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Coordinate functionals of a Schauder basis are bounded

Statement

Assume DC. If $(e_{n})_{n \geq 1}$ is a Schauder basis of a Banach space $X$ , then every coordinate functional $e_{n}^{*}$ and every partial-sum projection $P_{N}$ is bounded. Moreover

$K : = \sup_{N \geq 0} ∥ P_{N} ∥ < \infty .$

Facts & Assumptions

[A1]

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

[L1]

The coefficient space $E$ is Banach and its summation map $S : E \to X$ is a bounded linear bijection (The Schauder coefficient space is Banach).

[L2]

Under DC, a bounded linear bijection between Banach spaces has bounded inverse (Bounded inverse theorem).

[L3]

$P_{N} x = \sum_{n \leq N} e_{n}^{*} (x) e_{n}$ and the basis constant is the supremum of their norms (Partial-sum projections and basis constant).

Proof

technique · direct

Given: The objects and hypotheses in the Statement.

1.1A1L1L2

Apply [L2] to [L1]. The only choice use is [A1], through that bounded-inverse [given, L2, L1, A1] theorem. Thus $S^{- 1} : X \to E$ is bounded.

2.1givenL1L3step 1.1

Truncation $Q_{N} : E \to E$ , $(a_{n}) \mapsto (a_{1}, \dots, a_{N}, 0, \dots)$ , satisfies $∥ Q_{N} a ∥_{E} \leq ∥ a ∥_{E}$ , because every partial sum of $Q_{N} a$ is a partial sum of $a$ . Since $P_{N} = S Q_{N} S^{- 1}$ and $∥ S ∥ \leq 1$ ,

$∥ P_{N} ∥ \leq ∥ S^{- 1} ∥$

for every $N$ , including $N = 0$ . Hence $K < \infty$ . [L1, L3, step 1.1]

3.1givenL3step 2.1

For $n \geq 1$ ,

$e_{n}^{*} (x) e_{n} = (P_{n} - P_{n - 1}) x .$

Because $e_{n} \neq 0$ , taking norms gives $∣ e_{n}^{*} (x) ∣ \leq (∥ P_{n} ∥ + ∥ P_{n - 1} ∥) ∥ x ∥ / ∥ e_{n} ∥$ . Thus every $e_{n}^{*}$ is bounded. [L3, step 2.1] ∎

Depends on

Used by

Dependency tree · two levels

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Sources

Thomas Schlumprecht, Course Notes in Functional Analysis, Math 655 (standard reference, not scraped)