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?



Interpolation of a trilinear functional
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](http://www.math.chalmers.se/~bergh/Interpolation.pdf)), 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 $\int\_{[0,1]^2} T(f,g) \cdot h ~\mathrm{d}x~\mathrm{d}y \leq (C\_1 C\_2 C\_3)^{1/3} \|f\|\_{3}\|g\|\_3 \|h\|\_3$. 


