# Is there a noncompact operator from $\ell^\infty$ to a reflexive space?

It is well known that bounded operators from $c_{0}$ to a reflexive Banach space $X$ are in fact all compact. Indeed, since it can be shown that an operator is compact iff for any weakly convergent sequence in its domain, its image is convergent, one may consider the adjoint, which is an operator from $X$ to $\ell_{1}$, and use the Schur theorem (it is also known that an operator is compact iff its adjoint is)

I wonder if it can be done "the other way round": if I take an operator which is adjoint to an adjoint to one from $c_{0}$ to $X$, it is an operator from $\ell^{\infty}$ to $X$, and is compact. Hovere these aren't (are they?) all bounded operators.

I would be very grateful if someone provides an example of a noncompact operator from $\ell^{\infty}$ to an reflexive space. If such operators exists, what about ones to $\ell_{p}$, for $p\in [1,\infty)$, maybe at least they have to be compact?

• The answer for operators from $\ell_\infty$ to $\ell_p$ is that a noncompact operator from $\ell_\infty$ into $\ell_p$ exists if and only if $p\geq2$; see Remark 2 on p.211 of the Appendix of Rosenthal's article On quasi-complemented subspaces of Banach spaces, with an Appendix on compactness of operators from $L_p(\mu)$ to $L_r(\nu)$, J. Funct. Anal. 4, 176-214 (1969). There also exist noncompact operators from $\ell_\infty$ into Tsirelson's space $T^\ast$, which can be seen to follow from Proposition 5 of the Castillo-Sanchez paper Remarks on some basic proprties of Tsirelson's space Commented Jun 22, 2013 at 12:02

It is "well-known" that $\ell^\infty$ is isomorphic to $L^\infty[0,1]$. The standard embedding of $L^\infty[0,1]$ into $L^2[0,1]$ is not compact (e.g. the image of the unit ball contains an orthonormal basis).

• Thanks a lot! I have completely forgotten about this beeautiful and suprising result of Pelczynski. However, still, maybe one can provide a direct example... But I really do apperciate your solution. In fact, I'm facing an exam in Banach spaces today, and the above was one of the questions which arised during the lecture :).
– L_b
Commented Jun 21, 2013 at 7:06

Another way of producing such operators is using Khinchine's inequality. It follows from this inequality that Rademacher functions in $L_1$ span a copy of $\ell_2$. Let $T\colon \ell_2 \to L_1$ be an isomorphic embedding. Then $T^*\colon L_\infty \to \ell_2$ is a surjection.

Using some other probabilistic techniques you may find isometric copies of all $\ell_p$ ($p\in (1,2])$ in $L_1$ so each $\ell_q$ with $q\geqslant 2$ is a quotient of $L_\infty$.

There are even larger operators from $\ell_\infty$ onto non-separable Hilbert spaces. More specifically, for each infinite set $\Gamma$ there exists a surjective operator $$T\colon \ell_\infty(\Gamma)\longrightarrow \ell_2(2^{\Gamma}).$$

For the proof see, e.g., this thread.