Let $\ell_2:=\{ x:\mathbb{N} \to \mathbb{R}: \sum_{j=1}^\infty x_j^2<\infty\}$ and $c_0:=\{ x:\mathbb{N} \to \mathbb{R}: \lim_{j\to\infty}x_j=0,\, \sup_{j\in\mathbb{N}}|x_j|<\infty\}$ denote the usual Banach sequence spaces. Given Banach spaces $A,B$, let $L(A,B)$ denote the Banach space of bounded, linear operators from $A$ to $B$.

Question: is $L(\ell_2,\ell_2)\subseteq L(\ell_2,c_0)$ dense?

  • 2
    $\begingroup$ I don't think that you need to mention sup. $\endgroup$ – Włodzimierz Holsztyński Oct 15 '14 at 9:10
  • 1
    $\begingroup$ Hi Michael, we met at HLF two weeks ago. :) $\endgroup$ – Tomasz Kania Oct 15 '14 at 11:25

Choose finite sets $A_n\subset{\bf N}$ with $\max(A_n)<\min(A_{n+1})$, and then define $T:c_{00}\to c_0$ by $T(e_n)=\chi_{A_n}$.

It is easy to check that $T$ extends to a norm 1 linear map $\ell^2\to c_0$. On the other hand, I think that if you take $|A_n|\to\infty$ fast enough, then for each norm 1 $R:\ell^2\to\ell^2$ you can find $\xi\in c_{00}$ such that $\xi$ has 2-norm equal to 1 but $R\xi-T\xi$ has $c_0$-norm at least 1/3. Now use Riesz's lemma to conclude that the answer to your question is negative.

(I am writing in a hurry on an iThing, so apologies for lack of detail and pretty formatting. When I have breathing space at a keyboard I will try to elaborate.)

Added later:

As Mikael has pointed out in comments, one does not need any assumption on the growth rate of $|A_n|$, merely that $\sup_n |A_n|=+\infty$. Moreover, one doesn't need Riesz's lemma.

To give a little more detail: suppose $R: c_{00}\to \ell^2$ is any linear map such that $\Vert R - T \Vert_{2\to \infty} = \varepsilon < 1$. Then for every $k\in A_n$ we have $|(R e_n)_k| \geq 1-\varepsilon$, and thus $$ \Vert R e_n \Vert_2 \geq (1-\varepsilon) |A_n|^{1/2} $$ Provided that $\sup_n |A_n|=+\infty$ this shows that $R$ cannot extend to any continuous linear map $\ell^2\to\ell^2$.

Remark you can all skip: The reason I originally thought we might want growth conditions on the sequence $(|A_n|)_{n\geq 1}$ is because I was misremembering something I'd worked out years ago. Namely, if $1\leq p < \infty$ and we let $B$ be the subalgebra of $L(c_0,c_0)$ consisting of all operators that moreover map the dense subspace $\ell^p\subset c_0$ to itself, then $B$ is not dense in $L(c_0,c_0)$. (This rediscovery is what prompted me to ask this old MO question.) Explicit operators in $L(c_0,c_0)$ which can't be approximated by elements of $B$ can be provided by spreading shifts of the kind used in my answer here, but the proof they are not in the closure of $B$ seems to require more work. However, I'm not sure if this question is genuinely related to the one you asked, or whether they are just different questions involving some of the same symbols.

  • $\begingroup$ True, thanks for the fast answer! Is there a good way to characterize the closure of $L(\ell_2,\ell_2)$? $\endgroup$ – Michael Feischl Oct 15 '14 at 11:16
  • 2
    $\begingroup$ Actually the "fast enough" is not needed: as soon as $|A_n| \to \infty$, the open ball of radius $1$ around $T$ in $L(\ell_2,c_0)$ does not intersect $L(\ell^2,\ell^2)$. $\endgroup$ – Mikael de la Salle Oct 15 '14 at 11:55
  • $\begingroup$ @MikaeldelaSalle you are completely correct - I wrote my answer before hurrying to catch a bus, and while on the bus realized what you wrote. I will update my answer to reflect this. $\endgroup$ – Yemon Choi Oct 15 '14 at 15:42
  • $\begingroup$ @MichaelFeischl Your follow up question seems to be a duplicate of mathoverflow.net/questions/168304/… if I am not mistaken $\endgroup$ – Yemon Choi Oct 16 '14 at 0:47
  • $\begingroup$ (@YemonChoi, was the bus ride smooth enough to edit MO Answers and Questions?) $\endgroup$ – Włodzimierz Holsztyński Oct 16 '14 at 1:05

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.