Density of sets whose image is dense

Question

This is probably easy, but I can't think of an answer. Assume $X$ is a Banach space and $A$ is a (not assumed closed) subspace of $X$. Let $T:X \to X$ be a bounded linear operator, which is also injective. If $T(A)$ is dense in $X$, does it follow that $A$ is dense in $X$?

Answer 1

Here's an explicit example for what Bill's answer proposes.

Let $T_1$ be the operator which maps $e_n$ to $4^{-n} e_{n-1}$ for $n \ge 2$ and maps $e_1$ to $0$. That is, $T_1 \left(\sum_{n \ge 1} a_n e_n\right) = \sum_{n \ge 1} a_{n+1} 4^{-(n+1)} e_n$. Note $T_1$ is injective on $A$, and $T_1 A$ is dense (it contains $c_{00}$).

Let $y = \sum_{n} 2^{-n} e_n$. Note $y \notin T_1 A$. Define $T$ by $T e_n = 4^{-n} e_{n-1}$ for $n \ge 2$ and $T e_1 = y$. Now we still have that $TA = T_1 A$ is dense. To see $T$ is injective, suppose $Tx = 0$. Then we can write $x$ uniquely as $x = a_1 e_1 + u$ where $u \in A$. Thus $T x = a_1 y + T_1 u$. But $T_1 A$ is a vector space not containing $y$, so if $Tx=0$ we conclude that $a_1 = 0$ and $T_1 u = 0$. But $T_1$ was injective on $A$ so $u = 0$. Thus $x=0$.

Answer 2

No, not even if $X=\ell_2$. Le $A$ be the closed span of $(e_n)_{n=2}^\infty$ and map $A$ to a proper dense subspace and $e_1$ to a vector not in that subspace.