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Let $f,g,h\in L^2([0,1]^2)$ and let $K:\mathbb{R}^3\to \mathbb{C}$ be some smooth kernel with support containing $[0,1]^3$. Denote by $\|f\|_2$ the $L^2([0,1]^2)$ norm of $f$, and same with $g,h.$ If I have the three bounds $$\Bigg{|}\iiint_{\mathbb{R}^3} K(x,y,z)f(y,z)g(x,z)h(x,y)dxdydz\Bigg{|}\le \begin{cases} & C_1 \|f\|_2 \|g\|_4 \|h\|_4 \\ & C_2 \|f\|_4 \|g\|_2 \|h\|_4 \\ & C_3 \|f\|_4 \|g\|_4 \|h\|_2, \end{cases}$$ is there some way to interpolate between them? For example, is the integral bounded by $C\|f\|_3 \|g\|_3\|h\|_3$, where $C$ is a convex combination of $C_1, C_2, C_3$?

The only interpolation theorems I can find are for operators, not functionals, and I am not sure how to think of my integral as an operator in order to apply them. Note that I'm assuming the constants above are much better than 1, so I don't care about applying H\"older's inequality to get $some$ $C$.

If this is false, can you provide a counterexample?

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If you take the bilinear operator $T:(f,g) \mapsto \int K(x,y,z) f(y,z) g(x,z) ~\mathrm{d}z$, your three boundedness statements are equivalent to

  • $T: L^2 \times L^4 \to L^{4/3}$ with norm $C_1$
  • $T: L^4 \times L^2 \to L^{4/3}$ with norm $C_2$
  • $T: L^4 \times L^4 \to L^2$ with norm $C_3$

By multilinear interpolation (see Chapter 4, section 4.4 in Bergh and Löfström), the first two implies $T: L^{8/3} \times L^{8/3} \to L^{4/3}$ with norm $\sqrt{C_1 C_2}$.

Interpolating this with the third one gives $T: L^3 \times L^3 \to L^{3/2}$ with norm $(\sqrt{C_1C_2})^{2/3} C_3^{1/3} = (C_1 C_2 C_3)^{1/3}$.

This implies $\int_{[0,1]^2} T(f,g) \cdot h ~\mathrm{d}x~\mathrm{d}y \leq (C_1 C_2 C_3)^{1/3} \|f\|_{3}\|g\|_3 \|h\|_3$.

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    $\begingroup$ I should have just stopped by your office! Thanks Willie! $\endgroup$ – Maxim G. Jan 11 at 19:50

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