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Let $(X, \| \cdot \|)$ be a normed $\mathbb{C}$ - vector space. Define its dual

$$ X^* := \{l : X \to C \ | \ l \text{ linear } \} $$ with the weak-* topology.

Prove that the weak-* topology on $X^*$ corresponds to the subspace topology induced on the product topology $C^X$

We have $T_{X^*} = \{X^* \cap U | U \in T_{\mathbb{C}^X}$ } and $T_{weak}= \{e_x : X^* \to \mathbb{C}, \ e_x(l) = l(x), x \in X \ | \ e_x \text{ continuous } \}$.

But I do not know how to get started, showing that they are equal.

Henno Brandsma
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Olba12
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1 Answers1

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The weak$\ast$ topology is by its definition the smallest topology on $X^\ast$ that makes all $e_x$ continuous, where $e_x: X^\ast \to \mathbb{C}$ is given by $e_x(f) = f(x)$.

But if you look at the product topology on $\mathbb{C}^X$, these $e_x$ are precisely the restrictions of the projections $\pi_x(f) = f(x)$ (from $\mathbb{C}^X$ to $\mathbb{C}$) to $X^\ast$, and the product topology is precisely the smallest topology making all projections continuous.

$X^\ast$ embeds into $\mathbb{C}^X$ as a subspace and the subspace topology is the smallest topology that makes the inclusion mapping continuous. The transitive law for initial topologies , discussed in this answer implies that $X^\ast$ in the subspace topology wrt the product topology has exactly the initial topology wrt to the compositions of the $\pi_x$ and the inclusion map, i.e. the restricted $\pi_x$, the $e_x$.

So it's true on totally general principles.

Henno Brandsma
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