Double Orthogonal Complement - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T12:47:49Z http://mathoverflow.net/feeds/question/53330 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/53330/double-orthogonal-complement Double Orthogonal Complement Andre 2011-01-26T07:05:41Z 2012-11-28T19:52:12Z <p>Let $V$ be a complex inner product space. If $W$ is a closed subspace of $V$, we may define $W^\perp$ to be the subspace of all vectors $v \in V$ such that $\langle v | w\rangle =0$ for all $w \in W$. Repeating this construction a second time, we obtain the subspace $W^{\perp \perp}$, the double orthogonal complement of $W$. A moment's thought shows that $W \subseteq W^{\perp\perp}$, and $(W^{\perp\perp})^{\perp\perp}= W^{\perp\perp}$. If $V$ is complete, i.e., a Hilbert space, then $W = W^{\perp\perp}$. Presumably, this equality is not true in general, though I don't know of a counterexample. Thus, the equation $W = W^{\perp\perp}$ defines a class of closed subspaces of $V$.</p> <p>If $X$ is a set, let $\ell^1(X)$ denote the complex inner product space of absolutely summable functions $X \to \mathbb{C}$ with the usual inner product $\langle g |f\rangle= \sum_{x \in X}\overline{g(x)}f(x)$. If $\mathcal H$ is a Hilbert space, let $\mathcal B^1(\mathcal H )$ denote the complex inner product space of trace class operators on $\mathcal H$ with the usual inner product $\langle b | a \rangle = \mathrm{Tr}(b^*a)$.</p> <p>Which closed subspaces $W \subseteq \ell^1(X)$ satisfy $W = W^{\perp\perp}$? Which closed subspaces $W \subseteq \mathcal B ^1 (\mathcal H)$ satisfy $W = W^{\perp\perp}$? More formally, is it possible in each case to characterize the class of subspaces $W$ satisfying $W = W^{\perp\perp}$ in another natural or revealing way?</p> <p><strong>Edit.</strong> Here is a counterexample that shows that in general $W \neq W^{\perp\perp}$. Let $\ell^2$ be the usual Hilbert space of square summable sequences, and let $\ell^2_f$ be the subspace of sequences that are eventually zero. Let $s\in \ell^2$ be the sequence defined by $(s)_n = \frac 1 n$. Finally, let $W = \ell^2_f \cap {s}^\perp$, a closed subspace of $\ell^2_f$.</p> <p>For all natural numbers $n$, $W$ contains a sequence whose $n^{th}$ element is $-n$, whose $(n+1)^{st}$ element is $n+1$, and whose other elements are $0$. We deduce that $W^\perp \cap \ell^2_f = { 0 }$, so the double orthogonal complement of $W$ in $\ell^2_f$ is all of $\ell^2_f$. However, $W \neq \ell^2_f$ because $W$ does not contain the sequence $(1,0,0, \ldots)$. Thus $W$ is not equal to its double orthogonal complement in the complex inner product space $\ell^2_f$.</p> http://mathoverflow.net/questions/53330/double-orthogonal-complement/114807#114807 Answer by Chris Heunen for Double Orthogonal Complement Chris Heunen 2012-11-28T19:52:12Z 2012-11-28T19:52:12Z <p>In relation to the question: is it possible to characterize the class of subspaces $W$ satisfying $W=W^{\perp\perp}$ in another natural or revealing way? A <a href="http://dx.doi.org/10.2977/prims/1195195769" rel="nofollow">theorem of Amemiya and Araki</a> shows that the partially ordered set of such subspaces forms an orthomodular lattice if and only if the surrounding inner product space is a Hilbert space.</p>