Let $H$ be $\ell^2({\mathbb N})$ and let $S:H\to H$ be the unilateral forward shift, so that $S^*S=I\neq SS^*$. Then a bounded operator $T:H\to H$ is Hankel if and only if it satisfies $TS=S^*T$.

Let $V$ be the space of all bounded Hankel operators on $H$. Does there exists a bounded linear projection from $B(H)$ onto $V$?

I thought I'd seen something on this question mentioned in Nikolskii's books on operator theory, but am currently having no luck finding anything on MathSciNet, nor via Google. The only thing I can find is a 1980 survey article by Steve Power, in which he remarks that it is "unlikely" that $V$ is complemented in $B(H)$; but there the trail appears to go cold. Certainly, the naive attempt to construct a projection by "averaging" doesn't seem to work.

Perhaps this question is known to be open? It feels like something that people would have worked on, since there are various negative results showing that certain other natural subspaces of $B(H)$ are uncomplemented.

  • 2
    $\begingroup$ Yemon, I presume $T=M$? $\endgroup$ Nov 16, 2011 at 6:48

1 Answer 1


The answer is no: there is no bounded projection from $B(H)$ onto $V$. For a proof, see for example Theorem 5.12 in Peller's book Hankel operators and their applications.

If you replace $B(H)$ by the Schatten class $S^p$ with $1\leq p <\infty$, the answer becomes yes. For $1<p<\infty$, the natural averaging projection is bounded, and actually completely bounded and even regular (with norm of order $\sqrt p$ if $p>2$, and $1/\sqrt{p-1}$ if $p<2$). There is an interesting phenomenon at $p=1$: the natural projection is no longer bounded, but there is another one which is bounded. This is explained in chapter 5 of Peller's book.

  • $\begingroup$ Thanks Mikael! Good to know that my memory wasn't playing tricks on me. $\endgroup$
    – Yemon Choi
    Nov 16, 2011 at 20:49
  • $\begingroup$ Minor correction: the results are in Section 5 of Chapter 6 of Peller's book $\endgroup$
    – Yemon Choi
    Nov 17, 2011 at 3:47
  • $\begingroup$ Oops, sorry for the mistake. $\endgroup$ Nov 17, 2011 at 8:35
  • $\begingroup$ Ca ne fait rien. Interesting that while Peller attributes Theorem 5.12 to Kislyakov, Kislyakov's proof is quantitative (apparently) while Peller's argument is purely qualitative $\endgroup$
    – Yemon Choi
    Nov 17, 2011 at 19:28

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.