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A hyperplane near the centre separates opposite coordinate vertices

Statement

Let $E$ be a finite-dimensional coordinate Euclidean space of dimension $d > 0$ , let $T \subseteq E$ and $u \in {int}_{E} (T)$ , and choose an orthonormal basis $e_{1}, \dots, e_{d}$ and $ε > 0$ with $p_{i}^{\pm} = u \pm ε e_{i} \in T$ and $B_{E} (u, ε) \subseteq T$ . Every proper linear hyperplane $K \subset E$ meeting $B_{E} (u, ε / d)$ strictly separates $p_{i}^{+}$ and $p_{i}^{-}$ for some $i$ .

Here a linear hyperplane means a linear subspace $K$ with $\dim_{R} (E / K) = 1$ . Strict separation of $v, w \in E$ by $K$ means $a (v) a (w) < 0$ for a nonzero real linear functional $a : E \to R$ with $K = \ker a$ . Such a functional exists, and the sign condition is independent of its choice, as proved below.

Statement references

Coordinate vertices inside a relative open ball

Annihilators, relative hyperplanes and complexification

Linear hyperplane

Linear functionals and the algebraic dual V^*=L(V,F)

Real and complex inner product spaces, with the inner product linear in the first argument

Orthonormal families, complete orthonormal systems and Hilbert bases

Facts & Assumptions

[F1]

The coordinate-vertex construction supplies the orthonormal basis, positive $ε$ , vertices in $T$ and ball $B_{E} (u, ε) \subseteq T$ (Coordinate vertices inside a relative open ball).

[F2]

Balls and relative interior are taken in $E$ with its coordinate Euclidean metric (Annihilators, relative hyperplanes and complexification).

[F3]

A linear hyperplane $K$ is a linear subspace with $\dim_{R} (E / K) = 1$ (Linear hyperplane).

[F4]

The quotient $E / K$ has cosets $y + K$ , with $(y + K) + (z + K) = (y + z) + K$ and $t (y + K) = t y + K$ (The quotient vector space V/W and its canonical projection). The canonical projection $π (y) = y + K$ is linear and surjective, with $\ker π = K$ (Coset equality, well-defined quotient operations, and the canonical projection with kernel W).

[F5]

A real linear functional on $E$ is a real-linear map $E \to R$ (Linear functionals and the algebraic dual V^*=L(V,F)).

[F6]

The coordinate Euclidean inner product on $E$ is a real inner product: it is bilinear and positive definite (Real and complex inner product spaces, with the inner product linear in the first argument).

[F7]

The orthonormal family $e_{1}, \dots, e_{d}$ satisfies $⟨ e_{i}, e_{j} ⟩ = δ_{i j}$ , where $δ_{i j}$ is $1$ when $i = j$ and $0$ otherwise. The induced length is $∥ h ∥ = \sqrt{⟨ h, h ⟩}$ (Orthonormal families, complete orthonormal systems and Hilbert bases).

Proof

Given: $E, T, u, d, e_{i}, ε, p_{i}^{\pm}$ as in the statement, and a linear hyperplane $K$ meeting $B_{E} (u, ε / d)$ .

Goal: Construct a defining functional, verify that strict separation is independent of its choice, and establish strict separation of one opposite vertex pair.

Proof technique: direct.

1.1givenF3F4F5constructalgebra

By [F3], the real vector space $Q = E / K$ has a one-element basis ${q}$ . Thus every element of $Q$ is uniquely $t q$ for $t \in R$ . Define $λ (t q) = t$ . The quotient operations in [F4] give $λ (s t q + r t^{'} q) = s t + r t^{'}$ for real scalars $s, r, t, t^{'}$ , so $λ$ is linear and its kernel is ${0}$ . Set $a = λ \circ π$ . By [F4] and [F5], $a$ is a real linear functional with $\ker a = K$ ; it is nonzero because surjectivity of $π$ gives $z \in E$ with $π (z) = q$ and $a (z) = 1$ .

2.1step 1.1F5algebra

If $b : E \to R$ is another nonzero linear functional with kernel $K$ , then $y - a (y) z \in K$ for every $y \in E$ , so $b (y) = a (y) b (z)$ . The scalar $b (z)$ is nonzero, since $z \notin K$ . Consequently $b (v) b (w) = b (z)^{2} a (v) a (w)$ , so the product is negative for $b$ exactly when it is negative for $a$ . This proves the claimed independence of the strict-separation definition.

2.2givenF1F2F5F6F7step 1.1choosealgebra

Write $y = \sum_{j = 1}^{d} y_{j} e_{j}$ and set $a_{j} = a (e_{j})$ and $A = \max_{j} ∣ a_{j} ∣$ . Since $a$ is nonzero and the $e_{j}$ span $E$ , $A > 0$ . Choose $x \in K \cap B_{E} (u, ε / d)$ . Put $h = u - x = \sum_{j = 1}^{d} h_{j} e_{j}$ , where $h_{j} = u_{j} - x_{j}$ . By real bilinearity in [F6] and orthonormality in [F7], $∥ h ∥^{2} = ⟨ h, h ⟩ = \sum_{j, k = 1}^{d} h_{j} h_{k} ⟨ e_{j}, e_{k} ⟩ = \sum_{j = 1}^{d} h_{j}^{2} = \sum_{j = 1}^{d} ∣ u_{j} - x_{j} ∣^{2}$ . Each nonnegative summand is at most this sum, so $∣ u_{j} - x_{j} ∣ \leq ∥ u - x ∥$ . Using $a (x) = 0$ , we obtain $∣ a (u) ∣ = ∣ a (u - x) ∣ \leq \sum_{j = 1}^{d} ∣ a_{j} ∣ ∣ u_{j} - x_{j} ∣ \leq d A ∥ u - x ∥ < ε A$ .

3.1F1F5step 2.1step 2.2choosealgebra∎

Choose $i$ with $∣ a_{i} ∣ = A$ . By linearity and [F1], $a (p_{i}^{\pm}) = a (u) \pm ε a_{i}$ , so $a (p_{i}^{+}) a (p_{i}^{-}) = a (u)^{2} - ε^{2} a_{i}^{2} < 0$ by step 2.2. These two values therefore have opposite strict signs, proving strict separation of $p_{i}^{+}$ and $p_{i}^{-}$ in the stated, choice-independent sense.

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