We know that the Banach space $X$ is infinite-dimensional,
theconclusion we want to show is: then $X'$ is also infinite-dimensional.
$X'$: the space of linear bdd functions
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$\begingroup$ Interesting side remark, in some models where the axiom of choice fails, this is false. In particular it is consistent that there is a Banach space whose dual is trivial (both the algebraic and the topological dual). Of course, in that context "infinite dimensional" means just "not a finite dimension", rather than "there is an infinite [linear] basis". $\endgroup$– Asaf Karagila ♦Apr 20, 2014 at 14:10
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2 Answers
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Use contraposition: if $X'$ is finite dimensional, then $X''$ is as well. Since $X$ can be embedded isometrically in $X''$, it must be finite dimensional.
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Again, Hahn-Banach to the rescue : Choose an infinite basis $\mathcal{B}$ of $X$. Start with $v_1\in \mathcal{B}$, and choose $f_1 \in X'$ such that $\|f_1\| = 1$ and $f_1(v_1) = 1$.
Now choose $v_2 \in \mathcal{B}$ and use Hahn-Banach to produce $f_2 \in X'$ such that $f_2(v_1) = 0$ and $f_2(v_2) = 1$.
Thus proceeding, construct $\mathcal{D}_n := \{f_1, f_2,\ldots, f_n\} \subset X'$ such that $$ f_j(v_i) = 0 \quad\forall i<j, \text{ and } f_j(v_j) = 1 $$ Notice that the set $\mathcal{D}_n$ is linearly independent, and so $\text{dim}(X') \geq n$.
This is true for any $n\in \mathbb{N}$, and so $X'$ is infinite dimensional.
answered Apr 20, 2014 at 13:36
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$\begingroup$ One might want to add that this uses that finite-dimensional subspaces of a banach spach are always closed. Otherwise, applying Hahn-Banach in the first line wouldn't work. $\endgroup$– fgpApr 20, 2014 at 13:38