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Let $V$ be a vector space. This answer gives a nice explanation of why the "dual dual space" of $V$, i.e., $(V^*)^*$, is isomorphic to $V$ if $\dim V<\infty.$

  1. Can someone give an example where $\dim V=\infty$ and $(V^*)^*\ncong V$? I'm having a hard time imagining this.
  2. Additionally, must $V$ always be at least isomorphic to a subspace of $(V^*)^*$? It seems that the answer must be yes, since we can always define $\xi_{v∈ V}:V^*\to\mathbb R, \omega\mapsto\xi_v(\omega):=\omega(v).$
WillG
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3 Answers3

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Since you use the tag "functional analysis", I'll assume that you talk about continuous duals.

Let $V=c_0$, that is the space of complex sequences that converge to zero. It is well-known (and fairly easy) to show that $V^*=\ell^1(\mathbb N)$, and $V^{**}=\ell^\infty(\mathbb N)$.

In both cases the duality is $$\langle x,y\rangle=\sum_nx_ny_n.$$

The space $c_0$ is norm-separable, while $\ell^\infty(\mathbb N)$ isn't.

As for your second question, yes, you always have a canonical injection $V\to V^{**}$.

For completion's sake, the answer is similar if we consider the purely algebraic dual. Take $V=\mathbb R[x]$, the space of real polynomials. It has a countable basis, namely $\{x^n\}_{n\in\mathbb N\cup\{0\}}$. For each $t\in\mathbb R$ consider $\phi_t:\mathbb R[x]\to\mathbb R$ given by $\phi_t(p)=p(t)$. The uncountable set $\{\phi_t\}$ is linearly independent, so $\dim V^*>\dim V$. Then $\dim V^{**}\geq\dim V^*>\dim V$ and they cannot be isomorphic (the fact that the dimension of the dual is at least that of the space follows from considering dual basis).

Martin Argerami
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  • I'm a little bit confused. What does the $\mathbb{N}$ mean in $\ell^1(\mathbb{N})$? The range of the sequences is a subspace of $\mathbb{N}$? – Botond Oct 30 '19 at 19:49
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    @Botond You can generally consider $l^1(X)$ for any set X - this is the space of real or complex valued functions on $X$ which are integrable with respect to counting measure. – Slup Oct 30 '19 at 19:51
  • I see, thank you @Slup. And how do you identify the duals with those spaces? I can't really imagine the elements of a dual without an inner product (well, maybe the zero functional is an exception). – Botond Oct 30 '19 at 19:59
  • There is a hole industry in functional analysis that describes topological duals of various typical spaces. In the case above you can try to do it yourself (it is an interesting and nontrivial exercise) or look it up in a textbook. – Slup Oct 30 '19 at 20:02
  • @Slup I have a functional analysis course currently, but I don't see how can we identify $(c_0)^$ with $\ell^1$. On a Hilbert space $H$, it's clear that we can identify $H^$ with $H$ because of the Riesz representation theorem, but it does not work here. – Botond Oct 30 '19 at 20:06
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    Ok. Pick $(a_n)\in l^1$. Then it induces a functional $c_0\ni (b_n)\mapsto \sum_{n}a_nb_n\in \mathbb{R}$. You have to check that is bounded functional and that all functionals on $c_0$ are of this form. There are also other things to verify... – Slup Oct 30 '19 at 20:09
  • Thank you again, @Slup – Botond Oct 30 '19 at 20:11
  • @Slup this was one of the topics at today's class :) – Botond Nov 05 '19 at 11:27
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One can give a simple argument based on cardinality counting for the algebraic case.

Suppose that $V$ is a vector space of dimension $\mathrm{card}(I)$. Then $V$ is isomorphic to a direct sum $k^{\oplus{I}}$ of $I$-copies of $k$. Next

$$\left(k^{\oplus I}\right)^* = \mathrm{Hom}_k(k^{\oplus I},k) \cong k^{\prod I}$$

where the right hand side is the product of $I$ copies of $k$ (considered as a vector space over $k$). Suppose now that

$$\mathrm{card}(k) < \mathrm{card}(I)$$

Then

$$\mathrm{card}\left(k^{\prod I}\right) \geq \mathrm{card}\left(2^I\right) > \mathrm{card}(I) = \mathrm{card}\left(k^{\oplus I}\right)$$

This implies that under the assumption $\mathrm{card}(I) > \mathrm{card}(k)$ we have

$$\mathrm{card}(V^*) > \mathrm{card}(V)$$

In particular $\mathrm{dim}(V^*) > \mathrm{dim}(V) = \mathrm{card}(I)$. Now you can use exactly the same argument to show that

$$\mathrm{card}(V^{**}) > \mathrm{card}(V^*)$$

So in general $V$ and $V^{**}$ do not have the same cardinality.

Slup
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For the first question. For every vector space $V$ it is true that $$ \dim(V)\leq \dim(V^*) $$ with equality if and only if $V$ is finite dimensional (see Theorem 3.12 in Advanced Linear Algebra by Steven Roman). Thus, if $V$ is infinite dimensional, then $V\not\cong V^{**}$, so every infinite dimensional vector space serves you as an example.

For question 2: Yes.

Albert
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