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 $\ell_1$?

  • $\begingroup$ Sure. The zero operator. Maybe you meant to ask something else. $\endgroup$ – Bill Johnson Jun 23 '12 at 23:27
  • $\begingroup$ Of course, my mistake. Now edited. $\endgroup$ – Jan Vardøen Jun 23 '12 at 23:39


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.


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 $\ell_1$.

  • $\begingroup$ 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. $\endgroup$ – Philip Brooker Jun 24 '12 at 5:39
  • $\begingroup$ Phil, consider $x^* \otimes x$ where $x^*$ is in $\ell_2(2^{\aleph_0})$ and $x^*(x)=1$. $\endgroup$ – Bill Johnson Jun 24 '12 at 11:41
  • $\begingroup$ I will try to salvage my answer later when I find a spare moment to do so. $\endgroup$ – Philip Brooker Jun 25 '12 at 6:47
  • $\begingroup$ I removed the incorrect James tree space example from my answer, so hopefully my answer is now correct. $\endgroup$ – Philip Brooker Feb 21 '13 at 3:13

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