Let $(X,\mathscr{B},\mu)$ and $(Y,\mathscr{D},\nu)$ are two measure space, where $X$ is a compact metric space and $\mathscr{B}$ is the Borel $\sigma$- algebra on $X$. Let $\alpha:X\rightarrow Y$ be a measure preserving map(special map). Let $\mu=\int\mu_y\,d\nu(y)$ be the disintegration as described in Theorem 5.8. Define the measure $\mu\,\times_{Y}\,\mu$ on the product Borel $\sigma$-algebra of $X\,\times\,X$ by $\mu\,\times_{Y}\,\mu(A):=\int_{y}(\mu_y\,\times\,\mu_y)(A)\,d\nu(y)$. If $H\in L^2(X\,\times\,X,\mathscr{B}\times\mathscr{B},\mu\,\times_Y\,\mu)$ and $\phi\in L^2(X,\mathscr{B},\mu)$, the convolution $H*\phi$ is defined by $(H*\phi)(x)=\int H(x,x')\phi (x')\,d\mu_{\alpha(x)}(x')$.
Problem: Prove that $\left\lVert H*\phi\right\lVert_y\leq \left\lVert H\right\lVert_y\left\lVert\phi\right\lVert_y$, where $\left\lVert H*\phi\right\lVert_y$ is the $L^2$ norm w.r.t the measure $\mu_y$, and similarly others.
I got stuck in this while studying Compact extension from 'Recurrence in Ergodic Theory and Combinatorial Number Theory' by Harry Furstenberg. For the details discussion, see here.
Since $\alpha$ is a special map, $\mu_y$ is supported on $\alpha^{-1}(\{y\})$. Therefore,$$||H*\phi||_y^2 = \int_X |H*y|^2(x_0)d\mu_y(x_0) = \int_X \left| \int_X H(x_0,x)\phi(x)d\mu_{\alpha(x_0)}(x)\right|^2 d\mu_y(x_0)$$ $$ = \int_X \left|\int_X H(x_0,x)\phi(x)d\mu_y(x)\right|^2d\mu_y(x_0)$$ $$\le \int_X \left[\int_X |H(x_0,x)|^2d\mu_y(x)\right]\cdot\left[\int_X |\phi(x')|^2 d\mu_y(x')\right]d\mu_y(x_0)$$ $$ = \int_X \left[\int_X \int_X |H(x_0,x)|^2d\mu_y(x)d\mu_y(x_0)\right]|\phi(x')|^2d\mu_y(x')$$ $$ = ||H||_y^2 ||\phi||_y^2.$$