Stationary-phase decay for spherical surface measure

Reduction to the axis. For $\xi \mathrm{\ne }0$ write $\xi =r\nu$ with $r=\mathrm{\mid }\xi \mathrm{\mid }$ and $\nu \in S^{n-1}$, and choose an orthogonal map $R$ with $R\nu =e_n$. By the orthogonal invariance of [F2], $\widehat{\sigma }(\xi )=\widehat{\sigma }(r{e}_n)={\int }_{S^{n-1}}{e}^{-2\pi ir\omega \cdot e_n}d\sigma (\omega )$; the same identity with $\xi =0$ is trivial and is covered by the bound of [F5] for $r\le 1$. It therefore suffices to estimate $J(r):=\widehat{\sigma }(r{e}_n)$ for $r\ge 1$ and to add the trivial bound at small $r$.

Splitting with the chart partition. Let $({\chi }_j)$ be the finite smooth partition of [F1] subordinate to the hemisphere charts, so that $J(r)={\sum }_j{J}_j(r)$ with $J_j(r)={\int }_{S^{n-1}}{e}^{-2\pi ir\omega \cdot e_n}{\chi }_j(\omega )d\sigma (\omega )$. Each ${\chi }_j$ is supported in the image of one chart $X_i^{\epsilon }$, and by [F1] the chart formula writes $J_j(r)$ as an integral over the unit ball $B\subseteq {\mathbb{R}}^{n-1}$ of $e^{-2\pi ir\mathrm{\Phi }(y)}{b}_j(y)dy$ with $b_j:=({\chi }_j\circ X_i^{\epsilon })(1-\mathrm{\mid }y{\mathrm{\mid }}^2{)}^{-1\mathrm{/}2}$ and the smooth phase $\mathrm{\Phi }(y)=y_{n-1}$ if $i\mathrm{\ne }n$, or $\mathrm{\Phi }(y)=\epsilon \sqrt{1-\mathrm{\mid }y{\mathrm{\mid }}^2}$ if $i=n$ (the $n$-th coordinate of the chart being $\epsilon \sqrt{1-\mathrm{\mid }y{\mathrm{\mid }}^2}$). The support of ${\chi }_j\circ X_i^{\epsilon }$ is compact inside $B$, so $b_j$ extends by zero to $C_c^{\mathrm{\infty }}({\mathbb{R}}^{n-1})$. Coordinate-line product rules in [F6] justify its smoothness.

The non-stationary charts. If $i\mathrm{\ne }n$, then $\mathrm{\Phi }(y)=y_{n-1}$ has $\mathrm{\nabla }\mathrm{\Phi }=e_{n-1}\mathrm{\ne }0$ everywhere, so the first alternative of [F3] applied to the global phase $-y_{n-1}$ with $d=n-1$ and $\lambda =r$ gives $\mathrm{\mid }{J}_j(r)\mathrm{\mid }\le C_N{r}^{-N}$ for every $N$; such patches contribute negligibly for every $N$.

A global polar phase. For a polar amplitude $b_j$, choose $0<r_0<r_1<1$ with $\mathrm{supp}{b}_j\subset B_{r_0}(0)$. Put $\kappa (t)=1-s_0((t-r_0^2)\mathrm{/}(r_1^2-r_0^2))$ and define $a(t)=\kappa (t)(1-t{)}^{-1\mathrm{/}2}+1-\kappa (t)$ for $t<r_1^2$, and $a(t)=1$ for $t\ge r_1^2$. Rationalization gives $(\sqrt{u}{)}^{\mathrm{\prime }}=1\mathrm{/}(2\sqrt{u})$ for $u>0$; repeated product and quotient rules show that $(1-t{)}^{-1\mathrm{/}2}$ is smooth for $t<1$. Thus $a$ is a positive smooth function on $\mathbb{R}$: the first formula is smooth for $t<1$, and flatness of the smooth cutoff glues it to one at $r_1^2$. Set $A(s)=1-\frac{1}{2}{\int }_0^{s}a(t)dt$. By [F6], $A^{\mathrm{\prime }}=-a\mathrm{/}2$, so $A$ is smooth. On $s\le r_0^2$ its derivative and value at zero agree with $\sqrt{1-s}$, hence $A(s)=\sqrt{1-s}$. Thus $\tilde{\mathrm{\Phi }}(y)=\epsilon A(\mathrm{\mid }y{\mathrm{\mid }}^2)$ agrees with the original phase near $\mathrm{supp}{b}_j$, is smooth on all of ${\mathbb{R}}^{n-1}$, and satisfies $\mathrm{\nabla }\tilde{\mathrm{\Phi }}(y)=-\epsilon a(\mathrm{\mid }y{\mathrm{\mid }}^2)y$. Its only stationary point is zero, with Hessian $-\epsilon I$.

Application at the pole. Choose a bump $\beta$ equal to one near zero and supported inside $B$ by [F7], and a real $M>\sup \mathrm{\mid }{b}_j\mathrm{\mid }$. Both $b_j+M\beta$ and $M\beta$ have zero in the interior of their supports, since $\mathrm{Re}(b_j+M\beta )>0$ near zero. Apply the stationary alternative [F3] to each amplitude with the global phase $-\tilde{\mathrm{\Phi }}$ from step 2.2 and $\lambda =r$, then subtract their integrals. This gives $\mathrm{\mid }{J}_j(r)\mathrm{\mid }\le C_j{r}^{-(n-1)\mathrm{/}2}$ without any assumption that zero belongs to the original amplitude support.

Summation and the small-frequency bound. Summing the finitely many chart contributions of the non-stationary and polar-cap steps gives $\mathrm{\mid }\widehat{\sigma }(\xi )\mathrm{\mid }\le C_n^{\mathrm{\prime }}{r}^{-(n-1)\mathrm{/}2}$ for $r=\mathrm{\mid }\xi \mathrm{\mid }\ge 1$. For $\mathrm{\mid }\xi \mathrm{\mid }\le 1$, [F5] gives $\mathrm{\mid }\widehat{\sigma }(\xi )\mathrm{\mid }\le n{\lambda }_n(B_1^n)\le n{\lambda }_n(B_1^n)2^{(n-1)\mathrm{/}2}(1+\mathrm{\mid }\xi \mathrm{\mid }{)}^{-(n-1)\mathrm{/}2}$. Taking $C_n:=2^{(n-1)\mathrm{/}2}\max \{C_n^{\mathrm{\prime }},n{\lambda }_n(B_1^n)\}$ yields the asserted bound for every $\xi$. Since $\check{\sigma }(\xi )=\widehat{\sigma }(-\xi )$ and $\mathrm{\mid }-\xi \mathrm{\mid }=\mathrm{\mid }\xi \mathrm{\mid }$, the same bound holds for $\check{\sigma }$.

Conclusion. Step 4.1 proves $\mathrm{\mid }\widehat{\sigma }(\xi )\mathrm{\mid }\le C_n(1+\mathrm{\mid }\xi \mathrm{\mid }{)}^{-(n-1)\mathrm{/}2}$ for all $\xi$, and the identity $\check{\sigma }(\xi )=\widehat{\sigma }(-\xi )$ transfers it to $\check{\sigma }$. Countable Choice is inherited from the sphere chart and partition suppliers.