Let $X$ and $Y$ be Banach spaces and let $V\subseteq X$ be a dense subspace. Suppose $L:V\rightarrow Y$ is a bounded, injective and compact operator with $L(V)$ dense in $Y$. Can we conclude that the continuous extension $\bar{L}:X\rightarrow Y$ is also injective?
In my case I have that the extension is also compact but I am suspecting the compactness condition is irrelevant in this case? I have tried applying the closed graph theorem and I feel the denseness property of the image should imply injectivity of $\bar{L}$.
Let $X = Y = \ell^2$, $c_c$ are the finite sequences and $$ V = \left\{x \in c_c \;\middle|\; x_1 = \sum_{n=2}^\infty x_n \right\}.$$ Further, $$ L x := (x_2/2, x_3/3, x_4/4, \ldots).$$ It is clear that $L$ is compact and injective on $V$, but fails to be injective on $X$.
We just have to check that $V$ is dense. Every $x \in \ell^2$ can be approximated by $y \in c_c$, so we just have to check that $V$ is dense in $c_c$. For $y = (y_1,\ldots,y_n,0,\ldots)$ and $m \in \mathbb N$ we set $c = y_1 - \sum_{j=2}^n y_j$ and $$z = (y_1,\ldots,y_n, \underbrace{-c/m,\ldots,-c/m}_{\text{$m$ times}}).$$ Then, $z \in V$ and $\|y-z\|_{\ell^2}^2 = c^2/m \to 0$ as $m \to \infty$.