What can be said about the dual space of an infinite dimensional real vector space? Are both spaces isomorphic? Or shall we have something like the dual of the dual is isomorphic to the initial vector space? (Same as with the perpendicular subspace in a Hilbert space)

Many thanks

  • 1
    $\begingroup$ The dual space of an infinite-dimensional vector space is always strictly larger than the original space, so no to both questions. This was discussed on MO but I can't find the thread. $\endgroup$ – Qiaochu Yuan Apr 29 '11 at 6:42
  • $\begingroup$ Thanks Qiaochu for the result. $\endgroup$ – El Moro Apr 29 '11 at 6:46
  • 1
    $\begingroup$ The MathOverflow question Qiaochu was probably referring to is at mathoverflow.net/questions/13322/… $\endgroup$ – Jonas Meyer Apr 29 '11 at 19:43

In the abstract vector space case, where "dual space" is the algebraic dual (the vector space of all linear functionals), a vector space is isomorphic to its (algebraic) dual if and only if it is finite dimensional.

Bill Dubuque gives a nice argument in a sci.math post (see Google Groups or MathForum)

If $\mathbf{V}$ is an infinite dimensional vector space over $\mathbf{F}$ of dimension $d$, then the cardinality of $\mathbf{V}$ as a set is equal to $d|\mathbf{F}|=\max\{d,|\mathbf{F}|\}$, and $\mathbf{V}$ is isomorphic to $\mathbf{F}^{(d)}$ (functions from a set of cardinality $d$ to $\mathbf{F}$ with finite support), and the dual $\mathbf{V}^*$ is isomorphic to $\mathbf{F}^d$ (all functions from a set of cardinality $d$ to $\mathbf{F}$), so $|\mathbf{V}^*| = |\mathbf{F}|^d$.

If the dimension of $\mathbf{V}^*$ is $d'$, we want to show that $d'\gt d$. Note that, as with $\mathbf{V}$, we have $|\mathbf{V}^*|=d'|\mathbf{F}| = \max\{d',|\mathbf{F}|\}$.

Now let $\{\mathbf{e}_n\}$ be a countable linearly independent subset of $\mathbf{V}$, and extend to a basis. For each $c\in \mathbf{F}$, $c\neq 0$, define $\mathbf{f}_c\colon \mathbf{V}\to\mathbf{F}$ by $\mathbf{f}_c(\mathbf{e}_n) = c^n$, and making $\mathbf{f}_c$ equal to $0$ on the rest of the basis. Thet set of all $\mathbf{f}_c$, $c\neq 0$, is linearly independent, so we can conclude that the dimension if $\mathbf{V}^*$ must be at least equal to $|\mathbf{F}|$ (in the finite case, we know the dimension is at least $d\gt |\mathbf{F}|$).

That means that $$|\mathbf{V}^*| = d'|\mathbf{F}| = \max\{d',|\mathbf{F}|\} = d'.$$

But we also know that $|\mathbf{V}^*| = |\mathbf{F}|^d$. Since $d< |\mathbf{F}|^{d}$ (since $|\mathbf{F}|\geq 2$), then $d' = |\mathbf{F}|^d\gt d$, proving that the dimension of $\mathbf{V}^*$ is strictly larger (in the sense of cardinality) than that of $\mathbf{V}$.

The isomorphism in the finite dimensional case is standard.

So for the algebraic dual, there is never an isomorphism in the infinite dimensional case.

In the Hilbert space case (or in a Banach space, or more generally a topological vector space), one usually restricts to the continuous (or bounded) functionals, so that $\mathbf{V}^*$ denotes the bounded functionals rather than the regular functions. In that case, some spaces are topological-vector-space isomorphic to their double duals, and some not, as AD shows in his answer. The ones that are isomorphic are important enough to get their own name (reflexive). Hilbert spaces are always reflexive, and there are other classes of topological vector spaces that are always reflexive (see Wikipedia's page on reflexive spaces).


No, this is not the case in general. For example, take the real vector space of finite sequences $\{a_n\}$ of real numbers. Then every sequence $\{b_n\}$ of real numbers (finite or infinite) defines a linear function on this space via $\sum_n{a_nb_n}$, and the spaces of finite sequences and infinite sequences are not isomorphic.

  • $\begingroup$ Vielen danke joriki $\endgroup$ – El Moro Apr 29 '11 at 6:50
  • 1
    $\begingroup$ @El Moro: Actually, my initial answer contained an error since the cardinality of the spaces of finite and infinite sequences of reals is the same (namely that of the reals), just not the dimension -- I've deleted that part and will replace it with a more careful argument. $\endgroup$ – joriki Apr 29 '11 at 6:53

This depends on what you mean by the question. The algebraic dual on an infinite dimensional space is strictly larger. For the topological dual of a topological vector space $X$ we may very well have $X^{**}=X$.

Example 1. Sometimes we do have $X^{**}=X$ (read isometric to not equal to) for an infinite dimensional topological vector space, such space is said to be reflexive. The basic example is $X=L^p$ where $1<p<\infty$, in this case we have $(L^p)^*=L^q$ where $1/p+1/q=1$. (Here is a link to a wikipedia article on this also look at this ).

Example 2. Sometimes we have $X^{**}\ne X$ for an infinite dimensional topological vector space. The basic example is $L^1$, where we have $(L^1)^*=L^\infty$ while $(L^\infty)^*$ is a larger space of measures (containing $L^1$). (The same link as above applies).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.