Schwartz kernel theorem and order of distribution Let $\mathcal{S}(\mathbb{R}^N)$ be  a space of Schwartz functions and let $T: \mathcal{S}(\mathbb{R}^N)\times \mathcal{S}(\mathbb{R}^M) \to \mathbb{C}$ be a bilinear functional such that there exist $c_1,c_2>0$ and for every $g\in\mathcal{S}(\mathbb{R}^M)$ it holds
$$\forall_{f\in \mathcal{S}(\mathbb{R}^N)} |T(f,g)| \leq c_1 ~\|f\|_{D_1} $$
and for every $f\in\mathcal{S}(\mathbb{R}^N)$ it holds
$$\forall_{g\in \mathcal{S}(\mathbb{R}^M)} |T(f,g)| \leq c_2~\|g\|_{D_2} $$
where
$$ \|f\|_D = \sum_{|\alpha|,|\beta|<D}\|f\|_{\alpha\beta} $$ 
and
$$\|f\|_{\alpha,\beta}=\sup_{x\in\mathbf{R}^N} \left |x^\alpha D^\beta f(x) \right |.$$
It means that $T$ is a Schwartz distribution separately in the first and second argument. By the Schwartz kernel theorem $T$ is a distribution on $\mathcal{S}(\mathbb{R}^{N+M})$. Is it true that this distribution satisfies the following condition
$$\forall_{f\in \mathcal{S}(\mathbb{R}^N),g\in \mathcal{S}(\mathbb{R}^M)} |T(f,g)| \leq const ~\|f\|_{D_1}\|g\|_{D_2} $$
for some constant independent of $f$ and $g$.
 A: Consider $N=M=1$ and $$T(f,g)=\int_{-\infty}^\infty f'(x)g'(x)dx.$$ For each fixed $g$ integration by parts gives $$T(f,g)=fg'|_{-\infty}^\infty - \int_{-\infty}^\infty f(x)g''(x)dx =- \int_{-\infty}^\infty f(x)g''(x)dx $$
and hence $|T(f,g)|\le c(g) \|f\|_0$. Similarly, $|T(f,g)|\le c(f) \|g\|_0$.
However, there is no constant such that $|T(f,g)|\le c\|f\|_0\|g\|_0$ (otherwise, $T$ would extend to a continuous linear functional on the space of continuous functions on $\mathbb R^2$ with compact support and would have a representation by a measure on $\mathbb R^2$ which is not true).
A: A small complement to Jochen's answer. Separate continuity implies full continuity, essentially by the uniform boundedness principle. This then implies the existence of seminorms $||\cdot||_{E_1}$ and $||\cdot||_{E_2}$ so that
$$
|T(f,g)|\le cst\ ||f||_{E_1} ||g||_{E_2}
$$
What is impossible in general is to have the seminorms $||\cdot||_{E_1}$ and $||\cdot||_{E_2}$ be the same as the original ones $||\cdot||_{D_1}$ and $||\cdot||_{D_2}$, as shown by Jochen.
