Sequences and bounded injective linear operator Let $X$ be a Banach space and $L:X\to X$ be a bounded linear operator which is one-to-one and not surjective, but with dense range. Suppose $(x_n)$ is a sequence of elements of $X$ such that the sequence $(Lx_n)$ converges to 0. Does it follow that $(x_n)$ converges to $0$?  

It does not.  As an example, take $X = \ell^2$ and 
$$
L\left[(x(k))_{k \in \Bbb N}\right] = \left(\frac 1k x(k)\right)_{k \in \Bbb N}
$$
The image of $L$ is dense, since it is a subspace containing all sequence that terminate in $0$s.  However, consider the standard basis as a sequence, i.e.
$$
x_n(k) = \begin{cases}
1 & k=n\\
0 & \text{otherwise}
\end{cases}
$$
then $Lx_n \to 0$, but $(x_n)$ fails to converge.
