LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)

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A bounded-below operator has closed range

Statement

Assume Countable Choice. Let $X$ be a Banach space, $Y$ a normed space over the same field, and let $T \in B (X, Y)$ satisfy $∥ T x ∥ \geq c ∥ x ∥$ for all $x \in X$ and some $c > 0$ (A bounded operator that is bounded below, A bounded linear operator between normed spaces). Then $T$ is injective and its range $ran T$ is a closed linear subspace of $Y$ ; the inverse $ran T \to X$ is bounded with norm at most $1 / c$ . The proof uses Countable Choice only to pass from sequential closedness to closedness; the Cauchy-sequence step uses completeness of $X$ .

Facts & Assumptions

Given: A Banach space $X$ , a normed space $Y$ over the same field, and a bounded linear operator $T : X \to Y$ with $∥ T x ∥ \geq c ∥ x ∥$ for all $x \in X$ , for a constant $c > 0$ ; write $Z : = ran T$ .

[F1]

Bounded below and bounded: $T$ is linear and bounded, and $∥ T x ∥ \geq c ∥ x ∥$ for every $x \in X$ ; also $T 0 = 0$ and $T (u + v) = T u + T v$ , $T (λ u) = λ T u$ (A bounded operator that is bounded below, A bounded linear operator between normed spaces).

[F2]

$X$ is complete: every Cauchy sequence in $X$ converges in $X$ (Banach space, Complete metric space: every Cauchy sequence converges in the space, Cauchy sequence in a metric space).

[F3]

In a metric space every sequentially closed set is closed, and this direction spends Countable Choice once, precisely by manufacturing a sequence from an adherence point (A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F4]

Limits in a metric space are unique (A sequence in a metric space has at most one limit).

[F5]

A subset $W$ of a vector space is a linear subspace exactly when $0 \in W$ , $u + v \in W$ and $λ u \in W$ for all $u, v \in W$ and all scalars $λ$ (Linear subspace of a vector space, A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

[F6]

Suppose $x_{k} \to x$ in $X$ and $C \geq 0$ is a bound for $T$ ; then $∥ T x_{k} - T x ∥ = ∥ T (x_{k} - x) ∥ \leq C ∥ x_{k} - x ∥ \to 0$ , so $T x_{k} \to T x$ : this is continuity of $T$ in the sequential and in the $ε$ - $δ$ forms (A bounded linear operator between normed spaces, Metric continuity characterisations, with countable choice for the sequential converse, A norm on a real vector space, the induced metric, and the dictionary with the metric axioms, Vector addition and scalar multiplication are continuous in a normed space).

Proof

1.1F1

Injectivity: if $T x = 0$ then $0 = ∥ T x ∥ \geq c ∥ x ∥$ with $c > 0$ , so $∥ x ∥ = 0$ and $x = 0$ .

1.2F1F5

The range $Z$ is a linear subspace of $Y$ : $0 = T 0 \in Z$ ; if $z = T u$ and $w = T v$ then $z + w = T (u + v) \in Z$ ; and if $z = T u$ then $λ z = T (λ u) \in Z$ .

2.1F1F2F4F6step 1.1

Let $(z_{k}) \subseteq Z$ converge in $Y$ to some $y \in Y$ , say $z_{k} = T x_{k}$ with $x_{k}$ the unique preimage supplied by step 1.1. Then $∥ x_{k} - x_{m} ∥ \leq c^{- 1} ∥ T x_{k} - T x_{m} ∥ = c^{- 1} ∥ z_{k} - z_{m} ∥$ , so $(x_{k})$ is Cauchy in $X$ and hence converges to some $x \in X$ by completeness of $X$ . With any bound $C$ of $T$ , $∥ z_{k} - T x ∥ = ∥ T (x_{k} - x) ∥ \leq C ∥ x_{k} - x ∥ \to 0$ , so $z_{k} \to T x$ ; uniqueness of limits in $Y$ forces $y = T x \in Z$ . Thus $Z$ is sequentially closed in $Y$ .

2.2F1step 1.1algebra

Define $S : Z \to X$ by $S (y) : = x$ for the unique $x$ with $T x = y$ ; step 1.1 makes $S$ well defined with $T (S (y)) = y$ , and it is the inverse of $T$ viewed as a map onto $Z$ . For $y, z \in Z$ and scalars $a, b$ , applying $T$ to $a S (y) + b S (z)$ gives $a y + b z$ , so uniqueness gives $S (a y + b z) = a S (y) + b S (z)$ : the inverse is linear. For $y = T x \in Z$ we have $∥ S (y) ∥ = ∥ x ∥ \leq c^{- 1} ∥ T x ∥ = c^{- 1} ∥ y ∥$ , so $S$ is bounded with operator norm at most $1 / c$ .

3.1F3step 2.1

Since $Z$ is a sequentially closed subset of the metric space $Y$ , it is closed; this is the one step that uses Countable Choice, through the cited sequential-closure theorem.

4.1step 1.1step 1.2step 3.1step 2.2∎

Therefore $T$ is injective, $Z = ran T$ is a closed linear subspace of $Y$ , and the inverse map $S : Z \to X$ is bounded with norm at most $1 / c$ .

Depends on

Used by

Coercivity of the adjoint makes the form-operator range dense Lemma

Dependency tree · two levels

65 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

Richard S. Laugesen, Linear Analysis and Partial Differential Equations (University of Illinois, 2020, complete 158-page graduate notes) (standard reference, not scraped)
Leon Simon, Lectures on Partial Differential Equations (Stanford, complete 223-page author scan) (standard reference, not scraped)
Haim Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations (Springer Universitext, 2011, complete 614-page text) (standard reference, not scraped)