I am interested in the Banach space $\mathcal{K}=\mathcal{K}(\ell^2)$ of compact operators on $\ell^2$, however my questions can be stated for any $\mathcal{K}(E)$, where $E$ is an arbitrary Banach space. I think that everyone who tries to study "classical" operator spaces like $\mathcal{K}$, Schatten $p$-class operators etc. immediately discovers the similarity with "commutative" counterparts, i.e. $c_0$ and $\ell^p$. This phenomenon is visible when one uses (generalised) singular numbers for certain classes of operators. Again, I have got plenty of questions concerning this stuff, let me list at least two of them:

1) what are the complemented subspaces of $\mathcal{K}$? Is $\mathcal{K}$ complemented in $\mathcal{B}(\ell^2)$? Recently, Haydon and Argyros constructed an HI-space $E$ such that $\mathcal{K}(E)$ has codimension 1 $\mathcal{B}(E)$, thus complemented.

2) is every bounded operator from $p$-Schatten class to $\mathcal{K}$ compact?

What other properties $\mathcal{K}$ shares with $c_0$?

  • $\begingroup$ I've removed the "operator spaces" tag, since nowadays this usually refers to operator spaces in the sense of Effros, Ruan et al. $\endgroup$
    – Yemon Choi
    Commented Jun 8, 2011 at 19:24
  • $\begingroup$ What if you search MathSciNet for "operator spaces" to see what it commonly refers to in recent years? $\endgroup$ Commented Jun 8, 2011 at 20:17
  • $\begingroup$ @Gerald: Well, of the top 40 hits, all but 2 (maybe 3) seem to use the term in Yemon's sense. $\endgroup$ Commented Jun 8, 2011 at 20:33

3 Answers 3


It is easy to see that whenever a space has an unconditional basis then the space of diagonal operators of the basis is equivalent to $\ell_\infty$. If $c_0$ embeds in $K(X,Y)$ then $K(X,Y)$ is not complemented in $B(X,Y)$. One reference for this is: M. FEDER. On subspaces of spaces with an unconditional basis and spaces of operators. Illinois J. Math. 34 (1980), 196-205.

It is also a direct consequence of a result from a Studia paper of Tong and Wilken from 1971. Here they prove that if $Y$ has an unconditional basis then $K(X,Y)$ is uncomplemented in $B(X,Y)$ (assuming the spaces are not equal).

As far as I know the Argyros-Haydon space is the first example of a space for which it is known that $K(X)$ is complemented in $B(X)$.


Probably in (2) you meant to ask whether every bounded operator from $K$ into a Schatten $p$ class is compact, since every operator from $c_0$ into $\ell_p$, $p<\infty$, is compact. But no either way: $K$ and any Schatten $p$ class contain complemented subpspaces isometrically isomorphic to $\ell_2$ (e.g. operators whose matrix representation has zeroes except in the first column).


This post is not intended as an answer to a question that was asked & answered 10 years ago, but as an appendix for newcomers to provide background about the DPP in the context, which the question owner and the commentators (and the other experts in the field) already know about.

  1. Every reflexive complemented subspace of a C*-algebra is isomorphic to a Hilbert space.

  2. TFAE for a C*-algebra $A$:

  • $A^{\ast}$ has DPP (Dunford-Pettis property)
  • $A$ has DPP
  • $A$ contains no complemented copy of $\ell^2$.
  • $A^{\ast\ast}$ is a finite type I von Neumann algebra.

Perhaps the major difference between $c_0$ & $K(\ell^2)$, $\ell^1$ & trace class operators on $\ell^2$, is the lack of DPP for the latter.

  1. On the contrary, if a C*-algebra $A$ does not have DPP, then there exists a complete isometry $K(\ell^2)\to A$.

  2. If $E$ is reflexive, then

  • a. $K(E)$ is Arens regular. (so is $c_0$)
  • b. $K(E)$ contains no copy of $\ell^1$. (so does $c_0$)
  • c. $K(E)$ does not have DPP.

As a consequence of (b) and (c), if $X$ is a subspace of $K(E)$, then both $X$ and $K(E)/X$ cannot have DPP simultaneously.

  1. If $E$ is reflexive and has approximation property, then $K(E)^{\ast\ast}= B(E)$. Thus, $K(E)$ is an ideal in its bidual $B(E)$. Equivalently, every left & right multiplication operator $L_a,R_a:K(E)\to K(E)$, defined by $L_ax = ax$ & $R_ax = xa$ are weakly compact.

On the other hand, the left & right multiplication operators on $c_0$ are compact by DPP.

  • 1
    $\begingroup$ This is helpful. $\endgroup$
    – Nik Weaver
    Commented May 16, 2021 at 13:00

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