Some operators on spheres

Question

Let $S_2$ be the unit sphere in $\mathbb R^3$ equipped with normalized Haar measure. For a continuous function f and $\delta\in (-1,1)$ define $T_\delta f(x):=\int_{\{y:<x,y>=\delta\}}f(y)d_\delta(y)$ where $d_\delta(y)$ is the measure on the intersecting circle. Can it be shown that $T_\delta$ extends as a bounded self adjoint operator on $L_2(S_2)$ ?

Answer 1

Yes, it is true. It is more convenient to use the mean in the definition of $T_\delta f$, that is I divide by the lenght of the circle. Fix $\delta$ and for small $\epsilon>0$ let $E_\epsilon=\{(x,y) \in S_2 \times S_2: \delta \leq x\cdot y \leq \delta+\epsilon\}$. Let us introduce $$ T^\epsilon_\delta f(x)=\frac{1}{|E^x_\epsilon|} \int_{E^x_\epsilon} f(y) dy=\frac{1}{|E^x_\epsilon|} \int_{S_2} f(y) \chi_{E^x_\epsilon}(y)dy,$$ where $dy, | \cdot |$ denote measures on $S_2$ and $E_\epsilon^x=\{y \in S_2: (x,y) \in E_\epsilon\}$. Since the kernel of $T^\epsilon_\delta$ is symmetric, it is self-adjoint. It is also contractive in the sup-norm, then in the $L^1$ norm by duality, and in all $L^p$ norms (by interpolation or Young inequality). The final step consists in showing that $T^\epsilon_\delta f(x) \to T_\delta f(x)$ as $\epsilon \to 0$ for continuous $f$. This is checked using spherical coordinates (with $x$ the North pole).