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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 scalar multiple of the identity 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.

3 Corrected typo 5/5/10

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 $X_n^*$. 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 $. This is quite useful when dealing with spaces that fail the approximation property; see e.g. volume one of Lindenstrauss-Tzafriri and, for something recent, my papers with Bentuo Zheng, which you can download from my home page. 2 added 1572 characters in body EDIT 7/4/10: Once you get the biorthogonal sequence$(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.

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