If *Y* is weakly dense in *X*, is the unit ball in *Y* necessarily dense in the unit ball in *X*? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T19:59:12Z http://mathoverflow.net/feeds/question/32799 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/32799/if-y-is-weakly-dense-in-x-is-the-unit-ball-in-y-necessarily-dense-in-the-u If *Y* is weakly dense in *X*, is the unit ball in *Y* necessarily dense in the unit ball in *X*? Martin Raic 2010-07-21T14:05:06Z 2010-07-21T19:30:08Z <p>Let <em>X</em> be a normed space and denote by <em>X<sup>*</sup></em> the space of all bounded linear functionals on <em>X</em>. Take a linear subspace <em>G &le; X<sup>*</sup></em> which separates the elements of <em>X</em>, i.e., for each <em>x &isin; X</em>, there is an <em>f &isin; G</em> with <em>f(x) &ne; 0</em>. Denote by <em>B</em> the closed unit ball in <em>X</em>. Now consider a linear subspace <em>Y &le; X</em>. The question is:</p> <p>If <em>Y</em> is dense in <em>X</em> in the weak topology induced by <em>G</em>, is <em>Y &cap; B</em> necessarily dense in <em>X &cap; B</em> in that topology?</p> <h2>REMARKS, BACKGROUND AND MOTIVATION</h2> <p>Without the assumption that <em>G</em> separates points, there exists a trivial counter-example. Take <em>X</em> := &#8477;<sup>2</sup> with the supremum norm, i.e., &#8741;(<em>x</em>, <em>y</em>)&#8741; := max{|<em>x</em>|, |<em>y</em>|}. For <em>G</em>, take the linear span of the linear functional <em>f(x, y) := x + y</em>. Finally, take <em>Y</em> := {(<em>x</em>, 0) ; <em>x</em> &isin; &#8477;}. Then <em>Y</em> is <em>G</em>-dense in <em>X</em> because (<em>x</em>, <em>y</em>) and (<em>x</em> + <em>y</em>, 0) are indistinguishable in the <em>G</em>-topology. However, the element (1, 1) &isin; <em>B</em> is not in the closure of <em>Y &cap; B</em> because <em>f</em>(1, 1) = 2 and <em>f(x, 0) &le; 1</em> for each <em>x</em> with <em>(x, 0) &isin; B</em>.</p> <p>An interesting example is to take the space <em>G := L<sup>&infin;</sup>(S)</em>, the space of all bounded measurable functions on a measurable space <em>S</em>, equipped with the supremum norm. Take <em>X := G<sup>*</sup></em>, with the corresponding dual norm. The space <em>G</em> can be naturally considered as a subspace of <em>X<sup>*</sup></em>. Clearly, it separates the points in <em>X</em>, and the <em>G</em>-topology is exactly the weak *-topology.<br> An important subspace of <em>X</em> is <em>Y := M(S)</em>, the space of all real measures on <em>S</em> (with finite total variation). If <em>S</em> is large enough, let's say, &#8469;, then <em>Y</em> is a proper subspace of <em>X</em>. It is well-known that <em>Y</em> is weakly *-dense in <em>X</em>, but it is also interesting that <em>Y</em> is weakly *-complete by sequences (see Diestel: Sequences and Series in Banach spaces, Springer-Verlag, 1984).<br> By the Banach-Alaoglu theorem, <em>B</em> is weakly *-compact. One may wonder whether <em>X &cap; B</em> is also weakly *-compact. The answer is no. However, the argument that <em>Y</em> is weakly *-dense in <em>X</em> is insufficient; a sufficient argument is that that <em>Y &cap; B</em> is weakly *-dense in <em>X &cap; B</em>. 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 <em>Y</em> is dense in <em>X</em>.</p> <p>Many thanks in advance for any answer, reference or comment!</p> http://mathoverflow.net/questions/32799/if-y-is-weakly-dense-in-x-is-the-unit-ball-in-y-necessarily-dense-in-the-u/32821#32821 Answer by rpotrie for If *Y* is weakly dense in *X*, is the unit ball in *Y* necessarily dense in the unit ball in *X*? rpotrie 2010-07-21T17:17:13Z 2010-07-21T17:24:06Z <p>I am not sure if this counter example works, but I am pretty sure. It is related to the counterexample mentioned in the question.</p> <p>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 $i$. </p> <p>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.</p> <p>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. </p> <p>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 $v$.</p> <p>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.</p> http://mathoverflow.net/questions/32799/if-y-is-weakly-dense-in-x-is-the-unit-ball-in-y-necessarily-dense-in-the-u/32831#32831 Answer by Matthew Daws for If *Y* is weakly dense in *X*, is the unit ball in *Y* necessarily dense in the unit ball in *X*? Matthew Daws 2010-07-21T19:06:41Z 2010-07-21T19:06:41Z <p>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 <code>$X=G^*$</code>. Thus G induces the usual weak*-topology on X.</p> <p>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.</p> <p>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 <code>$\sum_k a_k e_k^* \in c_0$</code> 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).</p> http://mathoverflow.net/questions/32799/if-y-is-weakly-dense-in-x-is-the-unit-ball-in-y-necessarily-dense-in-the-u/32832#32832 Answer by Bill Johnson for If *Y* is weakly dense in *X*, is the unit ball in *Y* necessarily dense in the unit ball in *X*? Bill Johnson 2010-07-21T19:30:08Z 2010-07-21T19:30:08Z <p>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.</p>