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Knapp cap and dual tube volume calculation

Example

Assume Countable Choice, let $n \geq 2$ , and fix $0 < c \leq 1 / (100 \sqrt{n - 1})$ . For $δ \in (0, 1]$ compute the two quantities whose comparison yields the Knapp condition: $σ (C_{δ}) = \int_{∣ y ∣^{2} \leq 2 δ^{2} - δ^{4}} (1 - ∣ y ∣^{2})^{- 1 / 2} d y$ satisfies $c_{n} δ^{n - 1} \leq σ (C_{δ}) \leq C_{n} δ^{n - 1}$ , while the dual slab $T_{δ} = {ξ : ∣ ξ_{n} ∣ \leq c δ^{- 2}, ∣ ξ_{j} ∣ \leq c δ^{- 1} (j < n)}$ has volume $(2 c)^{n} δ^{- (n + 1)}$ . Hence $∥ 1_{C_{δ}} ∥_{L^{2} (σ)} ≍ δ^{(n - 1) / 2}$ and, by Cap wave packets concentrate on the dual tube, $∥ \hat{1_{C_{δ}} d σ} ∥_{q} ≳ δ^{n - 1 - (n + 1) / q}$ for every $1 \leq q \leq \infty$ ; comparing the two powers as $δ ↓ 0$ gives the necessary condition of Knapp necessary condition for spherical L2 restriction.

Verification

Given: Countable Choice, $n \geq 2$ , $δ \in (0, 1]$ , the cap $C_{δ} = {ω \in S^{n - 1} : 1 - ω \cdot e_{n} \leq δ^{2}}$ , the slab $T_{δ}$ with $0 < c \leq 1 / (100 \sqrt{n - 1})$ , and the extension $E$ of the spherical measure.

[F1] Cap and slab scales: in the graph chart $ω = (y, \sqrt{1 - ∣ y ∣^{2}})$ the cap is ${∣ y ∣^{2} \leq 2 δ^{2} - δ^{4}}$ , its measure satisfies $c_{n} δ^{n - 1} \leq σ (C_{δ}) \leq C_{n} δ^{n - 1}$ , and $λ_{n} (T_{δ}) = (2 c)^{n} δ^{- (n + 1)}$ . (Spherical cap and dual slab scales)

[F2] The extension is $E g (x) = \int e^{2 π i x \cdot ω} g (ω) d σ (ω)$ . Componentwise integration commutes with real parts, $Re e^{i θ} = \cos θ$ , and $\cos θ \geq 1 - ∣ θ ∣$ by the one-Lipschitz bound and $\cos 0 = 1$ . (Fourier restriction and adjoint extension operators, $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ , Sine and cosine are $1$ -Lipschitz on $R$ , The Lebesgue integral is linear on $L^{1} (μ)$ )

[F3] The comparison with the necessary condition: the extension estimate $E : L^{2} (S^{n - 1}) \to L^{q}$ holds only for $q \geq 2 (n + 1) / (n - 1)$ , the threshold forced by the cap family as $δ ↓ 0$ . (Knapp necessary condition for spherical L2 restriction)

1.1F1

The cap integral. With the equator omitted at $δ = 1$ as justified in [F1], the cap condition $1 - ω \cdot e_{n} \leq δ^{2}$ is equivalent to $\sqrt{1 - ∣ y ∣^{2}} \geq 1 - δ^{2}$ , that is $∣ y ∣^{2} \leq 2 δ^{2} - δ^{4}$ , and the chart density is $(1 - ∣ y ∣^{2})^{- 1 / 2}$ ; hence $σ (C_{δ}) = \int_{∣ y ∣^{2} \leq 2 δ^{2} - δ^{4}} (1 - ∣ y ∣^{2})^{- 1 / 2} d y$ , and by [F1] this is bounded between $c_{n} δ^{n - 1}$ and $C_{n} δ^{n - 1}$ .

1.2F1algebra

The slab volume. The slab is the box $[- c δ^{- 1}, c δ^{- 1}]^{n - 1} \times [- c δ^{- 2}, c δ^{- 2}]$ , a product of $n - 1$ intervals of length $2 c δ^{- 1}$ and one of length $2 c δ^{- 2}$ ; its volume is their product $λ_{n} (T_{δ}) = (2 c)^{n - 1} δ^{- (n - 1)} \cdot 2 c δ^{- 2} = (2 c)^{n} δ^{- (n + 1)}$ .

1.3F1F2givenalgebra

The explicit box concentration. For $x \in T_{δ}$ , $∣ x^{'} ∣ \leq \sqrt{n - 1} c δ^{- 1} \leq δ^{- 1} / 100$ , while on the cap $∣ ω^{'} ∣ \leq \sqrt{2} δ$ and $∣ ω_{n} - 1 ∣ \leq δ^{2}$ . Hence $∣ x \cdot (ω - e_{n}) ∣ \leq (\sqrt{2} + 1) / 100$ , since $∣ x_{n} ∣ \leq c δ^{- 2} \leq δ^{- 2} / 100$ . In particular $∣ 2 π x \cdot (ω - e_{n}) ∣ < 1 / 2$ . By [F2], $Re e^{2 π i x \cdot (ω - e_{n})} \geq 1 / 2$ . Integrating and removing the unit-modulus factor $e^{2 π i x_{n}}$ gives $∣ E 1_{C_{δ}} (x) ∣ \geq Re (e^{- 2 π i x_{n}} E 1_{C_{δ}} (x)) \geq σ (C_{δ}) / 2$ . This proves the required bound for every allowed $c$ , independently of the unspecified constant in the concentration lemma.

2.1F1step 1.1

The cap norm. By [F1], $∥ 1_{C_{δ}} ∥_{L^{2} (σ)} = σ (C_{δ})^{1 / 2}$ satisfies $c_{n}^{1 / 2} δ^{(n - 1) / 2} \leq ∥ 1_{C_{δ}} ∥_{2} \leq C_{n}^{1 / 2} δ^{(n - 1) / 2}$ , so the two quantities $∥ 1_{C_{δ}} ∥_{2}$ and $δ^{(n - 1) / 2}$ are comparable with constants depending only on $n$ .

2.2F1step 1.2step 1.3algebra

The extension lower bound. By step 1.3 the extension of the cap data satisfies $∣ E 1_{C_{δ}} (x) ∣ \geq \frac{1}{2} σ (C_{δ})$ for every $x \in T_{δ}$ , so for $1 \leq q < \infty$ $∥ \hat{1_{C_{δ}} d σ} ∥_{q} = ∥ E 1_{C_{δ}} ∥_{q} \geq \frac{1}{2} σ (C_{δ}) λ_{n} (T_{δ})^{1 / q} \geq \frac{1}{2} c_{n} (2 c)^{n / q} δ^{n - 1 - (n + 1) / q},$ and for $q = \infty$ the same lower bound reads $\geq \frac{1}{2} c_{n} δ^{n - 1}$ , which is the limiting value of the displayed exponent.

3.1F3step 2.1step 2.2algebra∎

The power comparison. Comparing the two powers of steps 2.1 and 2.2, the extension estimate with a constant uniform in $δ$ requires $δ^{(n - 1) - (n + 1) / q} ≲ δ^{(n - 1) / 2}$ as $δ ↓ 0$ , that is $(n - 1) / 2 \leq n - 1 - (n + 1) / q$ , or equivalently $q \geq 2 (n + 1) / (n - 1)$ ; this is exactly the necessary condition of [F3] and the conclusion of the Knapp example.

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Sources

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