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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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Yes it is possible. Let $K\in S'(\mathbb{R}^2)$ be the distribution acting on test functions $f(x,y)$ by $$ K(f)=\int_{|x-y|\ge 1} \frac{f(x,y)}{|x-y|}\ dx\ dy \ + \int_{|x-y|< 1} \frac{f(x,y)-f(x,x)}{|x-y|}\ dx\ dy\ . $$ What is impossible is to do this extension to the diagonal while preserving the degree -1 homogeneity. The only homogeneous distribution on $\mathbb{R}$ with this homogeneity is delta at the origin (up to scale).

A good reference for this type of questions is vol 1 of "Generalized Functions" by Gelfand and Shilov. This becomes interesting in the context of probability theory see: Squaring random Schwartz distributions

A related question is: can one interpret the energy field for the 2d Ising critical scaling limit as a random distribution?

Also a considerable amplification of this kind of methods for extending distributions to the diagonal is BPHZ renormalization. See, e.g., this lecture by Hairer: http://www.mathtube.org/lecture/video/bphz-theorem-stochastic-pdes-0

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(Edit: see below... I misread the question...) One particular way to regularize that integral, relating to meaningful things, is by viewing it as Hilbert transform applied to one Schwartz function, producing (at worst) a tempered distribution, which is then applied to the second Schwartz function.

Part of the point is that the Hilbert transform, or already the principal value integral attached to $1/x$ in one variable, is not a literal integral (against a locally $L^1$ function), although it is a perfectly fine tempered distribution (the derivative of the locally integrable $\log|x|$).

Another implicit aspect, mentioned by @Abdelmalek Abdesselam, is to think about such an extension problem, by (uniquely?!?) characterizing the restricted operator by symmetry/homogeneity, etc., and explicitly ask whether there exists an extension that preserves those features, and whether the extension is unique. (This can be usefully written as a homological issue, as I first saw in a paper of Casselman about extended notions of automorphic forms.) For example, the p.v. integral against $1/x$, rather than $1/x$ itself, is the unique distribution (up to multiples) with that homogeneity and parity, by the classification of distributions on $\mathbb R$ supported at $\{0\}$ and what amounts to an application of the snake lemma. (As usual, what can be usefully written as homological algebra, e.g., a snake lemma application, can also be done directly, in effect reproving a special case of the short-to-long business.)

EDIT: ooops, (thanks to Jean Duchon for noticing my error!): I dropped the absolute value in the question. Thus, this is not the Hilbert transform. In fact, in that vein, the symmetry and mildly homological extension argument proves that there is no distribution extending integration against $1/|x|$ (note the absolute values now), preserving the homogeneity and parity. (The positive-homogeneity is the same as $1/x$, but the parity is opposite. The existence of $\delta$, with the same positive-homogeneity and parity is equivalent to the non-extendability, in effect. The same arguments apply to Hadamard's finie-partie functionals, too.)

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    $\begingroup$ I think you forgot the |.|, your construction gives $\int\int\varphi_1(x)\varphi_2(y)/(x-y)\ dx\ dy$ $\endgroup$ – Jean Duchon Jun 29 '17 at 19:34
  • $\begingroup$ @JeanDuchon, oops! you're correct! So, the argument I sketched then proves the impossibility... Thanks for your observation! I will add a correction... $\endgroup$ – paul garrett Jun 29 '17 at 19:38
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You can try $\mathcal B(\varphi_1,\varphi_2)=\int \int\varphi'_1(x)\ \varphi_2'(y) (|x-y|\ln|x-y|)\ dx\ dy$ , i.e. $K$ is the second derivative $\partial_x\partial_y$ of (the distribution defined by) the locally integrable function $H(x,y)=|x-y|\ln|x-y|$ .

This extension is clearly non unique (since any distribution applied to $\varphi_1\varphi_2$ vanishes if the two functions have disjoint supports) but a somewhat standard solution (different from the one above) would use Schwartz's pseudo-function Pf. $r^m$ defined e.g. in Théorie des distributions, formula (II,3;4), a well-defined distribution whose restriction to $\{0\}^c$ coincides with the locally summable function $|x|^m$. This leads (via the Fourier transform of Pf. $r^{-1}$, formula (VII,7;18)) to $$\mathcal B(\varphi_1,\varphi_2)=-2[\int \hat{\varphi_1}(\xi)\ \hat{\varphi_2}(-\xi)\ln|\xi|\ d\xi\ +\ (\mathcal C+\ln(2\pi))\int\varphi_1\int\varphi_2]$$where $\mathcal C$ is Euler's constant. Or equivalently (through the definition of Pf. $r^{-1}$)$$\mathcal B(\varphi_1,\varphi_2)=\lim_{\varepsilon\to0}[\int\int_{|x-y|>\varepsilon}\frac{\varphi_1(x)\varphi_2(y)}{|x-y|}\ dx\ dy\ -\ I(\varepsilon)]$$where $I(\varepsilon)\propto \ln\varepsilon$ , i.e. the finite part of the possibly diverging integral.

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