4
$\begingroup$

Let $X$ be a Banach space and $1<p<\infty$. How to characterize $X$ such that any operator from $X$ to $l_{p}$ is compact? Are there any known or new results?

$\endgroup$
  • $\begingroup$ This seems very open-ended. Perhaps you could provide some examples of $X$ that you already know satisfy this property? $\endgroup$ – Yemon Choi Sep 23 '15 at 1:27
  • $\begingroup$ As far as I know the only characterization is basically trivial: There is a non compact operator from $\ell_p$ to $X$ iff there is a normalized basic sequence $(x_n)$ in $X$ and a $C$ s.t. for all finite sequence $(a_n)$ of scalars we have $\| \sum_n a_n x_n \|^p \le C \sum_n |a_n|^p$. $\endgroup$ – Bill Johnson Sep 23 '15 at 2:32
  • $\begingroup$ Any closed subspace of $l_{r}(p<r<\infty)$ satisfies this property. This is famous Pitt's Theorem. $\endgroup$ – Dongyang Chen Sep 23 '15 at 13:17
  • $\begingroup$ This paper might have some relevant results: Maslyuchenko, O. V.; Mykhaylyuk, V. V.; Popov, M. M. Asymptotic structure and the existence of noncompact operators between Banach spaces. (English summary) J. Funct. Anal. 253 (2007), no. 2, 550–560. $\endgroup$ – Kevin Beanland Sep 24 '15 at 0:32
1
$\begingroup$

NOTE: The following proof is valid if one defines an “operator” as a linear function, not necessarily assumed to be bounded, between two vector spaces.

I claim that if $X$ is as you described, then it must be finite-dimensional.

To see this, suppose, for the sake of contradiction, that $X$ is infinite-dimensional. Then, the dimension of $X$ must be at least of the cardinality of the continuum. (See here. In fact, one can use Baire’s category theorem to conclude that the dimension of $X$ must be uncountable, but the former result is stronger without assuming the continuum hypothesis. Why one needs the stronger result that $\operatorname{dim} X\geq\#\mathbb R$ will be clear below.) The dimension of $\ell^p$, on the other hand, is precisely $\#\mathbb R$ for any $p\in(1,\infty)$. (After all, the cardinality of all real sequences is $\#\mathbb R$.) Let’s take a Hamel basis $\mathscr H$ of $X$ and a Hamel basis $\mathscr L$ of $\ell^p$.

By the preceding arguments, there exists a surjective function $f:\mathscr H\to\mathscr L$. Using the fact that basic representations are unique, one can extend $f$ to a surjective linear function $F:X\to\ell^p$. By assumption, $F$ is a compact operator, so it is a fortiori continuous. Letting \begin{align*} U\equiv\{x\in X\,|\,\|x\|<1\},\\ C\equiv\{x\in X\,|\,\|x\|\leq 1\}, \end{align*} one has that $F(C)$ is precompact in $\ell^p$, so that $F(U)$ is also precompact. Invoking the open-mapping theorem, one can conclude that $F(U)$ is a non-empty, precompact open set. But this is impossible, since $\ell^p$, being an infinite-dimensional normed vector space, cannot be locally compact.

$\endgroup$
  • 6
    $\begingroup$ "Operator" means continuous linear map. $\endgroup$ – Jochen Wengenroth Sep 23 '15 at 7:43
  • 2
    $\begingroup$ In my question, "operator"means linear bounded mapping. I am sorry. $\endgroup$ – Dongyang Chen Sep 23 '15 at 13:05
  • 1
    $\begingroup$ I agree with @JochenWengenroth since the question is posed in the setting of Banach spaces by someone who works on Banach spaces. It is common practice for those who work in the areas of Banach spaces, Banach algebras, Cstar algebra, von Neumann algebras, etc to say "operator" as short-hand for "bounded linear map" $\endgroup$ – Yemon Choi Sep 23 '15 at 15:39
  • 1
    $\begingroup$ @DongyangChen In this case, I’m sorry my answer couldn’t help. Thank you for the clarification. $\endgroup$ – triple_sec Sep 23 '15 at 17:32
  • 1
    $\begingroup$ @triple_sec Anyway, I thank you for your consideration. $\endgroup$ – Dongyang Chen Sep 23 '15 at 22:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.