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For the finite-dimensional case, we have a canonical isomorphism between $\mathbf{V}$, a vector space with the usual addition and scalar multiplication, and $(\mathbf{V}^*)^*$, the "dual of the dual of $\mathbf{V}$." This canonical isomorphism means that the isomorphism is always the same, independent of additional choices.

We can define a map $I : \mathbf{V} \to (\mathbf{V}^*)^*$ by $$x \mapsto I(x) \in (\mathbf{V}^*)^* \ \text{ where } \ I(x)(f) = f(x) \ \text{for any } \ f \in \mathbf{V}^*$$

My Question: what can go wrong in the infinite-dimensional case? The notes I am studying remark that if $\mathbf{V}$ is finite-dimensional, then $I$ is an isomorphism, but in the infinite-dimensional case we can go wrong? How?

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1  
Another point worth noticing is that the map is always canonical (defined without a specific choice of a basis). It is natural in the sense that it induces a natural transformation between $1_{\mathbf{Vect}(k)}$ and $()^{\ast\ast}$, this being a natural isomorphism iff we restrict to the full subcategory of finitely-dimensional vector spaces. –  Chindea Filip Aug 6 '12 at 8:58

5 Answers 5

up vote 7 down vote accepted

Let $V$ be any vector space over a field $k$. Let $\{e_i\}_{i \in I}$ be a basis for $V$. For each $i \in I$, there is a unique linear functional $f_i: V \rightarrow k$ such that $f_i(e_j) = \delta_{ij}$: that is, $f_i(e_i) = 1$ and for every other basis element $e_j$, $f_i(e_j) = 0$.

CLAIM: The set $\{f_i\}_{i \in I}$ is linearly independent in $V^{\vee}$, and thus $\dim V \leq \dim V^{\vee}$. It is a basis if and only if $V$ is finite-dimensional (if and only if $V^{\vee}$ is finite-dimensional).

The linear independence is easy: if $a_1 f_{i_1} + \ldots + a_n f_{i_n} = 0$, then just by evaluating at $e_{i_1},\ldots,e_{i_n}$ we find $a_1 = \ldots = a_n = 0$.

In the finite-dimensional case -- say $I = \{1,\ldots,n\}$ -- we may write any linear $g: V \rightarrow k$ as

$g = g(e_1) f_1 + \ldots + g(e_n) f_n$,

which shows that $f_1,\ldots, f_n$ is a basis for $V^{\vee}$ (and implies $\dim V = \dim V^{\vee}$).

However, in the infinite-dimensional case the $\{f_i\}_{i \in I}$ do not form a basis..essentially because in an abstract vector space all our sums must be finite sums rather than infinite sums! Indeed the subspace spanned by the $f_i$'s is precisely the set of linear functionals which are zero at all but finitely many basis elements $e_i$, whereas to give a linear functional $f$ on $V$ the values $f(e_i)$ can be absolutely arbitrary. Concretely, the functional $f$ with $f(e_i) = 1$ for all $i \in I$ does not lie in the span of the $f_i$'s.

(Remark: In fact whenever $\dim V$ is infinite, we have $\dim V^{\vee} > \dim V$. That is, not only is $\{f_i\}_{i \in I}$ not a basis, there is no basis for the dual space of cardinality equal to that of $I$. This is actually not so easy to prove, and it is not needed to answer the question.)

Now we come back to the canonical map $I: V \rightarrow V^{\vee \vee}$ given by

$I(x): f \mapsto f(x)$.

CLAIM: a) $I$ is always injective.
b) $I$ is surjective if and only if $V$ is finite-dimensional.

To prove a), let $x$ be a nonzero element of $V$ and choose a basis $\{e_i\}_{i \in I}$ for $V$ in which $x$ is one of the basis elements, say $x = e_1$. Then $f_1$ is a linear functional which does not vanish at $x$, so $I(x)$ is a nonzero element of $V^{\vee \vee}$.

To prove b) we first use the fact that if $V$ is finite-dimensional, $\dim V = \dim V^{\vee} = \dim V^{\vee \vee}$. Thus $I: V \rightarrow V^{\vee \vee}$ is an injection between two vector spaces of the same finite dimension, so it must be an isomorphism.

Finally, if $I$ is infinite-dimensional, then one can see by choosing bases, dual sets and dual dual sets as above that $I$ is not surjective. (A good first step here is to confirm that in the finite-dimensional case, if we choose a basis $e_1,\ldots,e_n$ for $V$, a dual base $f_1,\ldots,f_n$ for $V^{\vee}$ and then a dual dual base $g_1,\ldots,g_n$ for $V^{\vee \vee}$, then the map $I$ is precisely the one which maps $e_i$ to $g_i$ for all $i$.) I can supply more details upon request. Note also that if we are willing to make use of the above parenthetical fact that $\dim V^{\vee} > \dim V$ when $\dim V$ is infinite, then we see that $\dim V^{\vee \vee} > \dim V^{\vee} > \dim V$, and thus not only is $I: V \rightarrow V^{\vee \vee}$ not an isomorphism, but moreover there is no isomorphism of vector spaces from $V$ to $V^{\vee \vee}$. Again though, this lies significantly deeper.

