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.

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 $D'$.