15
$\begingroup$

While digging through old piles of notes and jottings, I came across a question I'd looked at several years ago. While I was able to get partial answers, it seemed even then that the answer should be known and in the literature somewhere, but I never knew where to start looking. So I thought I'd ask here on MO if anyone knows of a reference for this observation.

Here's the notation and background for the question. Let $(E,\Vert\cdot\Vert)$ be a real, normed vector space (I think the complex case works out to be almost identical). We will see shortly that my question is vacuous unless $E$ is incomplete. Let $F$ be the completion of $E$.

Denote by $B(E,\Vert\cdot\Vert)$ the space of all linear maps $T:E\to E$ which ae bounded with respect to $\Vert\cdot\Vert$, i.e. there exists $C$ depending on $T$ such that $$ \Vert T(x)\Vert \leq C\Vert x\Vert \;\;\hbox{for all $x\in E$.} $$

Clearly each $T\in B(E,\vert\cdot\Vert)$ extends uniquely to a bounded linear operator $F\to F$, and we thus get an injective algebra homomorphism $\imath:B(E,\Vert\cdot\Vert)\to B(F)$. The question arises: when does $\imath$ have dense range?

  • It is not hard to show that if $E=c_{00}$ and $\Vert\cdot\Vert$ is the $\ell_\infty$ norm then $\imath$ does indeed have dense range.

  • On the other hand, if $E=\ell_1$ and $\Vert\cdot\Vert$ is the $\ell_\infty$ norm, then by considering "blocks" which have $\ell_\infty$-norm 1 and large $\ell_1$-norm, we can construct an isometry on $c_0$ which is not approximable by operators of the form $\imath(T)$; in particular, $\imath$ does not have dense range in this case. I can't find the piece of paper where I wrote down the details, but I seem to recall that one obtains the same answer if we replace $\ell_1$ by $\ell_p$ for $1\leq p < \infty$ and take $\Vert\cdot\Vert$ to be the $\ell_r$ norm for any $p<r\leq\infty$.

So here are two explicit questions. I haven't looked at them properly since about 2004/5, so they may well have straightforward solutions.

Q1. Let $\Vert\cdot\Vert$ be any norm on $c_{00}$, and let $F$ be the completion of $c_{00}$ in this norm. Does $\imath: B(c_{00},\Vert\cdot\Vert)\to B(F)$ have dense range?

Q2. Let $(F,\Vert\cdot\Vert)$ be a Banach space with an unconditional basis. Let $E$ be a proper dense subspace of $F$ which is a Banach space under some norm that dominates $\Vert\cdot\Vert$. Does $\imath: B(E,\Vert\cdot\Vert)\to B(F)$ always have non-dense range?

If the answers to these are known, does anyone know where I might find references to these in the literature?

Update 5th July 2010: Q1 has a positive answer, as given by Bill Johnson below (a simmilar approach was also elaborated by Pietro Majer). As pointed out (ibid.) the question can be rephrased/generalized to the following:

given a separable Banach space $F$ and a dense linear subspace $E$ of countable dimension, can every bounded operator on $F$ be approximated by operators which take $E$ to $E$?

I'd still be interested to know the answer to Q2, even in the special cases where $F=\ell_p$ for some $1\leq p < \infty$.

$\endgroup$
3
  • $\begingroup$ What is $c_{00}$ and $c_{0}$. I guess one are sequences $\mathbb{N} \to \mathbb{R}$, which converge to $0$. But the other? $\endgroup$
    – Helge
    Jul 2, 2010 at 22:17
  • 2
    $\begingroup$ $c_{00}$ is sequences which have only finitely many nonzero terms; $c_0$ is, as you guess, sequences which converge to $0$. $\endgroup$ Jul 2, 2010 at 23:17
  • $\begingroup$ Both Pietro and Bill's answers to Q1 are very helpful (not just for the sketched proofs but for their bakcground remarks). Since I can only accept one, I'm accepting Bill's since it was first and has now been updated to say something about Q2 also. $\endgroup$
    – Yemon Choi
    Jul 11, 2010 at 21:46

2 Answers 2

11
$\begingroup$

Q1: Yes. You ask ``If $X$ is a countable dimensional dense subspace of the Banach space $Y$, are the operators on $Y$ which leave $X$ invariant dense in the operators on $Y$?" Use Mackey's argument for producing quasi-complements (just a biorthogonalization procedure, going back and forth between a space and its dual) to construct a fundamental and total biorthogonal sequence $(x_n,x_n^*)$ for $Y$ with the $x_n$ in $X$; even a Hamel basis for $X$. Now use the principle of small perturbations to perturb an operator on $Y$ to a nearby one that maps each $x_n$ back into $X$. I am traveling now and so can't provide details or references, but I think that is enough for you, Yemon. The key point is that the biorthogonality makes the perturbation work--if $x_n$ were only a Hamel basis for $X$ it is hard to keep control.

I have my doubts whether this result appears in print even if oldtimers like me know the result as soon as the question is asked.

EDIT 7/4/10: Once you get the biorthogonal sequence $(x_n,x_n^*)$ with $x_n$ a Hamel basis for $X$, you finish as follows: WLOG $\|T\|=1$ and normalize the BO sequence s.t. $\|x_n^*\|=1$. Define the operator $S$ on $X$, the linear span of $x_n$, by $Sx_n=y_n$, where $y_n$ is any vector in $X$ s.t. $\|y_n-Tx_n\| < (2^{n}\|x_n\|)^{-1}\epsilon$. On $X$ you have the inequality $\|T-S\|<\epsilon$, so you get an extension of $S$ to $Y$ that satisfies the same estimate on $Y$. In checking the estimate you use the inequality $\|x\| \ge \sup_n |x_n^*(x)|$; i.e., biorthogonality is crucial.

To get the biorthogonal sequence, you take any Hamel basis $w_n$ for $X$ and construct the biorthogonal sequence by recursion so that for all $n$, span $(w_k)_{k=1}^n = $ span $(x_k)_{k=1}^n$. At step $n$ you choose any $x_n$ in span $(w_k)_{k=1}^n $ intersected with the intersection of the kernels of $x_k^*$, $1\le k < n$, and use Hahn-Banach to get $x_n^*$.

The Mackey argument I mentioned gives more. If you have any sequence $w_n$ with dense span in $Y$ and any $w_n^*$ total in $Y^*$, with a back and forth biorthogonalization argument you can build a biorthogonal sequence $(x_n,x_n^*)$ s.t. for all $n$, span $(x_k)_{k=1}^{2n}$ contains span $(w_k)_{k=1}^n $ and span $(x_k^*)_{k=1}^{2n}$ contains span $(w_k^*)_{k=1}^n $. This is quite useful when dealing with spaces that fail the approximation property; see e.g. volume one of Lindenstrauss-Tzafriri and, for something recent, my papers with Bentuo Zheng, which you can download from my home page.

EDIT 7/11/10: Getting a general positive answer to Q2 would be very difficult. Although not known to exist, it is widely believed that there is a Banach space with unconditional basis upon which every bounded linear operator is the sum of a diagonal operator and a compact operator. On such a space, the operators that map $\ell_1$ into itself would be dense in the space of all bounded linear operators.

$\endgroup$
3
  • $\begingroup$ Thanks, Bill, both for the outline and the closing remarks about this probably being folklore. Also, your rephrasing of the question is much less cumbersome and somehow more to the point. I will have a look at this argument in more detail. $\endgroup$
    – Yemon Choi
    Jul 4, 2010 at 1:12
  • $\begingroup$ And thanks for these latest remarks on Q2. I have since found my original notes on special cases of these questions -- still with me after five changes of house in as many years! -- and might have another look at the case of $\ell_p$. But I think your answer addresses both my underlying questions: namely whether the two questions are known folklore results or (variants of) known open problems. $\endgroup$
    – Yemon Choi
    Jul 11, 2010 at 21:44
  • $\begingroup$ Thanks for the update, Bill. I don't know what it says about me that I'd not checked this for my myself... $\endgroup$
    – Yemon Choi
    Oct 15, 2014 at 23:24
4
$\begingroup$

For a positive answer to Q1, I came to the same assumptions stated by Bill Johnson in his answer, so I'll adopt his notations.

Let $X$ be a dense, countably generated linear subspace of the separable Banach space $Y$ with unit ball $B_Y.$ Let $T\in L(Y)$ and $\epsilon>0$. Write $X$ as increasing union of a sequence $0=X_0\subset X_1\subset \dots$ of finite dimensional subspaces, with linear projectors $P_n:Y\to X_n$ (in particular, $P_0=0.$)

I think we can choose the projectors $P_n$ (depending on $T$) in such a way that, for every $k$ we have $(I-P_n)T_{|X_k}\to 0$ in the operator norm, as $n\to\infty$. As a consequence, there exists a natural number $n_k$ such that $$\| (I-P_{n_k})\, T\, (P_k-P_{k-1})\|\leq \| (I-P_{n_k})\, T_{|X_k}\|\, \|P_k-P_{k-1}\| \leq \epsilon\ 2^{-k}.$$ The sum $$\sum_{k=1}^\infty\ P_{n_k}\, T\, (P_k-P_{k-1})$$ is punctually finite on $X$, therefore it defines a linear map $T_{\epsilon}:X\to X$ (indeed, it takes $X_k$ into $X_{n_k}$ for every $k$). On the subspace $X$, the operator $T$ also writes in the form $$\sum_{k=1}^\infty\ T\, (P_k-P_{k-1})$$ and one has, on the subspace $X$ $$T-T_{\epsilon}=\sum_{k=1}^\infty\ (I-P_{n_k})\, T\, (P_k-P_{k-1}).$$

By the choice of the sequence $n_k$ the latter series is normally convergent to an operator of norm less than $\epsilon$. Therefore $T_{\epsilon}$ extends to a bounded operator on $Y$ with a distance less than or equal to $\epsilon$ from $T$ such that $T_{\epsilon}(X)\subset\,X.$

The claim should be proved as suggested below by Bill Johnson. Also, a suitable lemma for proving the claim could be stated as follows:

Given the subspaces $\{X_n\}_n$ as above and a countable subset $A\subset Y,$ there are linear projectors $P_n:X\to X_n$ such that $P_na\to a$ as $n\to\infty,$ for all $a\in A.$

Applying this to $A$ equals to the image of a Hamel basis of $X$ via $T$, one has $\|(I-P_n)T_{|X_k}\|=o(1)$ as $n\to\infty$ as we wanted.

Rmk. It seems to me that the statement gains something in generality and semplicity if one considers a different Banach space as codomain: if $T:F\to F'$ is a bounded linear operator; $D\subset F$ is a countable subset; $D'\subset F'$ is dense linear subspace; then $T$ can be approximated in operator norm by operators that map $D$ into $D'$ -hence of course $\mathrm{span}(D)$ into $D'$. This way one sees where the assumptions are needed: countability is only relevant for $D$, density is only relevant for $D'$.

$\endgroup$
6
  • $\begingroup$ uhm I've a little doubt about what I wrote, still seems ok. $\endgroup$ Jul 4, 2010 at 0:54
  • $\begingroup$ Thanks. I've been a bit busy but have been filling in the gaps in the outline given by Bill Johnson. Will give this a look too. $\endgroup$
    – Yemon Choi
    Jul 4, 2010 at 1:10
  • $\begingroup$ Pietro, I see no reason that such a $Q_k$ should exist. $\endgroup$ Jul 4, 2010 at 3:27
  • $\begingroup$ Exact, I see it no longer too... Actually something less is really needed to make the argument work, so maybe it can be fixed? I'll think a little more. $\endgroup$ Jul 4, 2010 at 6:27
  • 1
    $\begingroup$ Oh, that is OK, Pietro. $X_k$ has some fixed dimension $N=N_k$, so you can choose $X_{k+1}$ of dimension $2N$ which contains $X_k$ and a small perturbation of $TX_k$ and use any projection onto $X_{k+1}$ which has dimension at most $(2N)^{1/2}$. Knowing in advance a bound on the norm of the projection you will use is a big help. $\endgroup$ Jul 4, 2010 at 16:20

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.