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 $||S\_{d}||\leq||T||$
$\left(d\in\mathcal{D}\right)$?


I guess not, say $E=c\_{0}(I)$ and $T$ is a surjection, where $I$
has a "big" cardinality. But maybe I'm
wrong.


Any help will be highly appreciated.



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 ?
You have that $T^\*:(\ell^2)^\* \rightarrow E^{\*\*}$, 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 $T^\*$ as a map $\ell^2\rightarrow E^{\*\*}$.


By the Principle of Local Reflexivity (I've used a paper of Behrends in the past, which is overkill, but is freely available: <http://matwbn.icm.edu.pl/ksiazki/sm/sm100/sm10022.pdf> Or look in a book on Banach space theory) for each triple $i=(M,N,\epsilon)$, where $M\subseteq E^{\*\*}$ and $N\subseteq E^\*$ 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 $\phi\in N$.


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 $\phi\in E^\*$, $$S^\*(\phi)(a) = \phi(S\_i T^\* P\_n(a)) \rightarrow T^\*(a)(\phi) = T(\phi)(a),$$ as $i$ and $n$ increase.


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.


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.


