Sea $f_1,f_2,f_3:\mathbb{R}^2\to\mathbb{R}_{\ge 0}$ ser medible, acotada y con soporte compacto. Demostrar que
$$\int_{\mathbb{R}^3}f_1(y,z)f_2(x,z)f_3(x,y)d(x,y,z)\le\lVert f_1\rVert_{L^2(\mathbb{R}^2)}\lVert f_2\rVert_{L^2(\mathbb{R}^2)}\lVert f_3\rVert_{L^2(\mathbb{R}^2)}$$ Por el Teorema de Tonelli y Cauchy, \begin{align} \int_{\mathbb{R}^3}f_1(y,z)f_2(x,z)f_3(x,y)d(x,y,z) &=\int_{\mathbb{R}^2}\left(f_1(y,z)\int_{\mathbb{R}}f_2(x,z)f_3(x,y)dx\right)d(y,z)\\ &\le\sqrt{\int_{\mathbb{R}^2}f_1^2(y,z)d(y,z)\int_{\mathbb{R}^2}\left(\int_{\mathbb{R}}f_2(x,z)f_3(x,y)dx\right)^2d(y,z)} \\ &\le\lVert f_1\rVert_{L^2(\mathbb{R}^2)}\sqrt{\int_{\mathbb{R}^2}\left(\int_{\mathbb{R}}f_2^2(x,z)f_3^2(x,y)dx\right)d(y,z)} \\ &=\Vert f_1\Vert_{L^2(\mathbb{R}^2)}\sqrt{\int_{\mathbb{R}^3}f_2^2(x,z)f_3^2(x,y)d(x,y,z)} \\ &=\lVert f_1\rVert_{L^2(\mathbb{R}^2)}\sqrt{\int_{\mathbb{R}^2}\left(f_2^2(x,z)\int_{\mathbb{R}}f_3^2(x,y)dy\right)d(x,z)} \\ &\le\lVert f_1\rVert_{L^2(\mathbb{R}^2)}\sqrt[4]{\int_{\mathbb{R}^2}f_2^4(x,z)d(x,z)\int_{\mathbb{R}^2}\left(\int_{\mathbb{R}}f_3^2(x,y)dy\right)^2d(x,z)} \end{align} Como puedes ver, esto no parece ir a ninguna parte.
