mathscr{D}(X \times X)$?

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  • $\begingroup$ 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... $\endgroup$
    – Abdelmalek Abdesselam
    Commented Jan 23, 2020 at 15:14
  • $\begingroup$ ..might be enough. $\endgroup$
    – Abdelmalek Abdesselam
    Commented Jan 23, 2020 at 15:15
  • $\begingroup$ 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. $\endgroup$
    – Jochen Wengenroth
    Commented Jan 23, 2020 at 15:44
  • $\begingroup$ 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? $\endgroup$
    – Abdelmalek Abdesselam
    Commented Jan 23, 2020 at 16:19
  • $\begingroup$ 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? $\endgroup$
    – Abdelmalek Abdesselam
    Commented Jan 23, 2020 at 16:31

1 Answer 1

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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.

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  • $\begingroup$ 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 $\mathscr{D}'(\mathbb{R}\times X\times X)$? $\endgroup$
    – Abdelmalek Abdesselam
    Commented Jan 23, 2020 at 23:31
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    $\begingroup$ @ Abdelmalek Abdesselam $T_\lambda(x,y)$ belongs to $C^\infty(\mathbb R, \mathscr{D}^\prime(X \times X) = C^\infty(\mathbb R)\bar\otimes \mathscr{D}^\prime(X \times X)\subset \mathscr{D}^\prime(\mathbb R\times X \times X)$. $\endgroup$
    – Peter Michor
    Commented Jan 24, 2020 at 10:05

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