Algebraic Dual / Continuous Dual - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T04:35:38Z http://mathoverflow.net/feeds/question/13747 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/13747/algebraic-dual-continuous-dual Algebraic Dual / Continuous Dual Ady 2010-02-01T22:04:34Z 2010-02-05T18:11:45Z <p>Let $E$ be an infinite dimensional Banach space, let $E^{*}$ denote its continuous (i.e., Banach space) dual, and let $E^'$ be its algebraic dual. Clearly, $E^{*}$ is a proper vector subspace of $E'$. Now, let us suppose that $E^{*}$ and $E'$ are algebraically isomorphic (i.e., as vector spaces). Does it follow that $E$ contains an isomorph of the Banach space $\ell_{1}(\mathbb{R})$ ? </p> <p>[By "*isomorph of <em>X</em>" I mean a closed linear subspace both algebraically and topologically isomorphic to <em>X</em>.]</p> <p>P.S. This is under ZFC + CH.</p> <p>P.P.S. The answer is affirmative if $E$ is the dual of a separable [infinite-dimensional] Banach space. It would be interesting to see if it is also affirmative when $E$ is a "nice" space. For instant, a Banach <em>lattice</em>.</p> http://mathoverflow.net/questions/13747/algebraic-dual-continuous-dual/14048#14048 Answer by Ben for Algebraic Dual / Continuous Dual Ben 2010-02-03T20:42:56Z 2010-02-03T20:42:56Z <p>Maybe we can look at $\ell_2(\gamma)$, where $\gamma$ is an uncountable set. The topological dual is itself. Its algebraic dual seems to have the same cardinality as $\ell_2(\gamma)$.</p> http://mathoverflow.net/questions/13747/algebraic-dual-continuous-dual/14094#14094 Answer by Ilya Grigoriev for Algebraic Dual / Continuous Dual Ilya Grigoriev 2010-02-04T01:52:14Z 2010-02-04T01:52:14Z <p>I think that the cardinality of <em>E'</em> should always be greater than the cardinality of <code>$E^*$</code>, so they never will be isomorphic in any sense. Basically, as pointed out <a href="http://mathoverflow.net/questions/13322/slick-proof-a-vector-space-has-the-same-dimension-as-its-dual-if-and-only-if-it/13372#13372" rel="nofollow">here</a>, <code>$E^*$</code> is the space of all maps from a topological basis of <em>E</em> into a field. <em>E'</em> is analogously the space of all maps from an algebraic basis to a field. So, this boils down to the question:</p> <blockquote> <p>In an infinite-dimensional Banach space, does an algebraic basis ever have the same cardinality as a topological basis?</p> </blockquote> <p>I think the answer is no. For example, in $l^2$ (which is the smallest infinite-dimensional Banach space), the topological basis is countable. As Ady points out, the algebraic basis should have cardinality $2^{|\mathbb N|}$. </p> <p>I'll community wiki this because this is a guess rather than a proof. Feel free to edit &amp; improve. I seem to remember I knew a slick proof of Ady's statement; if I actually remember it, I'll put it in.</p> http://mathoverflow.net/questions/13747/algebraic-dual-continuous-dual/14136#14136 Answer by Pandelis Dodos for Algebraic Dual / Continuous Dual Pandelis Dodos 2010-02-04T11:51:46Z 2010-02-05T01:48:13Z <p>Ok Ady, since you like CH I will work with CH, and to make your life easier, I will work with GCH.</p> <p>Since I do not expect that everybody in MO is aware of various Banach space constructions, let me give some information on James tree spaces which are relevant to the question. </p> <p>A <em>tree</em> is a partially order set $(T,&lt;)$ such that for every $t$ in $T$ the initial segment $\{s\in T: s &lt; t\}$ is well-ordered under $&lt;$. A segment of $T$ is a subset $S$ of $T$ which is:</p> <ol> <li>linearly ordered under $&lt;$ and</li> <li>for all $s, t, w\in T$ if $s &lt; t &lt; w$ and $s, w \in S$ then $t\in S$.</li> </ol> <p>The <em>completion</em> of $T$, usually denoted by $c(T)$, is the collection of all initial segments of $T$ ordered by inclusion. Notice that $c(T)$ contains $T$ and is much larger than $T$. For instance, if $T$ is the tree of all finite sequences of natural numbers (usually called the Baire tree, which is clearly countable), then its completion is the Baire-tree together with its branches (i.e. the Baire space) and so it has the cardinality of the continuum.</p> <p>For every tree $T$ the corresponding James tree space $JT$ is defined to be the completion of $c_{00}(T)$ with the norm: $$\|v\| = \sup\{ (\sum_{i=1}^d (\sum_{t\in S_i} v(t) )^2 )^{1/2} \}$$ where the above supremum is taken over all finite families $(S_i)_{i=1}^d$ of pairwise disjoint segments of $T$. Basic facts (I can provide appropriate references to anyone who is interested):</p> <ul> <li>For every tree $T$ the space $JT$ is hereditarily $\ell_2$; that is, every infinite-dimensional subspace of $JT$ contains a copy of $\ell_2$.</li> <li>For every tree $T$ the second dual of $JT$ is linearly isometric to the James tree space of the completion $c(T)$ of $T$. In particular, neither $JT^*$ nor $JT^{**}$ contain a copy of $\ell_1$. </li> </ul> <p>Now we come to the specifics of the construction. Remember that we work with GCH. This implies, in particular, the following: if $X$ is a Banach space of cardinality $\kappa$, then the algebraic dual of $X$ has cardinality $\kappa^+$.</p> <p>Let $T$ be the tree of all countable subsets of $\omega_1$ equipped with the partial order of end-extension. We have GCH, hence, the tree is just all sequences of real numbers, and so, it has cardinality $\aleph_1$. The cardinality of the corresponding James tree space is also $\aleph_1$. </p> <p>The completion $c(T)$ of our tree $T$ is the set of all subsets of $\omega_1$. Hence it has cardinality $2^{\aleph_1}$ which is, under GCH, $\aleph_2$. It follows that the cardinality of $JT^{**}$ is $\aleph_2$.</p> <p>Now consider cases.</p> <p>Case 1: the topological dual $JT^*$ of $JT$ has cardinality strictly bigger than $\aleph_1$. Then we are done: our counterexample is $JT$.</p> <p>Case 2: the topological dual $JT^*$ of $JT$ has cardinality $\aleph_1$. We are also done: our counterexample is $JT^*$.</p>