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Ben McKay
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Does itthere exist a continuous bilinear form $\mathcal{B}$ on $\mathcal{S}(\mathbb{R})\times \mathcal{S}(\mathbb{R})$ such that \begin{equation} \mathcal{B}(\varphi_1, \varphi_2) =\int_{\mathbb{R}\times\mathbb{R}} \frac{\varphi_1(x) \varphi_2(y)}{\lvert x - y \rvert} \mathrm{d}x\mathrm{d}y \end{equation} for every $\varphi_1, \varphi_2 \in \mathcal{S}(\mathbb{R})$ with disjoint supports?

Context: A bilinear form $\mathcal{B}$ comes together with a kernel $K \in \mathcal{S}'(\mathbb{R}\times \mathbb{R})$ such that $\mathcal{B}(\varphi_1 ,\varphi_2) = \langle K , \varphi_1 \otimes \varphi_2 \rangle$ (kernelby the kernel theorem). The restriction to functions with disjoint supports allows us to avoid possible diagonal problems for the kernel $K$. It is possible to define a valid kernel of the form $K(x,y) = \lvert x - y \rvert^{-\lambda}$ outside the diagonal for $0<\lambda<1$. My guess is here that this is impossible for $\lambda = 1$, and therefore that it is impossible to find a bilinear form $\mathcal{B}$ satisfying the equation above when restricted to test functions with disjoint supports. Is this correct and if yes, how tocan one prove it?

Does it exist a continuous bilinear form $\mathcal{B}$ on $\mathcal{S}(\mathbb{R})\times \mathcal{S}(\mathbb{R})$ such that \begin{equation} \mathcal{B}(\varphi_1, \varphi_2) =\int_{\mathbb{R}\times\mathbb{R}} \frac{\varphi_1(x) \varphi_2(y)}{\lvert x - y \rvert} \mathrm{d}x\mathrm{d}y \end{equation} for every $\varphi_1, \varphi_2 \in \mathcal{S}(\mathbb{R})$ with disjoint supports?

Context: A bilinear form $\mathcal{B}$ comes together with a kernel $K \in \mathcal{S}'(\mathbb{R}\times \mathbb{R})$ such that $\mathcal{B}(\varphi_1 ,\varphi_2) = \langle K , \varphi_1 \otimes \varphi_2 \rangle$ (kernel theorem). The restriction to functions with disjoint supports allows to avoid possible diagonal problems for the kernel $K$. It is possible to define a valid kernel of the form $K(x,y) = \lvert x - y \rvert^{-\lambda}$ outside the diagonal for $0<\lambda<1$. My guess is here that this is impossible for $\lambda = 1$, and therefore that it is impossible to find a bilinear form $\mathcal{B}$ satisfying the equation above when restricted to test functions with disjoint supports. Is this correct and if yes, how to prove it?

Does there exist a continuous bilinear form $\mathcal{B}$ on $\mathcal{S}(\mathbb{R})\times \mathcal{S}(\mathbb{R})$ such that \begin{equation} \mathcal{B}(\varphi_1, \varphi_2) =\int_{\mathbb{R}\times\mathbb{R}} \frac{\varphi_1(x) \varphi_2(y)}{\lvert x - y \rvert} \mathrm{d}x\mathrm{d}y \end{equation} for every $\varphi_1, \varphi_2 \in \mathcal{S}(\mathbb{R})$ with disjoint supports?

Context: A bilinear form $\mathcal{B}$ comes together with a kernel $K \in \mathcal{S}'(\mathbb{R}\times \mathbb{R})$ such that $\mathcal{B}(\varphi_1 ,\varphi_2) = \langle K , \varphi_1 \otimes \varphi_2 \rangle$ (by the kernel theorem). The restriction to functions with disjoint supports allows us to avoid possible diagonal problems for the kernel $K$. It is possible to define a valid kernel of the form $K(x,y) = \lvert x - y \rvert^{-\lambda}$ outside the diagonal for $0<\lambda<1$. My guess is here that this is impossible for $\lambda = 1$, and therefore that it is impossible to find a bilinear form $\mathcal{B}$ satisfying the equation above when restricted to test functions with disjoint supports. Is this correct and if yes, how can one prove it?

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Goulifet
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No kernel of the form $\lvert x - y\rvert^{-1}$ on tempered distributions?

Does it exist a continuous bilinear form $\mathcal{B}$ on $\mathcal{S}(\mathbb{R})\times \mathcal{S}(\mathbb{R})$ such that \begin{equation} \mathcal{B}(\varphi_1, \varphi_2) =\int_{\mathbb{R}\times\mathbb{R}} \frac{\varphi_1(x) \varphi_2(y)}{\lvert x - y \rvert} \mathrm{d}x\mathrm{d}y \end{equation} for every $\varphi_1, \varphi_2 \in \mathcal{S}(\mathbb{R})$ with disjoint supports?

Context: A bilinear form $\mathcal{B}$ comes together with a kernel $K \in \mathcal{S}'(\mathbb{R}\times \mathbb{R})$ such that $\mathcal{B}(\varphi_1 ,\varphi_2) = \langle K , \varphi_1 \otimes \varphi_2 \rangle$ (kernel theorem). The restriction to functions with disjoint supports allows to avoid possible diagonal problems for the kernel $K$. It is possible to define a valid kernel of the form $K(x,y) = \lvert x - y \rvert^{-\lambda}$ outside the diagonal for $0<\lambda<1$. My guess is here that this is impossible for $\lambda = 1$, and therefore that it is impossible to find a bilinear form $\mathcal{B}$ satisfying the equation above when restricted to test functions with disjoint supports. Is this correct and if yes, how to prove it?