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Let $X$ be a finite-dimensional normed vector space. Let $X^*$ denote the space of linear dual space of $X$, i.e. $X^*=L(X,\mathbb{C})$ and let $X^{**}$ be the dual space of $X^*$. For each $x\in X$, define $\hat{x}:X^*\to \mathbb{C}$ by $\hat{x}(f)=f(x)$. Let $\hat{X}=\{\hat{x}:x\in X\}$. Verify that $X^{**}$ and $\hat{X}$ have the same dimension and therefore are the same.

It is a reformulation from a passage in Folland (p. 159). I know how to find the dimension of $X^{**}$ (In general if $\dim X=n$ and $\dim Y=m$, $\dim(L(X,Y))=nm$). But how can I know $\hat{X}$ has the same dimension?

ViktorStein
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stph
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    Let ${x_1,..,x_n}$ be a basis of $X$. Try to prove that ${\hat{x_1},...,\hat{x_n}}$ is a basis of $X^{**}$. – Nitrogen Sep 10 '15 at 23:37
  • @Nitrogen Thank you. I think the independence is obvious since $a_1\hat{x}_1 (f)+\cdots+a_n \hat{x}_n (f)=0\forall f\in X^*$, then we must have all the constants to be $0$. But to prove they span $X$ looks difficult. I will try. Thank you. – stph Sep 10 '15 at 23:55
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    An overkill proof: since all norms on finite dimensions are equivalent, the dual of your normed space is the same as the dual of the same linear space but equipped with an inner product instead. But then you have Riesz... – Ian Sep 11 '15 at 00:15
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    By Riesz I mean the Riesz representation theorem. – Ian Sep 11 '15 at 00:45
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    You should just note that the map $x \mapsto \hat{x}$ is injective. Once you know that, it follows $\dim X = \dim \hat{X}$ and you already know $\dim X =\dim X^\ast$ (right?). – PhoemueX Sep 11 '15 at 03:39

2 Answers2

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Let $\dim X = n$ then $\dim X'=\dim X"= n$. Since $J\colon X \to X"$ is one one linear isometry and dimension of $X$ and $X"$ are equal to $n$, this implies that J is surjective. Hence $X$ is reflexive.

Martin Sleziak
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user570344
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When $X$ is a vector space of finite dimension then $\mathrm{dim} X < \infty$. Let $\Phi$ the canonical mapping injection from $X$ to its bidual $X''$. The well-known equality $$ \underbrace{\mathrm{dim}\, \, \mathrm{Ker} \Phi}_{=0} + \mathrm{dim}\, \, \mathrm{Im} \Phi = \mathrm{dim} X. \qquad (1) $$ yelds, (since $\Phi$ is injective then $\mathrm{Ker} \Phi=\{0\}$ and hence $ \mathrm{dim}\, \mathrm{Ker} \Phi=0$.)
$$ \mathrm{dim}\, \, \mathrm{Im} \Phi = \mathrm{dim} X. $$ Which means that $\Phi$ is surjective. We conclude that $X$ is reflexive.

Furdzik Zbignew
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