LemmaStatement: AI-generatedProof: AI-generated

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Weighted cutoff maximum gives a uniform normal angle

Statement

Let $N$ be a smooth manifold with nonempty compact boundary $S$ , and let $c : S \times [0, ε) \to N$ be a supplied smooth boundary collar: $c$ is a smooth embedding, $c (s, 0) = s$ , and its image is an open neighborhood of $S$ in $N$ . Choose $0 < ℓ < ε$ , put $K = c (S \times [0, ℓ])$ , and supply a smooth real function $u$ on $K$ , smooth up to its boundary. The compact smooth graph collar in this lemma is $M = {(x, u (x)) : x \in K}$ , identified with $K$ by graph projection. Identify its inner boundary, the graph over $c (S \times {0})$ , with $S$ ; its artificial outer edge is the graph over $c (S \times {ℓ})$ . Let $h$ be a smooth Riemannian metric on $M$ , smooth up to the boundary. Use its gradient $\nabla = {grad}_{h}$ , squared norm $∣ \cdot ∣_{h}^{2}$ and positive-divergence Laplace–Beltrami operator $Δ = {tr}_{h} {Hess}_{h}$ . Suppose $v \geq 1$ is smooth up to the boundary, $Δ \log v \geq \frac{1}{4} ∣ \nabla \log v ∣_{h}^{2} - C v^{2}$ , and $v \leq B_{0}$ on $S$ . Let $χ$ be continuous on $M$ and smooth on ${χ > 0}$ , with the following assumed cutoff properties: $0 \leq χ \leq 1$ , $χ = 1$ on $S$ , $χ = 0$ on the outer edge, and on $χ > 0$ the bounds $∣ \nabla χ ∣_{h}^{2} \geq a v^{2} - b$ and $∣ Δ χ ∣ \leq C v^{2}$ . Here $a > 0$ , $b \geq 0$ and $C \geq 0$ are fixed constants; one common $C$ may be obtained by enlarging the constants in the two differential estimates. Then some integer $m > 4$ , determined only by these constants, satisfies $\sup_{χ \geq 1 / 2} v \leq 2^{m} \max {B_{0}, 1, \sqrt{2 b / a}}$ . No uniform candidate second-derivative estimate is assumed.

Statement references

Smooth collars of a manifold boundary

Facts & Assumptions

[F1]

The cutoff value, regularity, gradient and Laplacian properties in the Statement are assumptions of this lemma. For geometric applications, Projection estimates for a Gaussian collar cutoff supplies these properties under its own hypotheses: a smooth Gaussian tube with compact closure, smooth reference-boundary distance, a smooth spacelike maximal graph in the prescribed cone, the stated tube shrinking and cutoff-width restrictions.

[F2]

The logarithmic angle inequality in the Statement is an assumption of this lemma. For geometric applications, Logarithmic angle inequality through the full range v at least one supplies it for a smooth spacelike maximal hypersurface and a smooth future unit timelike field with the stated uniform observer-metric bounds on its first two covariant derivatives and the ambient Ricci tensor. Compactness here is only that of the chosen closed graph collar with its fixed outer edge.

[F3]

The metric is positive definite, including at a smooth boundary, as in Riemannian metric and riemannian manifold. The gradient is characterized by $h (\nabla f, X) = d f (X)$ as in The Riemannian gradient is the metric dual of the differential, and $Δ f$ is the metric trace of the Levi–Civita Hessian with the positive-divergence sign convention in Laplace–Beltrami operator as the trace of the Hessian.

[F4]

The Hessian formulas ${Hess}_{h} f (X, Y) = X (Y f) - (\nabla_{X}^{h} Y) f$ and $({Hess}_{h} f)_{i j} = \partial_{i} \partial_{j} f - Γ^{k}_{i j} \partial_{k} f$ are Gradient hessian and divergence connection formulas.

[F5]

A continuous real-valued function on a nonempty compact space attains its maximum, by claim 2 of A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism.

