Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Coordinate vectors converge weakly to zero in ell p

Example

In real or complex ℓp, 1<p<∞, the coordinate vectors en converge weakly to zero, although ∥en∥p=1.

Facts & Assumptions

[F1]

Put q=p/(p−1), so 1/p+1/q=1 (Conjugate exponents, including the endpoint conventions). The sequence norm is ∥x∥p=(∑k∣xk∣p)1/p (ℓp is the Lp space of counting measure); for complex sequences apply this to their real nonnegative moduli.

[F3]

Weak convergence tests every bounded scalar-linear functional (Weak convergence of nets and sequences).

Verification

Given: 1<p<∞ and a bounded scalar-linear F on ℓp.

1.1givenF1F2algebra

Put bk=F(ek). For a finite initial segment let vk=b‾k∣bk∣q−2 if bk≠0, and vk=0 if bk=0; set other coordinates zero. For real scalars conjugation fixes bk. Then bkvk=∣bk∣q and ∣vk∣p=∣bk∣(q−1)p=∣bk∣q. If AN=∑k<N∣bk∣q, linearity and boundedness give AN=F(v)≤∥F∥AN1/p. If AN=0 there is nothing to divide; otherwise AN1/q≤∥F∥.

2.1step 1.1F1F2F3

The nondecreasing partial sums AN are bounded by ∥F∥q, so their nonnegative series converges and bk→0. Indeed infinitely many ∣bk∣≥ε>0 would make arbitrarily large finite sums exceed that bound. For completeness finite Hölder bounds ∑k<N∣xkbk∣≤∥x∥p∥b∥q, so the coefficient pairing is absolutely convergent, consistently over both fields. In particular F(en)=bn→0 for every F, giving weak convergence. Directly ∥en∥p=(1p)1/p=1, so there is no norm convergence to zero. □

Depends on

Used by

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Dependency tree · two levels

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Sources