Fall 2011 - Problem 2

Solution.

1. Let $A \in \SO\p{3}$, so that $\abs{\det{A}} = 1$, $A^{-1} = A^\trans$, and $A$ is an automorphism of $\R^3$. In particular, $A^{-1}x \cdot \xi = x \cdot A\xi$, so by change of variables,

   $$\int_{S^2} e^{ix \cdot \xi} \,\diff\sigma\p{x} = \abs{\det{A^{-1}}} \int_{S^2} e^{iA^{-1}x \cdot \xi} = \int_{S^2} e^{ix \cdot A\xi} \,\diff{x}.$$

   In other words, this quantity is invariant under rotations, so it only depends on $\abs{\xi}$. Hence, it suffices to calculate the integral when $\xi = \p{0, 0, \abs{\xi}}$. By using spherical coordinates, we have

   $$\begin{aligned} \int_{S^2} e^{ix \cdot \xi} \,\diff\sigma\p{x} &= \int_{S^2} \cos\p{\abs{\xi}x_3} + i\sin\p{\abs{\xi}x_3} \,\diff\sigma\p{x} \\ &= \int_0^\pi \int_0^{2\pi} \p{\cos\p{\abs{\xi}\cos\phi} + i\sin\p{\abs{\xi}\cos\phi}} \sin\phi \,\diff\theta \,\diff\phi \\ &= 2\pi \int_{-1}^1 \cos\p{\abs{\xi}u} + i\sin\p{\abs{\xi}u} \,\diff{u} && (u = \cos\phi) \\ &= \frac{4\pi\sin{\abs{\xi}}}{\abs{\xi}}. \end{aligned}$$

2. Define

   $$L\p{f} = \int_{S^2} \int_{S^2} f\p{x + y} \,\diff\sigma\p{x} \,\diff\sigma\p{y}$$

   for $f \in C^\infty_c\p{\R^3}$. Since $C^\infty_c\p{\R^3}$ is dense in $L^2\p{\R^3}$ with respect to the $L^2$ norm, it suffices to show that there exists a constant $C > 0$ so that $\abs{L\p{f}} \leq C\norm{f}_{L^2}$ for any $f \in C^\infty_c\p{\R^3}$. The derivatives of such an $f$ are compactly supported as well, so $f$ is in the Schwartz space. In particular, we may apply the inverse Fourier transform and a change of variables to get

   $$f\p{x} = \int_{\R^3} e^{2\pi ix \cdot \xi}\hat{f}\p{\xi} \,\diff\xi = \int_0^\infty r^2 \int_{S^2} e^{2\pi ix \cdot r\xi}\hat{f}\p{r\xi} \,\diff\sigma\p{\xi} \,\diff{r}.$$

   Notice that by integrating by parts, we see that $\norm{\hat{f}}_{L^\infty\p{rS^2}}$ decays faster than any power of $r$, since $f^{\p{n}}$ has compact support. Thus, the integrand is absolutely convergent, and so we may apply Fubini's theorem in the following:

   $$\begin{aligned} L\p{f} &= \int_{S^2} \int_{S^2} f\p{x + y} \,\diff\sigma\p{x} \,\diff\sigma\p{y} \\ &= \int_{S^2} \int_{S^2} \int_0^\infty r^2 \int_{S^2} e^{2\pi i\p{x+y} \cdot r\xi}\hat{f}\p{r\xi} \,\diff\sigma\p{\xi} \,\diff{r} \,\diff\sigma\p{x} \,\diff\sigma\p{y} \\ &= \int_0^\infty \int_{S^2} r^2\hat{f}\p{r\xi} \p{\int_{S^2} e^{ix \cdot 2\pi r\xi} \,\diff\sigma\p{x}} \p{\int_{S^2} e^{iy \cdot 2\pi r\xi} \,\diff\sigma\p{y}} \,\diff\sigma\p{\xi} \,\diff{r} \\ &= \int_0^\infty \int_{S^2} r^2\hat{f}\p{r\xi} \p{\frac{4\pi\sin\abs{2\pi r\xi}}{\abs{2\pi r\xi}}}^2 \,\diff\sigma\p{\xi} \,\diff{r} \\ &= 4 \int_{\R^3} \hat{f}\p{\xi} \p{\frac{\sin 2\pi \abs{\xi}}{\abs{\xi}}}^2 \,\diff{\xi}. \end{aligned}$$

   The last equality comes from a change of variables. By Plancherel, $\norm{\hat{f}}_{L^2} = \norm{f}_{L^2}$. If we set $h\p{\xi} = \p{\frac{\sin 2\pi \abs{\xi}}{\abs{\xi}}}^2$, then

   $$\lim_{\xi\to0} h\p{\xi} = 2\pi,$$

   so $h$ is bounded near the origin. On the other hand,

   $$\abs{h\p{\xi}^2} \leq \frac{1}{\abs{\xi}^4},$$

   so $h \in L^2\p{\R^3}$. Thus, by Cauchy-Schwarz,

   $$\abs{L\p{f}} \leq 4\norm{h}_{L^2}\norm{\hat{f}}_{L^2} = 4\norm{h}_{L^2}\norm{f}_{L^2},$$

   which completes the proof.