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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.

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Incidentally, your entire first paragraph probably fits in a MathOverflow title. I don't know enough Banach theory to know what words are reasonable to leave out, but I recommend that you rewrite the title into a question. Even "Is a continuous linear operator of Banach spaces the strong-operator limit of a net of adjoint operators?" is probably reasonable, although clearly it drops / gets wrong a number of conditions. –  Theo Johnson-Freyd Apr 28 '10 at 3:03
@ Theo Johnson-Freyd I think a complete question would be too long as a title... "Strong operator limits of adjoint operators in Banach Spaces" would be acceptable ? If not, you can give any "fancy" title you want. Incidentally, did you read mathoverflow.net/questions/7751/… ? –  Ady Apr 28 '10 at 4:02

3 Answers 3

up vote 7 down vote accepted

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.

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As Matt said, the result is true with $\ell_2$ replaced by any dual Banach space $X^*$ 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 $Y^{**} = Y\oplus X^*$ with the projection $P$ onto $X^*$ having norm one, $Y^*$ has the MAP (even a monotone basis), and the embedding of $X^*$ into $Y^{**}$ is isometric and a weak$^*$ to weak$^*$ homeomorphism. Since $Y^*$ has the MAP, norm one operators into $Y^*$ are strongly approximable by norm one finite rank operators. Thus if $P$, considered as an operator from $Y^{**}$ to $X^*$, 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 $X^*$ weak$^*$ 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.

I don't know what can happen when $X$ has the MAP but $X^*$ fails the MAP.

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See, for the beginning, arXiv:1002.3902v1, section 3. I will prepare the more or less full answers to all questions from above.

Oleg Reinov, S. Petersburg University

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