TITLE: Smoothness of family of distributions
QUESTION [6 upvotes]: 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 $\Phi \in \mathscr{D}(X \times X)$?

REPLY [3 votes]: 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.