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Knapp necessary condition for spherical L2 restriction

Statement

Assume Countable Choice, let $n \geq 2$ , and take $p, q \in [1, \infty]$ . (a) If there is $C < \infty$ with $∥ \hat{f} ∥_{L^{2} (σ)} \leq C ∥ f ∥_{L^{p} (R^{n})}$ for all $f \in S (R^{n})$ , then $p \leq 2 (n + 1) / (n + 3)$ . Equivalently, if $E : L^{2} (S^{n - 1}) \to L^{q} (R^{n})$ satisfies $∥ E g ∥_{q} \leq C ∥ g ∥_{L^{2} (σ)}$ for all $g \in L^{2} (σ)$ , then $q \geq 2 (n + 1) / (n - 1)$ . (b) The necessity is exhibited by the cap data $g = 1_{C_{δ}}$ and the limit $δ ↓ 0$ .

Facts & Assumptions

Given: Countable Choice, $n \geq 2$ , $δ \in (0, 1]$ , the cap $C_{δ} = C_{δ} (e_{n}) = {ω \in S^{n - 1} : 1 - ω \cdot e_{n} \leq δ^{2}}$ , and the tube $T_{δ} = {ξ : ∣ ξ_{n} ∣ \leq c_{n} δ^{- 2}, ∣ ξ_{j} ∣ \leq c_{n} δ^{- 1} (j < n)}$ where $a_{n} > 0$ is a constant furnished by Cap wave packets concentrate on the dual tube and $c_{n} = a_{n} / \sqrt{n - 1}$ . For this box, $∣ (ξ_{1}, \dots, ξ_{n - 1}) ∣ \leq \sqrt{n - 1} c_{n} δ^{- 1} = a_{n} δ^{- 1}$ and $∣ ξ_{n} ∣ \leq c_{n} δ^{- 2} \leq a_{n} δ^{- 2}$ , so it lies in the cylindrical concentration tube.

[F1]

Duality: for $1 < p < \infty$ the restriction estimate $∥ \hat{f} ∥_{L^{2} (σ)} \leq C ∥ f ∥_{L^{p}}$ for all Schwartz $f$ is equivalent to the extension estimate $∥ E g ∥_{L^{p^{'}}} \leq C ∥ g ∥_{L^{2} (σ)}$ for all $g \in L^{2} (σ)$ , with the same least constant. (Restriction and extension estimates are dual, Conjugate exponents, including the endpoint conventions)

[F2]

Cap and tube scales: $c^{'} δ^{n - 1} \leq σ (C_{δ}) \leq C^{'} δ^{n - 1}$ and $λ_{n} (T_{δ}) = (2 c_{n})^{n} δ^{- (n + 1)}$ for $δ \in (0, 1]$ , with constants depending only on $n$ . (Spherical cap and dual slab scales)

[F3]

Cap concentration on the cylindrical tube of the supplier, and hence on the box specified in Given: for $g = 1_{C_{δ}}$ one has $∣ E g (x) ∣ \geq \frac{1}{2} σ (C_{δ})$ for every $x \in T_{δ}$ , hence $∥ E g ∥_{q}^{q} \geq (\frac{1}{2} σ (C_{δ}))^{q} λ_{n} (T_{δ})$ for every $1 \leq q < \infty$ ; and $∥ g ∥_{L^{2} (σ)} = σ (C_{δ})^{1 / 2}$ . (Cap wave packets concentrate on the dual tube, The nonnegative Lebesgue integral, Fourier pairing for a finite measure and Schwartz data, Explicit compactly supported smooth cutoffs, Monotone convergence for the integral, Differentiation under the integral sign)

[F4]

Positive-base real powers satisfy the exponent, product and iterated-power laws and are defined by $δ^{a} = \exp (a \log δ)$ . The logarithm is the inverse of the exponential, and $\exp t \to \infty$ as $t \to \infty$ , while $\exp t \to 0$ as $t \to - \infty$ . Thus if $δ^{a} \leq K δ^{b}$ for all $δ \in (0, 1]$ and $K > 0$ , then $a \geq b$ : otherwise put $δ = \exp (- k)$ for large $k$ to obtain $δ^{a - b} = \exp (k (b - a)) \to \infty$ , contradicting the bound. (The exponent, product, quotient, and iterated-power laws for positive real bases and real exponents, Real powers for positive bases, with the zero-base positive-exponent convention, The natural logarithm as the inverse of the exponential function, The exponential tends to $+ \infty$ at $+ \infty$ and to $0$ at $- \infty$ )

[F5]

The finite-measure pairing, smooth cutoffs, monotone convergence and compact-frequency differentiation are available. (Fourier pairing for a finite measure and Schwartz data, Explicit compactly supported smooth cutoffs, Monotone convergence for the integral, Differentiation under the integral sign)

Proof

technique · direct; evaluate the assumed extension bound on the cap data, insert the two exact scales, and let $\delta$ tend to $0$; the restriction form follows by duality

