Is any continuous linear operator from a dual Banach space to a separable Hilbert space the strong-operator limit of a net of adjoint operators of less or equal norm ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T16:35:30Z http://mathoverflow.net/feeds/question/22799 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22799/is-any-continuous-linear-operator-from-a-dual-banach-space-to-a-separable-hilbert Is any continuous linear operator from a dual Banach space to a separable Hilbert space the strong-operator limit of a net of adjoint operators of less or equal norm ? Ady 2010-04-28T02:44:01Z 2010-07-06T09:06:36Z <p>Let $E$ be an arbitrary Banach space and let $T:E^{*}\rightarrow\ell^{2}$ be a linear continuous operator. Is it true that $T$ must be the $so$-limit (i.e., limit w.r.t. the strong operator topology) of a net $(S_{d})^{*}$ $\left(d\in\mathcal{D}\right)$ of adjoint operators, with $S_{d}$:$\ell^{2}$ $\rightarrow$ $E$ and $||S_{d}||\leq||T||$ $\left(d\in\mathcal{D}\right)$?</p> <p>I guess not, say $E=c_{0}(I)$ and $T$ is a surjection, where $I$ has a "big" cardinality. But maybe I'm wrong.</p> <p>Any help will be highly appreciated.</p> http://mathoverflow.net/questions/22799/is-any-continuous-linear-operator-from-a-dual-banach-space-to-a-separable-hilbert/22822#22822 Answer by Matthew Daws for Is any continuous linear operator from a dual Banach space to a separable Hilbert space the strong-operator limit of a net of adjoint operators of less or equal norm ? Matthew Daws 2010-04-28T07:57:58Z 2010-04-28T07:57:58Z <p>You have that <code>$T^*:(\ell^2)^* \rightarrow E^{**}$</code>, so using that $(\ell^2)^*$ is isomorphic to $\ell^2$ (just in the linear sense, as we already have a co-ordinate system), we can regard <code>$T^*$</code> as a map <code>$\ell^2\rightarrow E^{**}$</code>.</p> <p>By the Principle of Local Reflexivity (I've used a paper of Behrends in the past, which is overkill, but is freely available: <a href="http://matwbn.icm.edu.pl/ksiazki/sm/sm100/sm10022.pdf" rel="nofollow">http://matwbn.icm.edu.pl/ksiazki/sm/sm100/sm10022.pdf</a> Or look in a book on Banach space theory) for each triple $i=(M,N,\epsilon)$, where <code>$M\subseteq E^{**}$</code> and <code>$N\subseteq E^*$</code> are finite-dimensional, and $\epsilon>0$, we can find an operator $S_i:M\rightarrow E$ such that $(1-\epsilon)\|x\| \leq \|S_i(x)\| \leq (1+\epsilon)\|x\|$ for $x\in M$, and with $\phi(S_i(x)) = x(\phi)$ for $x\in M$ and $\phi\in N$.</p> <p>So, let $P_n:\ell^2 \rightarrow \ell^2$ be the projection onto the first $n$ co-ords, let $M\supseteq T^*(P_n(\ell^2))$ and let $i=(M,N,\epsilon)$, so $S=S_i T^* P_n$ makes sense, and is a map $\ell^2\rightarrow E$. Then, for $a\in\ell^2$ and $\phi\in E^*$, <code>$$S^*(\phi)(a) = \phi(S_i T^* P_n(a)) \rightarrow T^*(a)(\phi) = T(\phi)(a),$$</code> as $i$ and $n$ increase.</p> <p>So we have found a bounded net $(S_d)$ (we can even choose it bounded by $\|T\|$ be rescaling a little) with $S_d^* \rightarrow T$ in the weak operator topology. But now a standard trick (take convex combinations, as the closure of a convex set is the same in the weak and norm topologies) allows us to find a net which converges SOT.</p> <p>I think the proof would work for any Banach space F replacing $\ell^2$, as long as we can find a bounded net of finite rank operators $(F_\alpha)$ with $F_\alpha\rightarrow 1$ SOT. That is, F should have the bounded approximation property.</p> http://mathoverflow.net/questions/22799/is-any-continuous-linear-operator-from-a-dual-banach-space-to-a-separable-hilbert/22887#22887 Answer by Bill Johnson for Is any continuous linear operator from a dual Banach space to a separable Hilbert space the strong-operator limit of a net of adjoint operators of less or equal norm ? Bill Johnson 2010-04-28T18:25:28Z 2010-04-28T18:25:28Z <p>As Matt said, the result is true with $\ell_2$ replaced by any dual Banach space <code>$X^*$</code> that has the metric approximation property (MAP). It is necessary that $X$ has the MAP. For simplicity, assume that $X$ is separable. Then there is a James-Lindenstrauss space $Y$ so that <code>$Y^{**} = Y\oplus X^*$</code> with the projection $P$ onto <code>$X^*$</code> having norm one, <code>$Y^*$</code> has the MAP (even a monotone basis), and the embedding of <code>$X^*$</code> into <code>$Y^{**}$</code> is isometric and a weak<code>$^*$</code> to weak<code>$^*$</code> homeomorphism. Since <code>$Y^*$</code> has the MAP, norm one operators into <code>$Y^*$</code> are strongly approximable by norm one finite rank operators. Thus if $P$, considered as an operator from <code>$Y^{**}$</code> to <code>$X^*$</code>, were strongly approximable by norm one dual operators, then $P$ would be weak$^*$ approximable by norm one finite rank dual operators and you would thus get the identity on <code>$X^*$</code> weak<code>$^*$</code> approximable by norm one finite rank dual operators, which is to say you would have norm one finite rank operators on $X$ that converge weakly to the identity on $X$. Pass to convex combinations of these to see that $X$ must then have the MAP.</p> <p>I don't know what can happen when $X$ has the MAP but <code>$X^*$</code> fails the MAP.</p> http://mathoverflow.net/questions/22799/is-any-continuous-linear-operator-from-a-dual-banach-space-to-a-separable-hilbert/30739#30739 Answer by Oleg Reinov for Is any continuous linear operator from a dual Banach space to a separable Hilbert space the strong-operator limit of a net of adjoint operators of less or equal norm ? Oleg Reinov 2010-07-06T08:34:42Z 2010-07-06T09:06:36Z <p>See, for the beginning, arXiv:1002.3902v1, section 3. I will prepare the more or less full answers to all questions from above.</p> <p>Oleg Reinov, S. Petersburg University</p>