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Weak and norm topologies differ on $ℓ^{1}$ despite identical convergent sequences

Statement

On each of the infinite-dimensional spaces $ℓ^{1} (R)$ and $ℓ^{1} (C)$ , the weak topology is strictly coarser than the norm topology. Nevertheless, a sequence converges weakly if and only if it converges in norm.

Facts & Assumptions

Given: A scalar field $K \in {R, C}$ and $X = ℓ^{1} (K)$ .

[F1]

The weak topology is generated by finite intersections of inverse images of scalar open sets under members of $X^{*}$ ; it is contained in the norm topology (Weak topology on a normed space).

[F2]

The norm induces the metric $d (x, y) = ∥ x - y ∥$ , and every metric-open set, including the open unit ball, is available as a norm-open set (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

[F3]

The coordinate sequences $e_{n}$ belong to $ℓ^{1} (K)$ and satisfy $∥ e_{n} ∥_{1} = 1$ ; finite coordinate sums have their usual $ℓ^{1}$ norm (Finite truncations approximate null and summable sequences).

[F4]

If a linear map has finite-dimensional domain, then its domain dimension is its nullity plus its rank (Rank-nullity: $\dim_{F} V = nullity T + rank T$ ).

[F5]

Both real and complex $ℓ^{1}$ have the Schur property: weak convergence of sequences implies norm convergence (Real and complex ell one have the Schur property).

Refutation

technique · Assume the norm-open unit ball were weakly open; a finite basic weak neighborhood inside it would contain a nonzero common kernel and all its scalar multiples

1.1

Let $B = {x \in X : ∥ x ∥_{1} < 1}$ . It is norm-open by [F2]. Suppose for contradiction that it is weakly open. Since $0 \in B$ , [F1] supplies a finite basic weak neighborhood $U = ⋂_{j = 1}^{m} f_{j}^{- 1} (V_{j})$ with $0 \in U \subseteq B$ , where $f_{j} \in X^{*}$ and each scalar-open $V_{j}$ contains zero. The case $m = 0$ means $U = X$ and already contradicts $U \subseteq B$ , since $2 e_{0} \notin B$ .

F1F2F3assume-contra

2.1

Assume $m \geq 1$ . Let $E = span {e_{0}, \dots, e_{m}}$ and define the linear map $A : E ⟶ K^{m}, A (x) = (f_{1} (x), \dots, f_{m} (x)) .$ The coordinate vectors are linearly independent by their explicit coordinates, so $\dim E = m + 1$ , whereas $rank A \leq m$ . Rank--nullity [F4] therefore gives a nonzero $v \in \ker A$ .

F3F4step 1.1

3.1

For every scalar $λ$ , each $f_{j} (λ v) = 0 \in V_{j}$ , so $λ v \in U$ . Since $v \neq 0$ , choose the positive real scalar $λ = 2 / ∥ v ∥_{1}$ . Absolute homogeneity gives $∥ λ v ∥_{1} = 2$ , hence $λ v \notin B$ . This contradicts $U \subseteq B$ .

step 1.1step 2.1F2discharge-contradiction: step 1.1

4.1

Thus the norm-open ball $B$ is not weakly open. Since [F1] says the weak topology is contained in the norm topology, the containment is strict over both scalar fields.

step 1.1step 3.1F1

5.1

Norm convergence implies weak convergence because the weak topology is coarser by [F1]. Conversely, [F5] turns every weakly convergent sequence in $X$ into a norm-convergent sequence. The two topologies therefore have exactly the same convergent sequences even though step 4.1 proves that they are different.

F1F5step 4.1

6.1

The witness uses $m + 1$ explicit coordinate vectors for an arbitrary finite list of $m$ functionals, so it also covers one functional and a list containing zero or repeated functionals. The empty list was handled in step 1.1. Only finite-dimensional rank--nullity and one formula-defined rescaling are used; there is no choice principle, and no assertion about nets having the same convergence behavior is made.

step 1.1step 2.1step 3.1step 5.1F3F4∎

Depends on

Used by

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Sources

Gerald Teschl, Topics in Real and Functional Analysis (standard reference, not scraped)

Weak and norm topologies differ on ℓ1ℓ1 despite identical convergent sequences