Uniform cutoff estimates at the CGO characteristic sphere

Rotate coordinates so the two given vectors are the first coordinate vectors. Write $\xi =(a,s{e}_2^{\mathrm{\prime }}+r\omega )$ with $a\in \mathbb{R}$, $r\ge 0$, and $\omega \in S^{d-2}$ in the remaining coordinates. Then $p_{\zeta }(\xi )=s^2-a^2-r^2+2isa$, and its zero set is $\mathrm{\Sigma }=\{a=0,r=s\}$. The exact distance to this sphere is $D(\xi )=\sqrt{a^2+(r-s{)}^2}$. For each fixed $K$, on $\mathrm{\mid }\xi \mathrm{\mid }\le Ks$ one has $\mathrm{\mid }{p}_{\zeta }(\xi )\mathrm{\mid }{\asymp }_{K}sD(\xi )$. To check the lower bound, if $\mathrm{\mid }a\mathrm{\mid }\ge D\mathrm{/}4$, use the imaginary part. Otherwise $\mathrm{\mid }r-s\mathrm{\mid }>D\mathrm{/}2$ and $\mathrm{\mid }a\mathrm{\mid }\le \mathrm{\mid }r-s\mathrm{\mid }\mathrm{/}2$; since $\mathrm{\mid }r-s\mathrm{\mid }\le r+s$, $\mathrm{\mid }{s}^2-r^2-a^2\mathrm{\mid }\ge (s+r)\mathrm{\mid }s-r\mathrm{\mid }-a^2\ge (3\mathrm{/}4)(s+r)\mathrm{\mid }s-r\mathrm{\mid }\ge (3\mathrm{/}8)sD$. The upper bound follows by expanding $s^2-a^2-r^2$ and using $\mathrm{\mid }a\mathrm{\mid }+r+s\le C_{K}s$. Also, if $\mathrm{\mid }\xi \mathrm{\mid }\ge 8s$, the triangle inequality gives $\mathrm{\mid }{p}_{\zeta }(\xi )\mathrm{\mid }\asymp \mathrm{\mid }\xi {\mathrm{\mid }}^2$. Therefore $s\mathrm{/}\mathrm{\mid }{p}_{\zeta }(\xi )\mathrm{\mid }\le C(1+D(\xi {)}^{-1})$ everywhere away from $\mathrm{\Sigma }$.

For $N>d+3$, the integral $\int ⟨\xi -\eta {⟩}^{-N}D(\xi {)}^{-1}d\xi$ is bounded uniformly in $s\ge 1$ and $\eta \in {\mathbb{R}}^d$, where $⟨x⟩=(1+\mathrm{\mid }x{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}$. Off $D<1\mathrm{/}4$ this follows from $D^{-1}\le 4$ and integrability of the bracket power. On $D<1\mathrm{/}4$, write $\eta =(b,s{e}_2^{\mathrm{\prime }}+t\nu )$ with $t\ge 0$ and any unit $\nu$ if $t=0$. The inequality $\mathrm{\mid }r\omega -t\nu \mathrm{\mid }\ge (r\mathrm{/}2)\mathrm{\mid }\omega -\nu \mathrm{\mid }$ follows from $(r-t{)}^2+rt\mathrm{\mid }\omega -\nu {\mathrm{\mid }}^2$ by minimizing in $t\ge 0$. Hence the angular integral of $⟨\xi -\eta {⟩}^{-N}$ is at most $C{r}^{-(d-2)}$ for $r\ge 3\mathrm{/}4$. Indeed, in polar angle $\theta$ from $\nu$, split at $\theta =1\mathrm{/}r$ and use $\mathrm{\mid }\omega -\nu \mathrm{\mid }=2\sin (\theta \mathrm{/}2)\ge 2\theta \mathrm{/}\pi$ and the spherical measure bound $C{\theta }^{d-3}d\theta$; the two integrals are bounded by $C{r}^{-(d-2)}$. This cancels the radial Jacobian $r^{d-2}$. The remaining integral over $\mathrm{\mid }a\mathrm{\mid },\mathrm{\mid }r-s\mathrm{\mid }<1\mathrm{/}4$ of $(a^2+(r-s{)}^2{)}^{-1\mathrm{/}2}$ is finite with an absolute bound by planar polar integration. This proves the uniform assertion, including all centers $\eta$.

Put $h=\xi -\eta$. The identity $p_{\zeta }(\eta )-p_{\zeta }(\xi )=\mathrm{\mid }\xi {\mathrm{\mid }}^2-\mathrm{\mid }\eta {\mathrm{\mid }}^2-2i\zeta \cdot h$ and the bound $\mathrm{\mid }\xi \mathrm{\mid }\le C(s+\sqrt{\mathrm{\mid }{p}_{\zeta }(\xi )\mathrm{\mid }})$ imply $s+\mathrm{\mid }{p}_{\zeta }(\eta )\mathrm{\mid }\le C(s+\mathrm{\mid }{p}_{\zeta }(\xi )\mathrm{\mid })⟨h{⟩}^2\mathrm{.}$ For the auxiliary bound, use $\mathrm{\mid }\xi {\mathrm{\mid }}^2\le \mathrm{\mid }{p}_{\zeta }(\xi )\mathrm{\mid }+2\sqrt{2}s\mathrm{\mid }\xi \mathrm{\mid }$ and solve this quadratic inequality. In the difference identity, the terms $s\mathrm{\mid }h\mathrm{\mid }$, $\sqrt{\mathrm{\mid }{p}_{\zeta }(\xi )\mathrm{\mid }}\mathrm{\mid }h\mathrm{\mid }$, and $\mathrm{\mid }h{\mathrm{\mid }}^2$ are absorbed by the displayed right side using $s\ge 1$ and $2ab\le a^2+b^2$. Since $\widehat{\varphi }$ is rapidly decreasing, steps 1.1–2.1 now give ${\sup }_{\eta }\int \mathrm{\mid }\widehat{\varphi }(\xi -\eta )\mathrm{\mid }\frac{s+\mathrm{\mid }{p}_{\zeta }(\eta )\mathrm{\mid }}{\mathrm{\mid }{p}_{\zeta }(\xi )\mathrm{\mid }}d\xi \le C_{\varphi }\mathrm{.}$ Indeed the integrand is bounded by $C_{\varphi }⟨\xi -\eta {⟩}^{-N}(1+D(\xi {)}^{-1})$ after choosing sufficiently many Schwartz powers.

