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The Stein-Tomas exponents on the circle

Example

Assume Countable Choice. For $n = 2$ the Stein-Tomas endpoints of Stein-Tomas spherical restriction theorem are $p_{0} = 2 (n + 1) / (n + 3) = 6 / 5$ and $q_{0} = 2 (n + 1) / (n - 1) = 6$ : on the circle $S^{1}$ , $∥ \hat{f} ∥_{L^{2} (S^{1})} ≲ ∥ f ∥_{L^{6 / 5} (R^{2})}$ and $∥ E g ∥_{L^{6} (R^{2})} ≲ ∥ g ∥_{L^{2} (S^{1})}$ . The pair is conjugate, $(6 / 5)^{'} = 6$ , and satisfies $1 / p_{0} - 1 / q_{0} = 2 / (n + 1) = 2 / 3$ , the exponent identity used by the fractional-integration step.

Verification

Given: Countable Choice, $n = 2$ , the circle $S^{1} \subset R^{2}$ , the Stein-Tomas endpoints $p_{0} = 2 (n + 1) / (n + 3)$ and $q_{0} = 2 (n + 1) / (n - 1)$ , and the conjugacy convention of Conjugate exponents, including the endpoint conventions.

[F1] The spherical restriction theorem holds for every $n \geq 2$ with the endpoint $p_{0} = 2 (n + 1) / (n + 3)$ and $q_{0} = 2 (n + 1) / (n - 1)$ : the restriction bound at $p_{0}$ and the extension bound at every $q \geq q_{0}$ . (Stein-Tomas spherical restriction theorem)

[F2] Conjugate exponents: $q$ is conjugate to $p$ when $1 / p + 1 / q = 1$ , and $(6 / 5)^{'} = 6$ because $5 / 6 + 1 / 6 = 1$ . (Conjugate exponents, including the endpoint conventions)

1.1F1algebra

The endpoint values. Substituting $n = 2$ into [F1] gives $p_{0} = 2 \cdot 3 / (2 + 3) = 6 / 5$ and $q_{0} = 2 \cdot 3 / (2 - 1) = 6$ .

2.1F1F2step 1.1algebra

Conjugacy. $1 / (6 / 5) + 1 / 6 = 5 / 6 + 1 / 6 = 1$ , so $q_{0} = p_{0}^{'}$ ; equivalently $(6 / 5)^{'} = 6$ , and the extension bound at $q_{0} = 6$ is the dual form of the restriction bound at $p_{0} = 6 / 5$ .

2.2step 1.1algebra

The exponent identity. $1 / p_{0} - 1 / q_{0} = 5 / 6 - 1 / 6 = 4 / 6 = 2 / 3$ , while $2 / (n + 1) = 2 / 3$ at $n = 2$ ; this is the identity $1 / p - 1 / p^{'} = 2 / (n + 1)$ used in the fractional-integration step of the endpoint proof.

3.1step 1.1step 2.1step 2.2algebra∎

Conclusion. On the circle the Stein-Tomas endpoints are $p_{0} = 6 / 5$ and $q_{0} = 6$ , the two are conjugate, and the fractional-integration identity reads $1 / p_{0} - 1 / q_{0} = 2 / 3$ .

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Sources

Mark Williams, Notes on harmonic analysis (standard reference, not scraped)