LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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A geometric median is close when a strict majority is close

Statement

Let $H$ be a real Hilbert space, let $k \geq 1$ be an integer, let $m_{1}, \dots, m_{k} \in H$ , and let $m$ minimize $F (z) = \sum_{j = 1}^{k} ∥ z - m_{j} ∥$ . For $0 < a < 1 / 2$ and $r > 0$ , put

C_{a} = \frac{1 - a}{\sqrt{1 - 2 a}} .

If $z \in H$ satisfies $∥ m - z ∥ > C_{a} r$ , then more than $a k$ of the points $m_{j}$ satisfy $∥ m_{j} - z ∥ > r$ .

Facts & Assumptions

Given: A real Hilbert space $H$ , an integer $k \geq 1$ , points $m_{1}, \dots, m_{k}$ , a minimizer $m$ of their geometric-median objective, $0 < a < 1 / 2$ , $r > 0$ , and $z \in H$ .

[F1]

A geometric median minimizes $F (w) = \sum_{j} ∥ w - m_{j} ∥$ (Geometric median of block means).

[F2]

The norm is induced by a real inner product; orthogonal decompositions satisfy Pythagoras. Cauchy–Schwarz follows by minimizing $∥ u - t v ∥^{2} \geq 0$ over real $t$ when $v \neq 0$ (Hilbert space).

Proof

technique · direct

Given: A real Hilbert space $H$ , an integer $k \geq 1$ , points $m_{1}, \dots, m_{k}$ , a geometric median $m$ , $0 < a < 1 / 2$ , $r > 0$ , and $z \in H$ .

1.1F2algebra

Suppose at most $a k$ points satisfy $∥ m_{j} - z ∥ > r$ . Since $C_{a}^{2} = 1 + a^{2} / (1 - 2 a) > 1$ , write $D = ∥ z - m ∥ > C_{a} r > r$ and $d = (z - m) / D$ . For each of the at least $(1 - a) k$ remaining points, $u_{j} = (m_{j} - m) / ∥ m_{j} - m ∥$ is defined and $⟨ u_{j}, d ⟩ \geq (D - ∥ m_{j} - z ∥) / ∥ m_{j} - m ∥ > 0$ . Put $\cos θ_{j} = ⟨ u_{j}, d ⟩$ . By [F2], $0 < \cos θ_{j} \leq 1$ ; writing $d = \cos θ_{j} u_{j} + w_{j}$ with $w_{j} ⊥ u_{j}$ , Pythagoras gives $∥ w_{j} ∥ = \sin θ_{j}$ . The distance from $z = m + D d$ to the line through $m$ and $m_{j}$ is $D ∥ w_{j} ∥ = D \sin θ_{j}$ , and is at most $∥ z - m_{j} ∥ \leq r$ . Therefore $\cos θ_{j} \geq \sqrt{1 - (r / D)^{2}} > \sqrt{1 - C_{a}^{- 2}} = a / (1 - a)$ .

2.1F1F2step 1.1algebra∎

For $m_{j} \neq m$ , rationalizing the difference quotient gives $\lim_{t ↓ 0} (∥ m + t d - m_{j} ∥ - ∥ m - m_{j} ∥) / t = ⟨ m - m_{j}, d ⟩ / ∥ m - m_{j} ∥ = - \cos θ_{j}$ ; if $m_{j} = m$ , the quotient is $∥ t d ∥ / t = 1$ . Thus the right directional derivative of $F (m + t d)$ is $- \sum_{m_{j} \neq m} \cos θ_{j} + # {j : m_{j} = m}$ . Each good point from step 1.1 contributes less than $- a / (1 - a)$ , and each other point contributes at most $1$ by [F2]. Since at least $(1 - a) k$ points are good, the derivative is strictly less than $k - (1 - a) k (1 + a / (1 - a)) = 0$ . At a minimizer every right difference quotient is nonnegative, so this negative derivative is impossible. Thus more than $a k$ points lie outside the radius- $r$ ball about $z$ .

Remarks

The nonempty-center condition $k \geq 1$ is necessary. If $H = R$ , $k = 0$ , $z = 0$ , $r = 1$ , $a = 1 / 4$ , and $m = 4$ , the empty objective is zero at every point, so $m$ is a minimizer. Here $C_{a} = 3 / (2 \sqrt{2}) < 4$ , but there are zero outliers, not more than $a k = 0$ . All conclusions above retain their stated constants for every positive center count.

Depends on

Used by

Dependency tree · two levels

8 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

Stanislav Minsker, Geometric Median and Robust Estimation in Banach Spaces, Lemma 2.1(a) (standard reference, not scraped)
Gábor Lugosi, Mean Estimation in High Dimension, §3 (standard reference, not scraped)