7
$\begingroup$

We consider a Banach space $X$ and its dual $X^*$.

Let $Q\colon X^\ast \to X^\ast$ be an idempotent operator. Question: Can we find an idempotent operator $P\colon X^\ast \to X^\ast$ which is weak${}^\ast$-to-weak${}^\ast$ continuous and with range isomorphic to range of $Q$ and $\mbox{im}P\subseteq \mbox{im}Q$? In fact, I am mostly interested in the case $\mbox{im }Q\cong \ell_p$ for $p\in [1,\infty)$.

Certainly, $P$ would have to be an adjoint to some idempotent on $X$. My feeling is that in general this is not the case but perhaps it might be true for some well-behaved class of Banach spaces $X$ like Banach lattices? Or Banach lattices without a complemented copy of $\ell_1$?

$\endgroup$
2
  • $\begingroup$ Sure. The zero operator. Maybe you meant to ask something else. $\endgroup$ Jun 23, 2012 at 23:27
  • $\begingroup$ Of course, my mistake. Now edited. $\endgroup$ Jun 23, 2012 at 23:39

2 Answers 2

6
$\begingroup$

In

Stegall, C. Banach spaces whose duals contain $\ell_1(\Gamma)$ with applications to the study of dual $L_1(\mu)$ spaces. Trans. Amer. Math. Soc. 176 (1973), 463–477

Stegall proved that $\ell_2$ is isometrically isomorphic to a norm one complemented subspace of $X^*$ when $X= (\sum_{n=1}^\infty \ell_2^n)_1$, yet $\ell_2$ does not embed into $X$.

It would be interesting to have an example of this phenomenon with $X^*$ separable.

ADDED 6/25/12:

Here is a more interesting example because the dual is separable, but the range of the projection is not $\ell_p$. Take any separable reflexive space $X$ that fails the approximation property and let $(E_n)$ be an increasing sequence of finite dimensional subspaces of $X$ whose union is dense in $X$. Let $c(E_n)$ be the space of sequences $(x_n)$ with $x_n$ in $E_n$ and $\lim_n x_n$ existing in $X$, normed by $\|(x_n)\|= \sup \|x_n\|$.
Define $c_0(E_n)$ to be $(\sum_{n=1}^\infty E_n)_0$.

Consider the short exact sequence (ses)

$0\to c_0(E_n) \to c(E_n) \to X\to 0$

where the second arrow is the inclusion mapping and the third is the quotient mapping $Q$ defined by $(x_n)\mapsto \lim x_n$. This sequence locally splits (with constant one), hence $Q^*$ maps $X^*$ onto a norm one complemented subspace of $c(E_n)^*$. The ses itself does not split because $ c(E_n)$ has the approximation property (even a finite dimensional decomposition).

This example is Proposition 2.4 in

Johnson, William B.; Oikhberg, Timur: Separable lifting property and extensions of local reflexivity, Illinois J. Math. 45 (2001), no. 1, 123–137.

The construction itself is due to W. Lusky

A note on Banach spaces containing $c_0$ or $C_\infty$, J. Funct. Anal. 62 (1985), no. 1, 1–7.

Lusky was interested in the case that $X$ has the bounded approximation property. Then the resulting ses splits.

$\endgroup$
3
$\begingroup$

We can find some counterexamples for the case $p=1$ be looking inside the class of $\mathcal{L}_\infty$ spaces.

For the first example, let $K$ be a compact Hausdorff space such that $C(K)$ is a Grothendieck space. Then $C(K)^\ast$ contains a complemented copy of $\ell_1$, but such a complemented subspace can never be the range of a projection that is an adjoint operator.

For the second example, let us consider the scenario that $X$ is indecomposable and $X^\ast$ is decomposable. Then a projection $Q$ on $X^\ast$ such that $Q$ and $I_{X^\ast}-Q$ both have infinite dimensional range provides a counterexample to the OP's question. The main example that comes to mind for me is the Argyros-Haydon space, which is indecomposable yet has dual isomorphic to the (separable) space $\ell_1$.

$\endgroup$
4
  • $\begingroup$ Of course, the point here is that any operator $P$ on $X^\ast$ whose range lies inside the summand $\ell_2(2^{\aleph_0})$ is an adjoint operator if and only if it is the zero operator; I have included the operator $Q$ in my answer to make it clear how this relates to the OP's question. $\endgroup$ Jun 24, 2012 at 5:39
  • $\begingroup$ Phil, consider $x^* \otimes x$ where $x^*$ is in $\ell_2(2^{\aleph_0})$ and $x^*(x)=1$. $\endgroup$ Jun 24, 2012 at 11:41
  • $\begingroup$ I will try to salvage my answer later when I find a spare moment to do so. $\endgroup$ Jun 25, 2012 at 6:47
  • $\begingroup$ I removed the incorrect James tree space example from my answer, so hopefully my answer is now correct. $\endgroup$ Feb 21, 2013 at 3:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.