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$ℓ^{\infty}$ is Banach for the supremum norm

Example

Let $ℓ^{\infty}$ be the space of bounded scalar sequences $x = (x_{n})_{n \geq 0}$ with norm $∥ x ∥_{\infty} : = \sup_{n \geq 0} ∣ x_{n} ∣ .$ Then $ℓ^{\infty}$ is a Banach space.

Facts & Assumptions

Given: A Cauchy sequence $(x^{(m)})$ in $ℓ^{\infty}$ for the supremum norm, with coordinates $x^{(m)} = (x_{n}^{(m)})_{n \geq 0}$ .

[L1]

A Banach space is a normed space complete for its norm metric (Banach space).

Verification

technique · direct

1.1

For each fixed coordinate $n$ , the scalar sequence $(x_{n}^{(m)})_{m}$ is Cauchy because $∣ x_{n}^{(m)} - x_{n}^{(ℓ)} ∣ \leq ∥ x^{(m)} - x^{(ℓ)} ∥_{\infty}$ . Let $x_{n} : = \lim_{m} x_{n}^{(m)}$ .

given

2.1

Since $(x^{(m)})$ is Cauchy, choose $M$ with $∥ x^{(m)} - x^{(ℓ)} ∥_{\infty} < 1$ for $m, ℓ \geq M$ . Fixing $ℓ = M$ and letting $m \to \infty$ coordinatewise gives $∣ x_{n} ∣ \leq ∥ x^{(M)} ∥_{\infty} + 1$ for every $n$ , so $x = (x_{n})$ is bounded and lies in $ℓ^{\infty}$ .

step 1.1given

2.2

Given $ε > 0$ , choose $M$ with $∥ x^{(m)} - x^{(ℓ)} ∥_{\infty} < ε$ for $m, ℓ \geq M$ . Letting $ℓ \to \infty$ coordinatewise yields $∣ x_{n}^{(m)} - x_{n} ∣ \leq ε$ for every $n$ , so $∥ x^{(m)} - x ∥_{\infty} \leq ε$ for $m \geq M$ .

step 1.1given

3.1

Thus every supremum-norm Cauchy sequence in $ℓ^{\infty}$ converges in $ℓ^{\infty}$ , so $ℓ^{\infty}$ is Banach by [L1].

step 2.1step 2.2L1∎

Depends on

Banach space

Used by

c₀ is Banach for the supremum norm Example

Dependency tree · two levels

3 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

Theo Buhler and Dietmar A. Salamon, Functional Analysis (standard reference, not scraped)
Andrew Lin and Casey Rodriguez, MIT 18.102 Introduction to Functional Analysis (standard reference, not scraped)

ℓ∞ℓ∞ is Banach for the supremum norm