For nonnegative weights $v,w$ positive almost everywhere and an integrable kernel $K$, Cauchy–Schwarz and Tonelli yield $\int w(\xi )\mathrm{\mid }(K\ast f)(\xi ){\mathrm{\mid }}^{2}d\xi \le \mathrm{\parallel }K{\mathrm{\parallel }}_1({\sup }_{\eta }\int \mathrm{\mid }K(\xi -\eta )\mathrm{\mid }\frac{w(\xi )}{v(\eta )}d\xi )\int v(\eta )\mathrm{\mid }f(\eta ){\mathrm{\mid }}^{2}d\eta \mathrm{.}$ To see this directly, bound $\mathrm{\mid }K\ast f{\mathrm{\mid }}^2$ by $\mathrm{\parallel }K{\mathrm{\parallel }}_1\int \mathrm{\mid }K(\xi -\eta )\mathrm{\mid }\mathrm{\mid }f(\eta ){\mathrm{\mid }}^{2}d\eta$, multiply by $w$, and interchange nonnegative integrals. Apply this with $K=(2\pi {)}^{-d\mathrm{/}2}\widehat{\varphi }$, $v=(s+\mathrm{\mid }{p}_{\zeta }\mathrm{\mid }{)}^{-1}$ and $w=\mathrm{\mid }{p}_{\zeta }{\mathrm{\mid }}^{-1}$. Step 3.1 and [F2] prove the negative-half cutoff bound. The locally integrable singularity proved in step 2.1 and the quadratic large-frequency growth show that every Schwartz function has finite negative-half norm.

The positive-half estimate is the dual of step 4.1. More explicitly, use the same negative-half bound for the cutoff $\bar{\varphi }$ and the Fourier $L^2$ pairing. Cauchy–Schwarz with reciprocal weights gives $\mathrm{\mid }(\varphi u,v{)}_{L^2}\mathrm{\mid }=\mathrm{\mid }(u,\bar{\varphi }v{)}_{L^2}\mathrm{\mid }\le \mathrm{\parallel }u{\mathrm{\parallel }}_{{\dot{X}}_{\zeta }^{1\mathrm{/}2}}\mathrm{\parallel }\bar{\varphi }v{\mathrm{\parallel }}_{{\dot{X}}_{\zeta }^{-1\mathrm{/}2}}\le C\mathrm{\parallel }u{\mathrm{\parallel }}_{{\dot{X}}_{\zeta }^{1\mathrm{/}2}}\mathrm{\parallel }v{\mathrm{\parallel }}_{X_{\zeta }^{-1\mathrm{/}2}}$. The supremum over Schwartz $v$ of the left side divided by its negative-half norm equals $\mathrm{\parallel }\varphi u{\mathrm{\parallel }}_{X_{\zeta }^{1\mathrm{/}2}}$. For this equality, truncate the weighted dual extremizer to a ball and away from $\mathrm{\Sigma }$, approximate it by smooth compactly supported frequency functions there, and use ordinary $L^2$ density where the weights are bounded above and below; monotone truncation recovers the full weighted integral. Thus the claimed supremum uses no unattained extremizer.

The positive-half weight is at least $s$, so step 5.1 immediately gives $\mathrm{\parallel }\varphi u{\mathrm{\parallel }}_2\le C{s}^{-1\mathrm{/}2}\mathrm{\parallel }u{\mathrm{\parallel }}_{{\dot{X}}_{\zeta }^{1\mathrm{/}2}}$. Split the gradient Fourier integral into $\mathrm{\mid }\xi \mathrm{\mid }\le 8s$ and its complement. The first part has norm at most $8s\mathrm{\parallel }\varphi u{\mathrm{\parallel }}_2$, hence at most $C{s}^{1\mathrm{/}2}\mathrm{\parallel }u{\mathrm{\parallel }}_{{\dot{X}}_{\zeta }^{1\mathrm{/}2}}$. On the complement $\mathrm{\mid }\xi {\mathrm{\mid }}^2\le C\mathrm{\mid }{p}_{\zeta }(\xi )\mathrm{\mid }$, so its norm is at most $C\mathrm{\parallel }\varphi u{\mathrm{\parallel }}_{X_{\zeta }^{1\mathrm{/}2}}\le C\mathrm{\parallel }u{\mathrm{\parallel }}_{{\dot{X}}_{\zeta }^{1\mathrm{/}2}}$. Since $s\ge 1$, their sum gives the gradient bound. A bounded map sends norm-Cauchy sequences to norm-Cauchy sequences, with the same bound; passing to equivalence classes therefore extends each map uniquely to the stated completions.