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Differentiated stationary end bounds imply the weighted Hölder interface

Statement

Fix a stationary chart with $X = \partial_{t}$ on $r = ∣ x ∣ > R$ , metric components $g_{μ ν} = η_{μ ν} + O_{k + 1} (r^{- α})$ , $k \geq 2$ , $0 < α \leq 1$ , and Minkowski matrix $η = diag (- 1, 1, 1, 1)$ . Here the asymptotic coefficient of $X$ means the constant $A$ in an expansion $X = (A + a) \partial_{t} + \sum_{i = 1}^{3} b^{i} \partial_{x^{i}}$ with $a, b^{i} = O_{0} (r^{- α})$ . Then after increasing $R$ , $g_{00} < - 1 / 2$ , $g_{i j} v^{i} v^{j} \geq ∣ v ∣^{2} / 2$ , $X$ is uniformly timelike and its asymptotic coefficient is $A = 1$ . Moreover for any $0 < λ \leq 1$ , every order- $k$ coordinate derivative $q$ of a metric component satisfies $∣ q (t, x) - q (τ, y) ∣ \leq C r^{- α - k - λ} (∣ t - τ ∣^{λ} + ∣ x - y ∣^{λ})$ when $∣ x - y ∣ \leq r / 2$ and $∣ t - τ ∣ \leq 1$ , where $r = ∣ x ∣$ . These are the local quantitative end inequalities; end gluing and global causality remain separate conditions.

Facts & Assumptions

[F1]

Stationary fields, charts and metric components have the specified geometric meanings. Stationary Lorentzian data and tensor conventions

[F2]

Differentiated decay controls Cartesian derivatives uniformly in angle. Differentiated asymptotic bounds and parameter uniformity

[F3]

If $f : [a, b] \to R$ is continuous on $[a, b]$ , differentiable on $(a, b)$ , $a < b$ , and $∣ f^{'} ∣ \leq M$ there for a finite $M \geq 0$ , then $∣ f (b) - f (a) ∣ \leq M (b - a)$ . The mean value inequality: if f : [a,b] → ℝᵐ is continuous and differentiable on (a,b) with ‖ f'‖₂ ≤ M, then ‖ f(b)-f(a)‖₂ ≤ M(b-a)

Proof

Given: The data and hypotheses in the Statement.

Goal: Establish the stated quantitative end inequalities and coefficient.

1.1givenF1F2

Fix a common radius $R_{1}$ beyond which the differentiated decay bounds hold, and increase the final $R$ so that $R > 2 R_{1}$ and every component of $g - η$ has absolute value at most $1 / 6$ for $r > R$ . Then $g_{00} \leq - 5 / 6 < - 1 / 2$ and $∣ \sum_{i j} (g_{i j} - δ_{i j}) v^{i} v^{j} ∣ \leq \frac{1}{6} (\sum_{i} ∣ v^{i} ∣)^{2} \leq ∣ v ∣^{2} / 2$ . Thus $g_{i j} v^{i} v^{j} \geq ∣ v ∣^{2} / 2$ . Since $X = \partial_{t}$ , its squared norm is $g_{00}$ , so it is uniformly timelike.

1.2givenF1F2

The local flow of $X = \partial_{t}$ translates $t$ and fixes $x$ . Its preservation of $g$ makes the metric components independent of $t$ ; hence derivatives containing a time derivative vanish, and purely spatial derivatives are independent of $t$ . The identity $X = \partial_{t}$ gives the defined expansion with $A = 1$ and $a = b^{i} = 0$ . In any such expansion, its time component gives $1 - A = a = O_{0} (r^{- α})$ ; since $α > 0$ , the right side tends to zero as $r \to \infty$ , so the constant $A$ is uniquely $1$ .

2.1givenF2F3step 1.1

Let $q$ be a purely spatial order- $k$ derivative of a metric component. Join $x$ to $y$ by changing their three coordinates one at a time. Every point $z$ on these coordinate segments satisfies $∣ z - x ∣ \leq ∣ y - x ∣ \leq r / 2$ , so $∣ z ∣ \geq r / 2 > R_{1}$ . The next differentiated bound gives $∣ \partial_{i} q (z) ∣ \leq C r^{- α - k - 1}$ on all three segments. On each nondegenerate coordinate interval the restriction of $q$ is $C^{1}$ , and its one-variable derivative is the corresponding partial derivative. Apply [F3] on that interval; zero-length intervals contribute zero. Telescoping the three differences yields $∣ q (x) - q (y) ∣ \leq C r^{- α - k - 1} \sum_{i = 1}^{3} ∣ y^{i} - x^{i} ∣ \leq \sqrt{3} C r^{- α - k - 1} ∣ x - y ∣$ .

3.1givenstep 1.2step 2.1∎

If $x \neq y$ , then for $0 < λ \leq 1$ one has $∣ x - y ∣^{1 - λ} \leq (r / 2)^{1 - λ}$ , and the preceding bound is at most $C^{'} r^{- α - k - λ} ∣ x - y ∣^{λ}$ . For $x = y$ the difference is zero. Purely spatial derivatives have no time dependence, and derivatives containing a time derivative are zero. Adding the nonnegative term $C^{'} r^{- α - k - λ} ∣ t - τ ∣^{λ}$ therefore gives the claimed estimate for every order- $k$ coordinate derivative when $∣ t - τ ∣ \leq 1$ .

Depends on

Used by · 2

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