[F6]

At an interior local maximum of a $C^{2}$ function on a Euclidean open set, its first derivatives vanish and its Euclidean Hessian is negative semidefinite. The Hessian is negative semidefinite at an interior local maximum The conversion to the metric trace on the collar is given below.

Proof

Given: The compact collar, Riemannian metric, regularity and inequalities in the Statement. Goal: Bound the angle on the positive inner part without a uniform candidate second-derivative estimate.

1.1F1F5given

For any integer $m > 4$ , $W = χ^{m} v$ is continuous on the nonempty compact collar, so [F5] gives a point $p$ where it attains its maximum. On the outer edge $W = 0$ , and on $S$ it is at most $B_{0}$ . If its maximum is zero the desired bound is immediate. Otherwise, if $p$ is interior, $χ (p) > 0$ and $F = \log W = \log v + m \log χ$ is smooth in a neighborhood of $p$ and has a local maximum there. Thus no differentiation across $χ = 0$ is needed.

2.1F3F4F6step 1.1algebra

At that interior positive maximum choose a smooth interior coordinate chart. Apply [F6] to the coordinate representative of $F$ to get $\partial_{i} F (p) = 0$ and a negative semidefinite matrix $(\partial_{i} \partial_{j} F) (p)$ . By [F4], the Christoffel term vanishes at $p$ , so the intrinsic Hessian has that same quadratic form. Its value on each $h$ -unit basis vector is nonpositive; tracing as in [F3] gives $Δ F (p) \leq 0$ . Also [F3] gives $\nabla F (p) = 0$ , hence $\nabla \log v = - m \nabla χ / χ$ at $p$ .

3.1F1F2F3F4step 2.1givenalgebra

For a positive smooth scalar $u$ , [F3] and [F4] give $\nabla \log u = u^{- 1} \nabla u$ and ${Hess}_{h} \log u = u^{- 1} {Hess}_{h} u - u^{- 2} d u \otimes d u$ by differentiating $\log u$ twice. Taking the metric trace gives $Δ \log u = (Δ u) / u - ∣ \nabla u ∣_{h}^{2} / u^{2}$ . Apply this identity to $χ$ near $p$ , and insert the assumed logarithmic angle inequality and $Δ χ \geq - C v^{2}$ . With step 2.1 this yields $0 \geq Δ F \geq (m^{2} / 4 - m) ∣ \nabla χ ∣_{h}^{2} / χ^{2} - C v^{2} (1 + m / χ)$ at $p$ .

4.1F1step 3.1givenchoosealgebra

Put $V_{0} = \max {1, \sqrt{2 b / a}}$ , which is real by $a > 0$ and $b \geq 0$ . If an interior positive maximum has $v (p) > V_{0}$ , then $∣ \nabla χ ∣_{h}^{2} \geq a v^{2} - b \geq (a / 2) v^{2}$ . Since $m > 4$ , the coefficient $m^{2} / 4 - m$ is positive. Multiplying step 3.1 by $χ^{2} / v^{2}$ and using $0 < χ \leq 1$ and $C \geq 0$ gives $0 \geq (a / 2) (m^{2} / 4 - m) - C (χ^{2} + m χ) \geq (a / 2) (m^{2} / 4 - m) - C (1 + m)$ . Choose an integer $m > \max {8, 32 C / a}$ . Then $m^{2} / 4 - m \geq m^{2} / 8$ and $(a / 2) (m^{2} / 4 - m) - C (1 + m) \geq a m^{2} / 16 - C (1 + m) > 0$ , a contradiction. This choice depends only on the stated constants.

5.1step 1.1step 4.1givenalgebra∎

For that fixed $m$ , a positive maximum therefore either lies on $S$ , where $W \leq B_{0}$ , or is interior with $v (p) \leq V_{0}$ , where $W (p) \leq V_{0}$ since $χ \leq 1$ . The zero-maximum case also obeys $\sup_{M} W \leq \max {B_{0}, V_{0}}$ . On $χ \geq 1 / 2$ , division by $χ^{m} \geq 2^{- m}$ now gives the stated bound. Only individual smoothness at a positive maximum was used, so no uniform candidate second-derivative estimate was assumed.

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