Is the limit of the dual spaces the dual space of the limit?

Note: Whenever I say "dual space" in this question, I mean the algebraic dual space.

Consider an infinite-dimensional vector space $V$. Since it is infinite-dimensional, its double-dual $V^{**}$ is strictly larger than $V$. On the other hand, there's a natural injection $\iota:V\to V^{**}$, so by identifying $v$ with $\iota(v)$, we get $V\subsetneq V^{**}$.

Of course $V^{**}$ again is an infinite dimensional vector space, so taking its double-dual gives again a larger vector space. So if we define $$V_0=V, V_{n+1}=V_n^{**}$$ and do the identification as above, we get an infinite sequence of vector spaces $$V_0\subsetneq V_1\subsetneq V_2\subsetneq\dots$$ We can now define the limit space as $$V_\infty = \bigcup_{n=0}^\infty V_n$$ It is not hard to verify that this is also a vector space: For every finite set of vectors in $V_\infty$ you find a $V_n$ which contains all of them, and thus also their linear combination, which therefore also is in $V_\infty$.

Now obviously also for the dual spaces we have that $V_{n+1}^*$ is the double dual of $V_n^*$, therefore we can also define the limit space of the duals: $$W_\infty = \bigcup_{n=0}^\infty V_n^*$$

Now my question is: Is $W_\infty=V_\infty^*$?

It seems intuitive that it should be, but on the other hand, each $V_n$ (except for $V_0$) is also the dual of $V_{n-1}^*$, and thus the same intuition gives that $V_\infty=W_\infty^*$. But both together cannot hold, since otherwise the double-dual of $V_\infty$ would be again $V_\infty$, which cannot be since $V_\infty$ is infinite-dimensional. So clearly the intuition doesn't help here. Indeed, that argument seems to indicate that both are not the same; however I can't see how to make that rigorous.

Now clearly $W_\infty\subseteq V_\infty^*$, so assuming they are not equal, is it possible to explicitly construct an element of $V_\infty^*\setminus W_\infty$?

• The double dual is not always strictly larger, even in infinite dimensions. There are a wealth of spaces, called reflexive spaces, where they are naturally isomorphic. $L^p$ for $1<p<\infty$ are classic examples. In Hilbert spaces the situation is even cleaner, since $H,H^*,H^{**}$ are all naturally isomorphic. There are also irreflexive spaces where they are not naturally isomorphic but they are "non-naturally" isomorphic (i.e. they are linearly isometric but not under the canonical embedding). – Ian Oct 4 '14 at 22:17
• (Cont.) As I recall the process you've described will actually terminate for this reason after a small number of steps. Unfortunately I can't find a reference. – Ian Oct 4 '14 at 22:17
• Oh, sorry, I think you are talking about the algebraic dual space in this discussion, rather than the continuous dual space of a topological vector space. It might be helpful to readers to make that explicit. – Ian Oct 4 '14 at 22:24
• As for continuous duals, I think it's quite the opposite of what you are stating, Ian: either $X=X^{**}$ or the chain will go on forever. Every space is a closed subspace of its own continuous double dual and a closed subspace of a reflexive space is reflexive... – wisefool Oct 4 '14 at 22:26
• @Ian: Thanks, I indeed meant the algebraic dual space. I'll add a note to this effect. Of course, the situation with continuous duals might also be interesting. – celtschk Oct 4 '14 at 22:33

I believe this is a possible answer for your question. Please check it.

First of all, since we want $V_i^*\subset V_{k}^*$, $i\leq k$, we must understand how to evaluate $f\in V_i^*$ on elements of $V_k$.

Let $k=i+1$. Since $V_{i+1}=(V_i^*)^*$ then $f(x):=x(f)$, for every $x\in V_{i+1}$. Let us suppose that we understand how to evaluate $f\in V_i^*$ on elements of $V_k$ $($thus we may assume that $f\in V_k^*)$ then $f(x):=x(f)$ for every $x\in V_{k+1}$.

Remark 1: Thus, if $f\in V_i^*$, $x\in V_k$ and $k>i$ then $f(x)=x(f)$. $($Am I right here?$)$

Assuming this is right, let us construct two sequences $(x_{2n+2})_{n\in\mathbb{N}}$ and $(y_{2n+1})_{n\in\mathbb{N}}$ .

Since $V_{2n+1}^*\setminus V_{2n}^*\neq \emptyset$ then exists $x_{2n+2}\in (V_{2n+1}^*)^*=V_{2n+2}$ such that

1. $x_{2n+2}|_{V_{2n}^*}\equiv 0$ and
2. for some $y_{2n+1}\in V_{2n+1}^*\setminus V_{2n}^*$ we have $x_{2n+2}(y_{2n+1})=1$.

Let us prove that $(x_{2n+2})_{n\in\mathbb{N}}$ is linear independent.

Assume $g=\sum_{i=1}^ta_ix_{2n_i+2}=0$ and $n_1<n_2<\ldots<n_t$.

Now, $0=g(y_{2n_1+1})=\sum_{i=1}^ta_ix_{2n_i+2}(y_{2n_1+1})$.

If $n_i>n_1$ then $y_{2n_1+1}\in V_{2n_1+1}^*\subset V_{2n_i}^*$ and $x_{2n_i+2}(y_{2n_1+1})=0$, by construction.

Notice that $x_{2n_1+2}(y_{2n_1+1})=1$, by construction. Thus, $a_1=0$. Analogously, we obtain $a_2=\ldots=a_t=0$. Thus, $(x_{2n+2})_{n\in\mathbb{N}}$ is linear independent.

Now, let $f\in W_\infty$, thus $f\in V_{2n}^*$ for some $n$.

Thus, if $k\geq n$ then $x_{2k+2}\in V_{2k+2}$ and $2k+2>2n$. Therefore, by remark 1, $f(x_{2k+2})=x_{2k+2}(f)$.

Next, by construction, $x_{2k+2}(f)=0$, since $V_{2n}^*\subset V_{2k}^*$. Therefore if $k\geq n$ then $f(x_{2k+2})=0$.

Remark 2:Thus, for every $f\in W_\infty$, exists $n$ such that for $k\geq n$, $f(x_{2k+2})=0$.

Now since $(x_{2n+2})_{n\in\mathbb{N}}$ is linear independent, we can extend this set to a basis of $V_{\infty}$. Let $g\in V_{\infty}^*$ be such that $g(x_{2n+2})=1$ for every $n$. Thus, $g\in V_{\infty}^*\setminus W_{\infty}$, by remark 2.

• Very nice. As far as I can tell, your argumentation is correct. – celtschk Oct 5 '14 at 20:46