# Interpolation of a trilinear functional

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?

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

• I should have just stopped by your office! Thanks Willie! – Max G. Jan 11 at 19:50