Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let X be a normed space and denote by X* the space of all bounded linear functionals on X. Take a linear subspace G ≤ X* which separates the elements of X, i.e., for each x ∈ X, there is an f ∈ G with f(x) ≠ 0. Denote by B the closed unit ball in X. Now consider a linear subspace Y ≤ X. The question is:

If Y is dense in X in the weak topology induced by G, is Y ∩ B necessarily dense in X ∩ B in that topology?

REMARKS, BACKGROUND AND MOTIVATION

Without the assumption that G separates points, there exists a trivial counter-example. Take X := ℝ2 with the supremum norm, i.e., ∥(x, y)∥ := max{|x|, |y|}. For G, take the linear span of the linear functional f(x, y) := x + y. Finally, take Y := {(x, 0) ; x ∈ ℝ}. Then Y is G-dense in X because (x, y) and (x + y, 0) are indistinguishable in the G-topology. However, the element (1, 1) ∈ B is not in the closure of Y ∩ B because f(1, 1) = 2 and f(x, 0) ≤ 1 for each x with (x, 0) ∈ B.

An interesting example is to take the space G := L(S), the space of all bounded measurable functions on a measurable space S, equipped with the supremum norm. Take X := G*, with the corresponding dual norm. The space G can be naturally considered as a subspace of X*. Clearly, it separates the points in X, and the G-topology is exactly the weak *-topology.
An important subspace of X is Y := M(S), the space of all real measures on S (with finite total variation). If S is large enough, let's say, ℕ, then Y is a proper subspace of X. It is well-known that Y is weakly *-dense in X, but it is also interesting that Y is weakly *-complete by sequences (see Diestel: Sequences and Series in Banach spaces, Springer-Verlag, 1984).
By the Banach-Alaoglu theorem, B is weakly *-compact. One may wonder whether X ∩ B is also weakly *-compact. The answer is no. However, the argument that Y is weakly *-dense in X is insufficient; a sufficient argument is that that Y ∩ B is weakly *-dense in X ∩ B. Though this is not difficult to prove in our particular case, it might by a non-trivial issue in more general cases. If the answer to my initial question is yes, it will be sufficient to only prove that Y is dense in X.

Many thanks in advance for any answer, reference or comment!

share|improve this question
add comment

3 Answers 3

up vote 2 down vote accepted

If I understand the question correctly, then maybe you are after special cases, as well as a general comment. So, as one of your examples suggests, one special case is to let G be a Banach space, considered as sitting inside its own bidual, and let $X=G^*$. Thus G induces the usual weak*-topology on X.

So an example of a positive answer is furnished by the Kaplansky Density Theorem: here G would be the predual of a von Neumann algebra, X would be a von Neumann algebra, and we let Y be any self-adjoint subalgebra which is weak*-dense. Then Kaplansky Density tells us that indeed the unit ball of Y is weak*-dense in the unit ball of X. This is an incredibly useful tool in Operator Algebra theory.

This then suggests that the result is unlikely to be true in general. Indeed, I think rpotrie's counter-example works! But here's an easier variant. Let $G=c_0$ and $X=\ell^1$, and let Y be the span of vectors $e_n+ne_{n+1}$. To see that this is weak*-dense, suppose that $\sum_k a_k e_k^* \in c_0$ annihilates all of Y. Then $a_n + na_{n+1}=0$ for all $n$, so $a_1 + a_2=0$ and $0 = a_2+2a_3 = 2a_3 - a_1$ and $0=a_3+3a_4 = (1/2)a_1+3a_4$, so $a_1 = -a_2 = a_3/2 = -a_4/3$ and an easy induction shows $a_1 = (-1)^{n-1}a_n/n$. Thus $|a_n| = n|a_1|$ for all $n$, but as $|a_n|\rightarrow 0$, it follows that $a_1=0$, and so actually $a_n=0$ for all $n$. Hence Y is weak*-dense. However, $e_1$ is in the closed unit ball of X, but it's pretty clear that we can't approximate it by norm one elements in Y (to do this without a tedious calculation defeats me right now).

share|improve this answer
add comment

You can derive a complete answer to your question and more from W. Davis, J. Lindenstrauss ``On total nonnorming subspaces", PAMS 31 no. 1 109-111 (1972), although their theorem is stated in the dual space.

share|improve this answer
add comment

I am not sure if this counter example works, but I am pretty sure. It is related to the counterexample mentioned in the question.

Consider in $\ell^1(\mathbb{Z})$ the usual ``base'' formed by the vectors $e_i$ given by the sequence having a $1$ in the position $i$.

We consider $G$ to be the subspace of $(\ell^1)^\ast = \ell^\infty$ generated by the dual of those vectors $e_i^\ast$ which clearly separates.

Now, consider $Y$ be the linear subspace generated by finite linear combinations of the vectors: $e_1$, $e_2 + 2e_3$, $e_3+ 2e_4+ 3e_5$, ....., $e_n + 2e_{n+1} + \ldots +(n-1) e_{2n-1}$, etc.

This subspace is dense since given any vector $v\in \ell^1$, we can arrange to construct a vector in $Y$ which coincides upto any finite number of coordinates with $v$.

However, if we consider for example the vector $x= \sum_i \frac 1 {i^2} e_i$, we can only aproach it with vectors in $Y$ of arbitrarily large norm.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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