Added: Let me say a bit about the more ambitious approach of showing $\operatorname{dim}_k V^{\vee} > \operatorname{dim}_k V$ for any infinite-dimensional vector space $V$. Let $I$ be a basis for $V$, so $V \cong \bigoplus_{I} k$. As mentioned above, to give a linear functional on $V$ it is necessary and sufficient to assign to each basis element an arbitrary value in $k$, whence an isomorphism $V^{\vee} \cong k^I = \prod_{I} k$. Thus dualization replaces a direct sum over $I$ with a direct product over $I$. When $I$ is finite there is no difference, so we recover $V \cong V^{\vee}$. However, when $I$ is infinite I claim that

$$ \operatorname{dim}_k V^{\vee} = \operatorname{dim}_k k^I = \# k^{\# I} \geq 2^{\# I} > \# I = \operatorname{dim}_k V.$$

I know of almost no standard texts which include a proof of this result, and indeed some cleverness / real ideas seem to be required (unlike the above discussion of the non-surjectivity of $I$ in the infinite-dimensional case which is, while somewhat lengthy to write out in detail, really very straightforward). However, by coincidence I just found on the web a very nice proof of this result which deduces it from Dedekind's Linear Independence of Characters. Please see this note of France Dacar.

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As a complement to the addition, here is a complete thread about it: math.stackexchange.com/questions/58548/… –  Asaf Karagila Aug 6 '12 at 10:47

The 'double dual' can be much 'larger' than the original space.

Here is a concrete example: Take $c_0$ the space of sequences that converge to zero with the $\sup$ norm. Then it turns out that $c_0^* = l_1$, the space of summable sequences, and furthermore, $l_1^* = l_{\infty}$, the space of bounded sequences. Hence $c_0^{**} = l_{\infty}$, and $l_{\infty}$ is vastly larger than $c_0$.

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There are two things that can go wrong in the infinite-dimensional (normed) case.

First you could try to take the algebraic dual of $V$. Here it turns out that $V^{**}$ is much larger than $V$ for simple cardinality reasons as outlined in the Wikipedia article.

On the other hand, if $V$ is a normed linear space and you take the continuous dual, $V''$, then $V'$ (and thus also $V''$) will always be a Banach space. But! While $I$ as a map from $V$ to $V^{**}$ is obviously well-defined, it is not entirely obvious that $I(x)$ is in fact continuous for all $x$. In fact it is a consequence of the Hahn–Banach theorem which (roughly) states that there are "enough" continuous linear maps from $V$ to the base field in order for $V'$ and $V''$ to be interesting, e.g. the map $I$ is injective.

If $V$ is not a normed linear space, then things are more complicated and better left to a more advanced course in functional analysis.

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You don't need Hahn-Banach to see that $\lVert I(x) \rVert \leq \lVert x\rVert$. That's just unwinding the definition of the operator norm: $\lvert I(x)\varphi\rvert = \lvert \varphi(x)\rvert \leq \lVert \varphi\rVert \lVert x\rVert$, so I'd say it is obvious that $I(x)$ is in $V\prime\prime$ and thus $I\colon V \to V\prime\prime$ is well-defined and contractive. What's not obvious and does need Hahn-Banach is that $I$ is injective and in fact isometric, which is probably what you mean to say in the third paragraph. –  t.b. Aug 6 '12 at 11:00

(This is just a bit too longer to be a comment)

It is important to remark that the fact that for an infinite dimensional space $\bf V$ is not isomorphic to $\bf V^{**}$ requires the axiom of choice. Furthermore, the counterexample is not some pathological space.

In some models without the axiom of choice a peculiar thing happens: every linear operator from a Banach space into a normed space is continuous. One example of such model is Solovay's model in which all sets of reals are Lebesgue measurable. We remark that in this model the principle of Dependent Choice holds, which is enough to develop most classical analysis.

In such model $\ell_2$, a Banach space, has only continuous linear functionals, so the algebraic dual is the same as the topological dual. The fact that $\ell_2$ is a self-dual (in the topological sense) does not require much of the axiom of choice, not more than we have in Solovay's model anyway.

Now we need to verify that the evaluation map is a linear isomorphism, it is that and more. It is an isometry. This is not a very hard exercise in applying basic functional analysis theorems.

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Further, if I recall correctly, in the absence of AC there is a vector space $V$ with $V^{\vee} = 0$. And of course the very first step of my long answer was to choose a basis for $V$. So, as usual for infinite constructions in algebra, the story without AC is a completely different one. –  Pete L. Clark Aug 6 '12 at 10:58
    
Just a nitpick: automatic continuity is in fact an entire field where the term is interpreted in a bit more general way than the property you mention. It basically says that map satisfying some functional equation(s) (and maybe some decency conditions) is continuous: $\ast$-homomorphisms between $C^\ast$-algebras, derivations on Banach algebras, measurable homomorphisms between Polish groups, etc. The closed graph theorem itself is also an automatic continuity result. –  t.b. Aug 6 '12 at 11:07
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See Andres Caicedo's answer here for some facts from that field that you might find interesting. –  t.b. Aug 6 '12 at 11:14
    
@Pete: Indeed. There are such spaces, those are slightly more pathological in nature (although in a model where $\ell_2$ is self-dual, $\ell_\infty/c_0$ has no functionals at all). –  Asaf Karagila Aug 6 '12 at 11:15
    
@t.b.: I see. I did not know that. I'll edit slightly. –  Asaf Karagila Aug 6 '12 at 11:16

In the infinite dimensional case, $V^{**}$ will always have dimension strictly greater than that of $V.$ Let $\mathcal{A}$ be a basis for $V$ over $F.$ By the universal property for direct products, there exists a linear map $f: \hat{V} \rightarrow \displaystyle\prod_{a\in \mathcal{A}} F_a $ satisfying $\pi_A \circ f (v) = \epsilon_a (v)$ for all $v\in V^*,$ where $\epsilon_a $ is the evaluation at $a$ map and $\pi_a $ projects the range of $f$ onto its $a$th coordinate. It isn't too difficult to show that $f$ is an isomorphism. All that is left to show is that $\displaystyle\prod_{a\in \mathcal{A}} F_a$ has dimension stirctly larger than that of $V$ (for this, note that $V \cong \bigoplus_{a\in \mathcal{A}} F_a$ and give a cardinality-based argument)

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