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 $1\leq p < \infty$.



Approximating operators on Banach spaces by bounded operators on a proper dense subspace
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.


