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Under Dependent Choice, a bounded operator between Banach spaces is bounded below exactly when it is injective with closed range

Statement

Assume the Axiom of Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

Let $X$ and $Y$ be Banach spaces over the same scalar field, and let $T : X \to Y$ be a bounded linear operator. Then $T$ is bounded below if and only if it is injective and has closed range.

Facts & Assumptions

Given: Banach spaces $X$ and $Y$ , and a bounded linear operator $T : X \to Y$ .

[L0]

Dependent Choice is assumed (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

[L1]

Being bounded below means that some $c > 0$ satisfies $∥ T x ∥ \geq c ∥ x ∥$ for every $x \in X$ (A bounded operator that is bounded below).

[L2]

A Banach space is complete, and a closed subspace of a Banach space is Banach (Banach space, A closed subspace of a Banach space is Banach).

[L3]

In a nonempty complete metric space, a countable union of closed sets with empty interior cannot be the whole space (Under Dependent Choice, a nonempty complete metric space is not a countable union of closed sets with empty interior).

[L4]

A map is injective when equal outputs force equal inputs (Injection, surjection, bijection).

Proof

technique · direct

1.1

Assume $T$ is bounded below, with constant $c > 0$ from [L1]. If $T x = 0$ , then $0 = ∥ T x ∥ \geq c ∥ x ∥$ , so $∥ x ∥ = 0$ and $x = 0$ . Thus $T$ is injective.

L1L4

1.2

Let $(T x_{n})$ be a sequence in $ran (T)$ converging to $y \in Y$ . Then $∥ x_{n} - x_{m} ∥ \leq c^{- 1} ∥ T x_{n} - T x_{m} ∥$ by [L1], so $(x_{n})$ is Cauchy in $X$ and hence converges to some $x \in X$ by [L2]. If $C$ is any bound for $T$ , then $∥ T x_{n} - T x ∥ \leq C ∥ x_{n} - x ∥ \to 0$ , so $T x_{n} \to T x$ . Since limits are unique in normed spaces, $y = T x$ . Therefore $ran (T)$ is closed.

L1L2choose

1.3

Conversely, assume $T$ is injective and $ran (T)$ is closed. Then $Z : = ran (T)$ is Banach by [L2], and $T : X \to Z$ is a bounded linear bijection.

L2L4

1.4

Let $B : = {x \in X : ∥ x ∥ < 1}$ . Since $Z = ⋃_{n \geq 1} \overline{T (n B)}$ , [L0] and [L3] yield an integer $N \geq 1$ such that $\overline{T (N B)}$ has nonempty interior in $Z$ . So there exist $z_{0} \in Z$ and $r > 0$ with $B_{Z} (z_{0}, r) \subseteq \overline{T (N B)}$ . Because $0 \in \overline{T (N B)}$ as well, subtraction gives $B_{Z} (0, r) \subseteq \overline{T (2 N B)}$ .

L0L2L3choose

2.1

We claim that every $y \in Z$ with $∥ y ∥ < r$ has a preimage $x$ with $T x = y$ and $∥ x ∥ \leq 4 N$ . Start with $e_{0} : = y$ . If $∥ e_{k} ∥ < r 2^{- k}$ , then $∥ 2^{k} e_{k} ∥ < r$ , so step 1.4 gives $v_{k} \in 2 N B$ with $∥ 2^{k} e_{k} - T v_{k} ∥ < r / 2$ . Put $u_{k} : = 2^{- k} v_{k}$ and $e_{k + 1} : = e_{k} - T u_{k}$ . Then $∥ u_{k} ∥ \leq 2 N 2^{- k}$ and $∥ e_{k + 1} ∥ < r 2^{- k - 1}$ . Inductively this constructs $(u_{k})$ with $∥ e_{k} ∥ < r 2^{- k}$ for every $k$ . The series $\sum_{k} u_{k}$ is absolutely convergent because $\sum_{k} 2 N 2^{- k} < \infty$ , so [L2] gives $x : = \sum_{k} u_{k} \in X$ with $∥ x ∥ \leq 4 N$ . Also $y - T (\sum_{j < m} u_{j}) = e_{m} \to 0$ , hence $T x = y$ .

step 1.4L2chooseconstruct

3.1

Now let $z \in Z$ with $z \neq 0$ . Put $λ : = r / (2 ∥ z ∥)$ , so $∥ λ z ∥ = r / 2 < r$ . Step 2.1 gives $u \in X$ with $T u = λ z$ and $∥ u ∥ \leq 4 N$ . Then $x : = λ^{- 1} u$ satisfies $T x = z$ and $∥ x ∥ \leq (8 N / r) ∥ z ∥$ . The same inequality is trivial at $z = 0$ , so the inverse $T^{- 1} : Z \to X$ is bounded by $8 N / r$ .

step 2.1algebra

4.1

Applying step 3.1 to $z = T x$ gives $∥ x ∥ \leq (8 N / r) ∥ T x ∥$ for every $x \in X$ , that is, $∥ T x ∥ \geq (r / 8 N) ∥ x ∥$ . Therefore $T$ is bounded below.

step 3.1L1

5.1

Step 1.2 proves that bounded below implies injective with closed range, and steps 1.3 through 4.1 prove the converse.

step 1.1step 1.2step 1.3step 4.1∎

Depends on

Used by

Dependency tree · two levels

19 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

Haim Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations (standard reference, not scraped)
Gerald Teschl, Topics in Real and Functional Analysis (standard reference, not scraped)