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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 $g$.

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  • $\begingroup$ The suprema you write are (except for trivialities) infinite. You probably mean: For all $g\in\mathcal S(\mathbb R^M)$ there is $c$ such that $|T(f,g)| \le c\|f\|_{D_1}$ and similarly in the other variable. Abstract theory tells you that separately continuous bilinear forms on Frechet spaces are automatically jointly continuous. This, however, does not immediately answer the question because you have to change $D_1,D_2$. $\endgroup$
    – Jochen
    Feb 21 '18 at 13:19
  • $\begingroup$ @Jochen Thanks for the comment. You are right. I corrected the question. $\endgroup$
    – user72829
    Feb 21 '18 at 14:16
  • $\begingroup$ Your question is vague because you did not put quantifiers on $D_1$ and $D_2$. $\endgroup$ Mar 7 '18 at 17:09
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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).

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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.

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