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I am trying to investigate the relation between Uniformly Convexity and existence of Schauder Basis for a Banach space.

I read in a Handbook article that $B(H)$ (the algebra of all bounded operators on a Hilbert space $H$) and $B(H)/K(H)$ (the Calkin algebra for the same Hilbert space) are both naturally occurring examples of space which fails the Approximation Property.

I am wondering if we know:

a) if the above algebras are separable Banach spaces;

b) the convexity of them as Banach spaces (e.g. strictly convex, uniformly convex, ...?)

I apologize if this is really trivial functional analysis problem. Thank you!

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  • $\begingroup$ I can't help mentioning that the Theorem: "B(H) fails the AP" is a highly non-trivial result of Szankowski (to which a whole Acta 1981 paper is devoted). I think I've heard specialists would still like to understand this better and find an easier proof. $\endgroup$
    – Julien
    Commented Jul 4, 2013 at 4:49

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$B(H)$ is not separable since it contains isometric copy of $\ell_\infty$ which is non-separable. $B(H)$ is not strctly convex since it contains isometric copy of $\ell_\infty$ which is not strictly convex. A fortiori $B(H)$ is not uniformly convex because every uniformly convex space is stricly convex.

The same result holds for $C(H):=B(H)/K(H)$ since it contains isometric copy of $\ell_\infty/c_0$ which is not separable and not strictly convex.

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  • $\begingroup$ Thanks Norbert - do you have a reference for the above results? I would like to get acquainted with such kinds of results~ $\endgroup$ Commented Jul 3, 2013 at 21:10
  • $\begingroup$ @FanFeiChong some of them I just remember but I'll try to give references $\endgroup$
    – Norbert
    Commented Jul 3, 2013 at 21:11
  • $\begingroup$ @FanFeiChong I've added new referrences. If you wat I can write the proof of non separability of $\ell_\infty/c_0$ but in a separate qestion. $\endgroup$
    – Norbert
    Commented Jul 3, 2013 at 22:03
  • $\begingroup$ Thank you! It is very helpful! Just one more question, how does one see that there is an isometric copy of $l_\infty$ in $B(H)$? $\endgroup$ Commented Jul 4, 2013 at 2:45
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    $\begingroup$ @FanFeiChong You can assume $H$ is infinite-dimensional separable. Then fixing an orthonormal basis, the bounded diagonal operators in that basis are isometric to $\ell^\infty$. And the compact ones are isometric to $c_0$. $\endgroup$
    – Julien
    Commented Jul 4, 2013 at 4:57

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