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Let X be a normed $\mathbb K$-linear space .Is it true that if $X^*$ is finite dimensional then $(X^*)^*$ is also finite dimensional and then $X$ is finite dimensional?

Here $X^*$ is the dual space of $X$.

Actually it is a part in a proof of a theorem. Here $X^*=span(\cup_{n\in \displaystyle{ \mathbb N} }f_n)$ and from here we conclude that $X^*$ is of finite dimensional as any Banach space is either finite or uncountable dimensional. Then the above question is written in the proof.

Please someone give some hints. to prove the above question.

Thank you.

Mini_me
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2 Answers2

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If $X^*$ denotes the algebraic dual, or if $X^*$ is a locally convex space, it is an easy exercise that if $Y$ is finite-dimensional, then so is $Y^*$ (by constructing a dual basis). So in your case you know that $X^{**}$ is finite-dimensional.

You also know that $X$ embeds as a subspace of $X^{**}$. As subspaces of finite-dimensional spaces are finite-dimensional, the conclusion is that $X$ is finite-dimensional.

If $X^*$ denotes in general the continuum dual of a topological vector space, the result is not true. For example, it is well known that there exist infinite dimensional topological vector spaces with trivial dual.

Martin Argerami
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  • The second part may not be true if $X$ is not a Banach space (the natural function from $X$ to $X^{**}$ need not be injective). – tomasz Oct 15 '17 at 16:31
  • Your first two sentences address the converse of the asker's statement to prove. – amWhy Oct 15 '17 at 16:31
  • @tomasz: yes. Because the question doesn't specify it, I assume that $X$ was simply a vector space. The rest of the question and the tags imply that $X$ is a normed space, maybe a Banach space. It would be nice if the OP clarifies that. – Martin Argerami Oct 15 '17 at 16:45
  • @amWhy: how come? – Martin Argerami Oct 15 '17 at 16:45
  • $X$ embedding into $X^{**}$ also needs Hahn Banach and a norm on $X$, doesn't it? I think this question should either want $X$ to be a more general class of topological VS or does not want to assume Hahn Banach. – s.harp Oct 15 '17 at 17:50
  • @s.harp: I guess the point is what the meaning of $X^*$ is. My answer assumes it is the algebraic dual. – Martin Argerami Oct 15 '17 at 18:56
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Given a basis $e_1, \dots, e_n$ for the vector space $V$, define a set $\epsilon_1, \dots, \epsilon_n$ of elements of $V^*$ by $\epsilon_i(e_{ij}) := \delta_{ij}$. ($\delta_{ij}$ is equal to $1$ if $i=j$, and $0$ otherwise.)

Show that this set is independent, and that it spans $V^*$.

(Basically I've just taken the transpose of the fact that $e_1, \dots, e_n$ form a basis of $V$.)

Patrick Stevens
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