Stein-Tomas TT-star bound from fractional integration

The slice superposition. The bounded kernel and $f\in L^1$ make the full integral absolutely convergent; Fubini therefore licenses splitting the variable $y=(y^{\mathrm{\prime }},s)\in {\mathbb{R}}^{n-1}\times \mathbb{R}$ and writing $x=(x^{\mathrm{\prime }},t)$, the defining convolution and [F1] give, at every point, $f\ast \check{\mu }(x^{\mathrm{\prime }},t)={\int }_{\mathbb{R}}({\int }_{{\mathbb{R}}^{n-1}}f(x^{\mathrm{\prime }}-y^{\mathrm{\prime }},t-s)K(y^{\mathrm{\prime }},s)d{y}^{\mathrm{\prime }})ds={\int }_{\mathbb{R}}U(s)f_{t-s}(x^{\mathrm{\prime }})ds\mathrm{.}$

Minkowski in the slice variable. Fix $t$. By [F2] applied in the space variable $x^{\mathrm{\prime }}$ with $\nu$ the Lebesgue measure in $s$, $\parallel {\int }_{\mathbb{R}}U(s)f_{t-s}ds{\parallel }_{L^{p^{\mathrm{\prime }}}({\mathbb{R}}^{n-1})}\le {\int }_{\mathbb{R}}\mathrm{\parallel }U(s)f_{t-s}{\mathrm{\parallel }}_{L^{p^{\mathrm{\prime }}}({\mathbb{R}}^{n-1})}ds\mathrm{.}$ For Schwartz $f$, $g(\tau )=\mathrm{\parallel }{f}_{\tau }{\mathrm{\parallel }}_p$ decays faster than any prescribed power (bound $\mathrm{\mid }f(x\mathrm{\prime },\tau )\mathrm{\mid }$ by $C_N(1+\mathrm{\mid }x\mathrm{\prime }\mathrm{\mid }{)}^{-N}(1+\mathrm{\mid }\tau \mathrm{\mid }{)}^{-N}$). Thus the slice estimate of [F1] bounds the right side by $C\int g<\mathrm{\infty }$, licensing [F2] for every $t$; this is the Minkowski inequality for the complex-valued measurable integrand of [F4].

The slice decay. For the exponent $p=2(n+1)\mathrm{/}(n+3)$ one has $1\mathrm{/}p-1\mathrm{/}2=1\mathrm{/}(n+1)$, so [F1] gives $\mathrm{\parallel }U(s)f_{t-s}{\mathrm{\parallel }}_{L^{p^{\mathrm{\prime }}}}\le C_a⟨s{⟩}^{-(n-1)\mathrm{/}(n+1)}\mathrm{\parallel }{f}_{t-s}{\mathrm{\parallel }}_{L^p({\mathbb{R}}^{n-1})}$. Substituting into step 2.1 and substituting $\tau =t-s$, $\parallel {\int }_{\mathbb{R}}U(s)f_{t-s}ds{\parallel }_{L^{p^{\mathrm{\prime }}}({\mathbb{R}}^{n-1})}\le C_a{\int }_{\mathbb{R}}⟨t-\tau {⟩}^{-\beta }g(\tau )d\tau ,g(\tau ):=\mathrm{\parallel }{f}_{\tau }{\mathrm{\parallel }}_{L^p({\mathbb{R}}^{n-1})},\beta :=\frac{n-1}{n+1}\mathrm{.}$ The function $g$ is measurable and nonnegative and, by Tonelli, $\mathrm{\parallel }g{\mathrm{\parallel }}_{L^p(\mathbb{R})}^p={\int }_{\mathbb{R}}{\int }_{{\mathbb{R}}^{n-1}}\mathrm{\mid }f(x^{\mathrm{\prime }},\tau ){\mathrm{\mid }}^{p}d{x}^{\mathrm{\prime }}d\tau =\mathrm{\parallel }f{\mathrm{\parallel }}_{L^p({\mathbb{R}}^n)}^p$, so $g\in L^p(\mathbb{R})$ and $\mathrm{\parallel }g{\mathrm{\parallel }}_p=\mathrm{\parallel }f{\mathrm{\parallel }}_p$.

Fractional integration. Let $\alpha :=1-\beta =2\mathrm{/}(n+1)\in (0,1)$ and compare kernels: $⟨t-\tau {⟩}^{-(1-\alpha )}\le \mathrm{\mid }t-\tau {\mathrm{\mid }}^{-(1-\alpha )}=K_{\alpha }(t-\tau )$ for $\tau \mathrm{\ne }t$ by [F1]. Hence the function $h(t):=\mathrm{\parallel }(f\ast \check{\mu })(\cdot ,t){\mathrm{\parallel }}_{L^{p^{\mathrm{\prime }}}({\mathbb{R}}^{n-1})}$ satisfies $h(t)\le C_a{2}^{1-\alpha }{\int }_{\mathbb{R}}{K}_{\alpha }(t-\tau )g(\tau )d\tau =C_a{2}^{1-\alpha }{I}_{\alpha }g(t)$ for almost every $t$; the single diagonal point $\tau =t$ has zero measure and does not affect the comparison. The HLS hypotheses hold: $0<\alpha <1$ because $n\ge 2$, and $1<p<1\mathrm{/}\alpha =(n+1)\mathrm{/}2$ because $p=2(n+1)\mathrm{/}(n+3)$ and $n>1$. By [F3] applied in dimension one, $\parallel {\int }_{\mathbb{R}}U(s)f_{t-s}ds{\parallel }_{L^{p^{\mathrm{\prime }}}({\mathbb{R}}^n)}=\mathrm{\parallel }h{\mathrm{\parallel }}_{L^{p^{\mathrm{\prime }}}(\mathbb{R})}\le C_a{2}^{1-\alpha }{C}_{\alpha ,p}\mathrm{\parallel }g{\mathrm{\parallel }}_{L^p(\mathbb{R})}=C_a^{\mathrm{\prime }}\mathrm{\parallel }f{\mathrm{\parallel }}_{L^p({\mathbb{R}}^n)},$ with $C_a^{\mathrm{\prime }}$ depending on $a,h,n$.

Conclusion. Steps 1.1–4.1 show that for the endpoint exponent $p=2(n+1)\mathrm{/}(n+3)$ the convolution $f\ast \check{\mu }$ lies in $L^{p^{\mathrm{\prime }}}({\mathbb{R}}^n)$ with norm controlled by $\mathrm{\parallel }f{\mathrm{\parallel }}_p$, the constants depending only on $a,h,n$. The exponent identity $1\mathrm{/}p-1\mathrm{/}{p}^{\mathrm{\prime }}=2\mathrm{/}(n+1)=1-\beta$ used above is exactly the conjugacy of $p$ and $p^{\mathrm{\prime }}$.