ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13

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Left shift powers converge in sot not operator norm

Example

On real or complex $ℓ^{2}$ indexed by $k \geq 0$ , let $L (x_{0}, x_{1}, \dots) = (x_{1}, x_{2}, \dots)$ . Then $L^{n} \to 0$ in SOT, while $∥ L^{n} ∥ = 1$ for every $n \geq 0$ .

Facts & Assumptions

[F1]

SOT is convergence in norm on each fixed vector (Strong and weak operator topologies).

[F2]

The operator norm is the supremum of image norms on the unit ball (The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

Verification

Given: $ℓ^{2} = {x : \sum_{k \geq 0} ∣ x_{k} ∣^{2} < \infty}$ with square-sum norm and the left shift $L$ .

1.1

The shift and its powers are scalar-linear and $∥ L^{n} x ∥_{2}^{2} = \sum_{k \geq 0} ∣ x_{k + n} ∣^{2} = \sum_{j \geq n} ∣ x_{j} ∣^{2} \leq ∥ x ∥_{2}^{2}$ . These tails tend to zero for every fixed $x$ by convergence of its nonnegative series. Therefore each $L^{n}$ is bounded and $L^{n} \to 0$ in SOT.

givenF1

2.1

Step 1.1 gives $∥ L^{n} ∥ \leq 1$ . But $L^{n} e_{n} = e_{0}$ and both coordinate vectors have norm one, so F2 gives $∥ L^{n} ∥ \geq 1$ . Thus the operator norms remain exactly one and cannot converge to zero. The witness varies with $n$ , which is compatible with convergence on every fixed vector. $□$

step 1.1F2

Depends on

Used by

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Dependency tree · two levels

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Sources

Bühler–Salamon, Functional Analysis (2017); exact harvest in batch coverage (standard reference, not scraped)
Teschl, Topics in Real and Functional Analysis (2017); exact harvest in batch coverage (standard reference, not scraped)