LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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Dual ball as a closed subset of a product

Statement

Let $X$ be a normed space over $K \in {R, C}$ and let $B_{X^{*}} = {f \in X^{*} : ∥ f ∥ \leq 1}$ . For each $x \in X$ put

D_{x} = {z \in K : ∣ z ∣ \leq ∥ x ∥} .

The evaluation map

$E : B_{X^{*}} ⟶ \prod_{x \in X} D_{x}, E (f) = (f (x))_{x \in X} .$

is a homeomorphism onto a closed subspace of the product. This includes $X = {0}$ .

Facts & Assumptions

Given: A normed space $X$ over $R$ or $C$ .

[F1]

The weak-star topology on $X^{*}$ is the initial topology of the evaluations $f \mapsto f (x)$ (The weak-star topology from finite evaluations).

[F2]

The product topology is the initial topology of the coordinate projections, and an empty product is a one-point space (The product set $\prod_{i \in I} X_{i}$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space).

[F3]

Elements of $X^{*}$ are bounded linear functionals and $∥ f ∥ = \sup_{∥ x ∥ \leq 1} ∣ f (x) ∣$ (The dual space X^* of a normed space and its dual norm).

Proof

technique · direct

1.1

If $f \in B_{X^{*}}$ , then $∣ f (x) ∣ \leq ∥ x ∥$ , so $E (f)$ belongs to the displayed product; evaluations separate functionals, so $E$ is injective. By the two initial-topology descriptions, the subspace topology pulled back by $E$ is exactly $σ (X^{*}, X)$ on the ball.

F1F2F3

1.2

Inside the product let $C$ be the set of all $z = (z_{x})_{x \in X}$ satisfying

$z_{x + y} = z_{x} + z_{y}, z_{λ x} = λ z_{x} (x, y \in X, λ \in K) .$

Each equality defines a closed set: it is the inverse image of ${0}$ under a continuous finite linear combination of coordinate projections. Hence $C$ , their intersection, is closed. [F2]

2.1

Every $E (f)$ lies in $C$ . Conversely, if $z \in C$ , then $f_{z} (x) = z_{x}$ is linear and its coordinate bound gives $∣ f_{z} (x) ∣ \leq ∥ x ∥$ for every $x$ . Thus $f_{z}$ is bounded with $∥ f_{z} ∥ \leq 1$ , so $z = E (f_{z})$ . Consequently $E [B_{X^{*}}] = C$ .

F3step 1.2

3.1

Steps 1.1 and 2.1 show that $E$ is a homeomorphism onto the closed subspace $C$ . When $X = {0}$ , both the ball and the product are one-point spaces and the same argument applies.

F2step 1.1step 2.1∎

Depends on

Used by

Banach–Alaoglu Theorem

Dependency tree · two levels

15 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

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