TT-star reduces extension to convolution with the surface-measure transform

The $T{T}^{\ast }$ identity. Applying the convolution identity [F2] with the given $F$ gives $(\widehat{F}\mu {)}^{\vee }=F\ast \check{\mu }$ as functions on ${\mathbb{R}}^n$. Since $E_{\mu }^{\ast }F=\widehat{F}{\mathrm{\mid }}_{\mathrm{supp}\mu }$ and $E_{\mu }(\widehat{F}{\mathrm{\mid }}_{\mathrm{supp}\mu })=(\widehat{F}\mu {)}^{\vee }$, this reads $E_{\mu }{E}_{\mu }^{\ast }F=F\ast \check{\mu }$, which is the first assertion.

The squared norm identity. The $L^2(\mu )$ norm is computed by [F3]: $\mathrm{\parallel }\widehat{F}{\mathrm{\parallel }}_{L^2(\mu )}^2=\int \mathrm{\mid }\widehat{F}{\mathrm{\mid }}^{2}d\mu =\int \widehat{F}\bar{\widehat{F}}d\mu$. Applying the pairing identity [F2] with $G=\widehat{F}$ (a Schwartz function) gives $\int (\widehat{F}\mu {)}^{\vee }\bar{F}dx=\int \widehat{F}\bar{\widehat{F}}d\mu$, that is $\int (\widehat{F}\mu {)}^{\vee }\bar{F}dx=\mathrm{\parallel }\widehat{F}{\mathrm{\parallel }}_{L^2(\mu )}^2$. By step 1.1 the left side equals $⟨F\ast \check{\mu },F⟩$, so the first two equalities of the statement hold.

The Hölder bound. By [F3] for the $L^p$–$L^{p^{\mathrm{\prime }}}$ pairing with $h=F\ast \check{\mu }$ and $k=F$, $\mid ⟨F\ast \check{\mu },F⟩\mid =\mid \int (F\ast \check{\mu })\bar{F}dx\mid \le \mathrm{\parallel }F\ast \check{\mu }{\mathrm{\parallel }}_{p^{\mathrm{\prime }}}\mathrm{\parallel }F{\mathrm{\parallel }}_p$ when $F\ast \check{\mu }\in L^{p^{\mathrm{\prime }}}$; if that norm is infinite, the inequality is automatic. The pairing itself is absolutely convergent since $F\in L^1$ and $F\ast \check{\mu }$ is bounded. Combining with step 2.1, $\mathrm{\parallel }\widehat{F}{\mathrm{\parallel }}_{L^2(\mu )}^2=⟨F\ast \check{\mu },F⟩\le \mathrm{\parallel }F{\mathrm{\parallel }}_p\mathrm{\parallel }F\ast \check{\mu }{\mathrm{\parallel }}_{p^{\mathrm{\prime }}}$.

Specialization to a hypersurface. Let $S$ be a compact hypersurface with surface measure $\sigma$; then $\sigma$ is a finite Borel measure, $\check{\sigma }=\widehat{\sigma }(-\cdot )$ obeys $\mathrm{\mid }\check{\sigma }\mathrm{\mid }\le \sigma (S)$, and the identities of steps 1.1–3.1 hold with $\mu =\sigma$. Consequently, if there is $C<\mathrm{\infty }$ with $\mathrm{\parallel }F\ast \check{\sigma }{\mathrm{\parallel }}_{L^{p^{\mathrm{\prime }}}}\le C\mathrm{\parallel }F{\mathrm{\parallel }}_{L^p}$ for every $F\in \mathcal{S}({\mathbb{R}}^n)$, then $\mathrm{\parallel }\widehat{F}{\mathrm{\parallel }}_{L^2(\sigma )}^2\le C\mathrm{\parallel }F{\mathrm{\parallel }}_p^2$, that is $\mathrm{\parallel }\widehat{F}{\mathrm{\parallel }}_{L^2(\sigma )}\le C^{1\mathrm{/}2}\mathrm{\parallel }F{\mathrm{\parallel }}_{L^p}$: a bound on the convolution with $\check{\sigma }$ bounds the restriction estimate.

Conclusion. Step 1.1 proves $E_{\mu }{E}_{\mu }^{\ast }F=F\ast \check{\mu }$; steps 2.1 and 3.1 prove the chain $\mathrm{\parallel }\widehat{F}{\mathrm{\parallel }}_{L^2(\mu )}^2=⟨(\widehat{F}\mu {)}^{\vee },F⟩=⟨F\ast \check{\mu },F⟩\le \mathrm{\parallel }F{\mathrm{\parallel }}_p\mathrm{\parallel }F\ast \check{\mu }{\mathrm{\parallel }}_{p^{\mathrm{\prime }}}$ for every $1\le p\le \mathrm{\infty }$; step 4.1 records the hypersurface specialization. Countable Choice is inherited from the finite-measure pairing and transform suppliers.