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The L^2 unit sphere is not weakly sequentially closed in infinite dimensions

Statement refuted

Refuted: that the unit sphere $S = {u \in H : ∥ u ∥ = 1}$ of an infinite-dimensional real Hilbert space $H$ is weakly sequentially closed — equivalently, that norm closedness and norm boundedness of a subset of a Hilbert space force weak sequential closedness (Weak convergence of nets and sequences).

Assume the Axiom of Choice (The Axiom of Choice). The witness works in every infinite-dimensional real Hilbert space, for instance $H = ℓ^{2} (N; R)$ (Hilbert space). There $S$ is norm closed and norm bounded, yet an orthonormal sequence $(e_{j}) \subseteq S$ satisfies $e_{j} ⇀ 0$ while $0 \notin S$ , so $S$ is not weakly sequentially closed. In particular the direct method for minimisation cannot be applied to the unit sphere by weak closedness alone; the repair used for the eigenvalue problems below is the strong $L^{2}$ compactness of Weak H^1 convergence plus Rellich preserves the L^2 unit normalisation. By Weak closure of the unit sphere is the closed unit ball (which assumes the Hahn–Banach extension principle, available under our Axiom of Choice hypothesis through Hahn-Banach dominated extension theorem for real vector spaces) the weak closure of $S$ is exactly the closed unit ball of $H$ .

Facts & Assumptions

Given: An infinite-dimensional real Hilbert space $H$ (assumed to admit no ordered basis of finite length), with the Axiom of Choice available.

[A1]

The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ ), AC implies DC implies countable choice: the Axiom of Choice implies Countable Choice, so the countable-selection and maximal-family suppliers below apply.

[F1]

Existence of a maximal orthonormal family, and maximality as completeness: $H$ contains an orthonormal set maximal under inclusion, and an orthonormal set is maximal exactly when it is complete, that is, when its closed linear span is $H$ (Orthonormal families, complete orthonormal systems and Hilbert bases).

[F2]

A finite-dimensional normed subspace is closed, Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis: the linear span of a finite set is finite dimensional and closed in $H$ , so if a complete orthonormal set were finite, its span would already be the closed linear span and $H$ would admit an ordered basis of finite length.

[F3]

The infinite set $B$ admits a sequence of distinct elements under the Axiom of Choice; the finite-tuple recursion establishing this fact is given in step 3.1.

[F4]

Orthonormal families, complete orthonormal systems and Hilbert bases: a subset of an orthonormal family is orthonormal; in particular $∥ e_{j} ∥ = 1$ and $⟨ e_{i}, e_{j} ⟩ = 0$ for $i \neq j$ .

[F5]

The Bessel inequality for an arbitrary orthonormal family: for every $y \in H$ the family $(∣ ⟨ y, e_{j} ⟩ ∣^{2})_{j}$ has finite sum $\sum_{j} ∣ ⟨ y, e_{j} ⟩ ∣^{2} \leq ∥ y ∥^{2}$ .

[F6]

Riesz representation for Hilbert spaces: every bounded linear functional $f$ on $H$ has the form $f (x) = ⟨ x, y_{f} ⟩$ for a unique $y_{f} \in H$ .

[F7]

Weak convergence of nets and sequences: $e_{j} ⇀ 0$ means $f (e_{j}) \to 0$ for every bounded linear functional $f$ .

[F8]

The reverse triangle inequality in a normed space, For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f (\overline{A}) \subseteq \overline{f (A)}$ : $∣ ∥ x ∥ - ∥ y ∥ ∣ \leq ∥ x - y ∥$ , so the norm is continuous and preimages of closed sets under it are closed.

Counterexample

technique · direct

Given: An infinite-dimensional real Hilbert space $H$ and the unit sphere $S = {u \in H : ∥ u ∥ = 1}$ .

1.1A1F1

By [F1] choose a maximal, equivalently complete, orthonormal set $B \subseteq H$ .

2.1step 1.1F1F2

The set $B$ is infinite: were $B$ finite, its linear span would be finite dimensional and closed by [F2], and completeness would force it to equal the closed linear span of $B$ , namely $H$ ; then $H$ would admit an ordered basis of finite length, contrary to the hypothesis.

3.1step 2.1A1F3F4

Let $T$ be the nonempty set of all finite tuples of distinct elements of $B$ , including the empty tuple. Every tuple has an extension by one new element because its range is finite and $B$ is infinite. The Axiom of Choice in [A1] selects one such extension for each tuple in $T$ . Starting with the empty tuple, iterate this fixed extension function recursively over $N$ ; the successive appended elements give distinct $e_{j} \in B$ . This proves [F3] locally. By [F4], $(e_{j})$ is orthonormal, so $∥ e_{j} ∥ = 1$ and $e_{j} \in S$ for every $j$ .

4.1step 3.1F5F6F7

We claim $e_{j} ⇀ 0$ . Fix $y \in H$ ; by Bessel's inequality [F5] the series $\sum_{j} ∣ ⟨ e_{j}, y ⟩ ∣^{2}$ has finite sum, so its terms tend to $0$ , that is $⟨ e_{j}, y ⟩ \to 0$ . Given a bounded linear functional $f$ , write $f (x) = ⟨ x, y_{f} ⟩$ by [F6]; then $f (e_{j}) = ⟨ e_{j}, y_{f} ⟩ \to 0$ , which by the definition of weak convergence [F7] is exactly $e_{j} ⇀ 0$ .

5.1step 3.1step 4.1F8F4∎

The set $S$ is norm closed, because $S$ is the preimage of the closed singleton ${1}$ under the continuous norm [F8], and it is norm bounded because $∥ u ∥ = 1$ for every $u \in S$ . Since $e_{j} \in S$ for every $j$ by step 3.1 while $0 \notin S$ because $∥ 0 ∥ = 0 \neq 1$ , and $e_{j} ⇀ 0$ by step 4.1, the sphere $S$ is not weakly sequentially closed; the refuted claim is therefore false.

Remarks

The weak closure is much larger than $S$ : by Weak closure of the unit sphere is the closed unit ball it is the closed unit ball $B = {u : ∥ u ∥ \leq 1}$ , which contains $0$ and every point of the open unit ball. The maximal orthonormal family used above exists in every Hilbert space under the Axiom of Choice; on the concrete space $ℓ^{2} (N; R)$ the standard basis itself is the orthonormal sequence (The standard basis of $ℓ^{2} (N)$ ), and no maximal-family argument is needed.
Why compactness replaces closedness. A bounded sequence in an infinite-dimensional Hilbert space need not have a strongly convergent subsequence, but Weak H^1 convergence plus Rellich preserves the L^2 unit normalisation shows that weak $H_{0}^{1}$ convergence plus Rellich compactness nevertheless preserves the $L^{2}$ normalisation along a subsequence, which is the substitute used in the eigenvalue problems of this page.

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Sources

Gerald Teschl, Partial Differential Equations: From Classical to Modern (2026 author manuscript; complete 392-page archived text) (standard reference, not scraped)