1.1F3algebra

If $q = \infty$ , the necessary inequality $q \geq 2 (n + 1) / (n - 1)$ is automatic. For $1 \leq q < \infty$ , assume $∥ E g ∥_{q} \leq C ∥ g ∥_{L^{2} (σ)}$ for all $g \in L^{2} (σ)$ and let $g = 1_{C_{δ}} \in L^{2} (σ)$ . By [F3], $∥ g ∥_{L^{2} (σ)} = σ (C_{δ})^{1 / 2}$ and $∥ E g ∥_{q}^{q} \geq (\frac{1}{2} σ (C_{δ}))^{q} λ_{n} (T_{δ})$ . Combining with the assumed bound, $\frac{1}{2} σ (C_{δ}) λ_{n} (T_{δ})^{1 / q} \leq ∥ E g ∥_{q} \leq C σ (C_{δ})^{1 / 2} .$

2.1F2step 1.1algebra

Inserting the scales. Dividing by $σ (C_{δ})^{1 / 2}$ and inserting [F2], $\frac{1}{2} c^{' 1 / 2} (2 c_{n})^{n / q} δ^{(n - 1) / 2 - (n + 1) / q} \leq \frac{1}{2} σ (C_{δ})^{1 / 2} λ_{n} (T_{δ})^{1 / q} \leq C .$ Thus there is a constant $K$ depending only on $n, q$ with $δ^{a} \leq K$ for all $δ \in (0, 1]$ , where $a : = (n - 1) / 2 - (n + 1) / q$ .

3.1F4step 2.1algebra

Forcing the exponent. Applying [F4] with $b = 0$ to the inequality $δ^{a} \leq K δ^{0}$ gives $a \geq 0$ , that is $(n - 1) / 2 \geq (n + 1) / q$ , hence $q \geq 2 (n + 1) / (n - 1)$ : no extension bound can hold for smaller $q$ .

4.1F1F2F3F5step 3.1algebra

Suppose the restriction estimate holds at exponent $p$ . For $1 < p < \infty$ , [F1] and step 3.1 imply $p^{'} \geq 2 (n + 1) / (n - 1)$ , hence $p \leq 2 (n + 1) / (n + 3)$ . At $p = 1$ the necessary inequality is automatic. At $p = \infty$ , apply the pairing identity [F5] with the finite positive measure $μ = 1_{C_{δ}} σ$ and a Schwartz cutoff equal to one near $S^{n - 1}$ as its frequency factor. Thus an $L^{\infty}$ restriction bound would imply $∣ \int E g \overline{F} ∣ \leq C ∥ g ∥_{2} ∥ F ∥_{\infty}$ on compactly supported smooth $F$ . For $g = 1_{C_{δ}}$ , $h = E g$ is smooth (differentiate its finite compact-frequency integral), and the tests $F = χ_{R} h / (∣ h ∣^{2} + ϵ^{2})^{1 / 2}$ have supremum at most one. Choose $0 \leq χ_{R} \leq 1$ equal to one on the radius- $R$ ball. The nonnegative pairing integrand therefore bounds the integral over that ball by $C ∥ g ∥_{2}$ . Let the balls increase to $R^{n}$ , then let $ϵ ↓ 0$ , by monotone convergence to obtain $∥ E g ∥_{1} \leq C ∥ g ∥_{2}$ , contradicted by the cap lower bound at $q = 1$ . Thus $p = \infty$ is impossible too.

4.2F2F3F4step 2.1step 3.1

The exhibited family. The data witnessing the necessity are exactly the functions $g_{δ} = 1_{C_{δ}}$ , $δ \in (0, 1]$ : for any $q < 2 (n + 1) / (n - 1)$ the quotient $∥ E g_{δ} ∥_{q} / ∥ g_{δ} ∥_{L^{2} (σ)} \geq \frac{1}{2} σ (C_{δ})^{1 / 2} λ_{n} (T_{δ})^{1 / q}$ is unbounded along $δ = \exp (- k) ↓ 0$ by [F4] and steps 1.1–3.1, so no finite constant works. This is clause (b).

5.1step 1.1step 3.1step 4.1step 4.2∎

Conclusion. Steps 1.1–4.1 prove that any $L^{2} \to L^{q}$ extension bound forces $q \geq 2 (n + 1) / (n - 1)$ , step 4.1 transfers this to the restriction form $p \leq 2 (n + 1) / (n + 3)$ through the duality lemma, and step 4.2 exhibits the cap family and the limit $δ ↓ 0$ .

Depends on

Used by

Knapp rules out extension below the Tomas exponent Counterexample
Knapp cap and dual tube volume calculation Example
The general Fourier restriction problem remains open Remark
Stein-Tomas spherical restriction theorem Theorem

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Sources

K. Merz, Some notes on restriction theory (standard reference, not scraped)