LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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A continuous strongly monotone map of finite-dimensional space is bijective

Statement

Let $d \geq 1$ , $κ > 0$ , and let $H : R^{d} \to R^{d}$ be continuous. Suppose

$⟨ H (x) - H (y), x - y ⟩ \geq κ ∥ x - y ∥_{2}^{2} (x, y \in R^{d}) .$

Then $H$ is bijective. Its inverse satisfies

$∥ H^{- 1} (u) - H^{- 1} (v) ∥_{2} \leq κ^{- 1} ∥ u - v ∥_{2} (u, v \in R^{d}) .$

Facts & Assumptions

Given: $d \geq 1$ , a continuous map $H : R^{d} \to R^{d}$ , a constant $κ > 0$ , and the strong-monotonicity inequality in the Statement. The inner product and norm are those of $R^{d}$ .

[F1]

The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ : $⟨ x, y ⟩$ is the Euclidean inner product on $R^{d}$ , and $∥ x ∥_{2} = \sqrt{⟨ x, x ⟩}$ .

[F2]

Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ .

[F3]

Brouwer fixed point theorem: every continuous map from the closed unit ball in $R^{d}$ to itself has a fixed point.

[A1]

The strong-monotonicity inequality in the Statement holds for every pair $x, y \in R^{d}$ , and $κ > 0$ .

Proof

1.1A1F1F2choosealgebra

Fix $b \in R^{d}$ and put $F (x) = H (x) - b$ . Applying [A1] with $y = 0$ and using [F2] gives $⟨ F (x), x ⟩ = ⟨ H (x) - H (0), x ⟩ + ⟨ H (0) - b, x ⟩ \geq κ ∥ x ∥_{2}^{2} - ∥ H (0) - b ∥_{2} ∥ x ∥_{2}$ . Choose $R > ∥ H (0) - b ∥_{2} / κ$ ; then $⟨ F (x), x ⟩ > 0$ whenever $∥ x ∥_{2} = R$ .

1.2F1F3construct

Let $P_{R} (v) = v$ when $∥ v ∥_{2} \leq R$ , and $P_{R} (v) = R v / ∥ v ∥_{2}$ when $∥ v ∥_{2} > R$ . The two formulas agree on the sphere, so $P_{R}$ is continuous and maps $R^{d}$ into the closed ball ${\overline{B}}_{R}$ . Thus $T (x) = P_{R} (x - F (x))$ is a continuous self-map of ${\overline{B}}_{R}$ ; rescaling by $x \mapsto x / R$ and applying [F3] gives a fixed point $x = T (x)$ .

2.1step 1.1step 1.2algebra

If $∥ x ∥_{2} < R$ , the radial projection $P_{R}$ can equal $x$ only when its input equals $x$ , hence $x - F (x) = x$ and $F (x) = 0$ . If $∥ x ∥_{2} = R$ , the equality $P_{R} (x - F (x)) = x$ implies $x - F (x) = α x$ for some $α \geq 1$ ; therefore $⟨ F (x), x ⟩ = (1 - α) R^{2} \leq 0$ , contradicting step 1.1. So the fixed point is interior and $H (x) = b$ . Since $b$ was arbitrary, $H$ is onto.

3.1A1F1F2step 2.1algebra∎

If $H (x) = H (y)$ , [A1] gives $0 \geq κ ∥ x - y ∥_{2}^{2}$ , hence $x = y$ ; thus $H$ is one-to-one. For $u = H (x)$ and $v = H (y)$ , [A1] and [F2] give $κ ∥ x - y ∥_{2}^{2} \leq ∥ u - v ∥_{2} ∥ x - y ∥_{2}$ . Dividing when $x \neq y$ , with equality of inverse values when $x = y$ , proves the inverse bound and completes bijectivity.

Depends on

Used by

Backward Euler has a globally defined nonexpansive resolvent (contractive when strongly dissipative) Theorem

Dependency tree · two levels

23 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Miller, Algebraic Topology I lecture notes, Theorem 10.7 and proof (standard reference, not scraped)