LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22

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A minimizing sequence in a convex set is Cauchy

Statement

Let $C$ be a nonempty convex subset of a real or complex inner-product space, let $x$ be a vector and put $d = \inf_{c \in C} ∥ x - c ∥$ . If $c_{n} \in C$ is a sequence with $∥ x - c_{n} ∥ \to d$ , then $(c_{n})$ is a Cauchy sequence.

Facts & Assumptions

[A1]

A subset $C$ is convex when $(1 - t) u + t v \in C$ for all $u, v \in C$ and $0 \leq t \leq 1$ (Convex sets and continuous real-hyperplane separation in a normed space).

[A2]

If $d = \inf S$ then $d \leq s$ for every $s \in S$ (Greatest lower bound (infimum)).

[A3]

Every inner-product norm satisfies the parallelogram law (The parallelogram law).

[A4]

The norm is induced by the pairing, in particular $∥ u - v ∥$ is the distance between $u$ and $v$ and $∥ u + v ∥ = ∥ v + u ∥$ (Real and complex inner-product spaces and their induced length).

Proof

technique · direct

Given: A nonempty convex set $C$ , a vector $x$ , the number $d = \inf_{c \in C} ∥ x - c ∥$ and a sequence $c_{n} \in C$ with $∥ x - c_{n} ∥ \to d$ .

1.1A1A2A4

Put $u_{n} = x - c_{n}$ ; for all $m, n$ the midpoint $(c_{m} + c_{n}) / 2$ lies in $C$ by convexity, so $∥ \frac{1}{2} (u_{m} + u_{n}) ∥ = ∥ x - \frac{1}{2} (c_{m} + c_{n}) ∥ \geq d$ because $d$ is a lower bound of the distances from $x$ to points of $C$ .

2.1step 1.1A3algebra

The parallelogram law applied to $u_{m}, u_{n}$ gives $∥ u_{m} - u_{n} ∥^{2} = 2 ∥ u_{m} ∥^{2} + 2 ∥ u_{n} ∥^{2} - 4 ∥ \frac{1}{2} (u_{m} + u_{n}) ∥^{2} \leq 2 ∥ u_{m} ∥^{2} + 2 ∥ u_{n} ∥^{2} - 4 d^{2}$ .

3.1step 2.1algebra∎

Given $ε > 0$ , convergence $∥ u_{n} ∥ \to d$ provides $N$ with $2 ∥ u_{n} ∥^{2} - 2 d^{2} < ε^{2} / 2$ for all $n \geq N$ , and then step 2.1 gives $∥ c_{m} - c_{n} ∥^{2} = ∥ u_{m} - u_{n} ∥^{2} < ε^{2}$ for all $m, n \geq N$ ; hence $(c_{n})$ is Cauchy.

A minimizing sequence in a convex set is Cauchy

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources