Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13
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Right shift powers converge in wot not sot

Example

On real or complex 2 indexed by k0, let R(x0,x1,)=(0,x0,x1,). Then Rn0 in WOT but not in SOT.

Facts & Assumptions

[F1]

SOT tests pointwise norm convergence; WOT tests each bounded scalar functional on each fixed vector (Strong and weak operator topologies).

[F2]

Sequence norms use square sums (p is the Lp space of counting measure). Finite real Cauchy–Schwarz bounds sums of products of nonnegative coordinate moduli (The Cauchy-Schwarz inequality for finite sums).

Verification

Given: the right shift R and a bounded scalar-linear F on 2.

1.1

Put bk=F(ek) and test F on v=k<Nbkek. By F3, BN=F(v)=k<Nbk2FBN, giving BNF2, also when BN=0. Thus b2. Truncations converge in the square-sum norm, so F(x)=limNk<Nxkbk=kxkbk, with absolute convergence: F2 bounds every finite sum k<Nxkbk by x2b2, and the nonnegative partial sums converge to their finite supremum. The same finite inequality applied to tails gives the infinite tail bound used below.

givenF2F3
2.1

Consequently F(Rnx)=k0xkbk+n and F(Rnx)x2(jnbj2)1/20. This proves WOT convergence. But Rnx22=kxk2=x22, so at x=e0 the norms remain one. Thus SOT convergence to zero fails.

step 1.1F1F2

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources