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 $1\leq p < \infty$.

  • $\begingroup$ 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
    Commented 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 $0$. $\endgroup$ Commented 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
    Commented Jul 11, 2010 at 21:46

2 Answers 2


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.

  • $\begingroup$ 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
    Commented 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
    Commented 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
    Commented Oct 15, 2014 at 23:24

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 $D'$.

  • $\begingroup$ uhm I've a little doubt about what I wrote, still seems ok. $\endgroup$ Commented 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
    Commented Jul 4, 2010 at 1:10
  • $\begingroup$ Pietro, I see no reason that such a $Q_k$ should exist. $\endgroup$ Commented 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. $\endgroup$ Commented 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. $\endgroup$ Commented Jul 4, 2010 at 16:20

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