On a trilinear singular integral form with determinantal kernel (Q2809201)

The authors study a highly symmetric trilinear form which is a singular integral variant of the determinant functionals considered by \textit{P. T. Gressman} [Proc. Am. Math. Soc. 139, No. 7, 2473--2484 (2011; Zbl 1231.28006)] and \textit{S. I. Valdimarsson} [Rev. Mat. Iberoam. 28, No. 1, 25--55 (2012; Zbl 1248.44006)]. Define the trilinear form NEWLINE\[NEWLINE \begin{aligned} \Lambda(f,g,h) : &= \text{ p.v. } \int_{( {\mathbb R}^2)^3} f(x) g(y) h(z) \delta (x+y+z) \det \begin{pmatrix} 1 & 1 & 1 \\ x & y & z \end{pmatrix}^{-1} dxdydz \\ &= \frac{1}{3}\text{ p.v. } \int_{( {\mathbb R}^2)^3} \widehat{f}(\xi) \widehat{g}(\eta) \widehat{h}(\zeta) \det \begin{pmatrix} 1 & 1 & 1 \\ \xi & \eta & \zeta \end{pmatrix}^{-1} d\xi d\eta d\zeta. \end{aligned}NEWLINE\]NEWLINE Note that \(x,y,z\) are elements of \({\mathbb R}^2\), written as \(2 \times 1\) columns, and thus the determinant is taken of a \(3 \times 3\) matrix. The authors prove the following: if \(2<p,q,r < \infty\) and \(1/p + 1/q + 1/r =1\), then NEWLINE\[NEWLINE | \Lambda(f,g,h)| \leq C \| f \|_{L^p} \| g \|_{L^q} \| h \;Vert_{L^r}. NEWLINE\]NEWLINE The range of exponents is sharp.