Closure of range of injective compact operator on a Hilbert space Let $\mathcal{H}$ be a separable Hilbert space and $C$ a compact operator on $\mathcal{H}$. Assume that $C$ is injective. Is it then true that the closure of the range of $C$ is $\mathcal{H}$?
 A: Here's a counterexample. Let $C : l^2 (\mathbf N) \to l^2 (\mathbf N)$ be defined as
$$ (x_1, x_2, x_3, x_4, ...) \mapsto (0, x_1, \tfrac 1 2 x_2, \tfrac 1 3 x_3, \tfrac 1 4 x_4, \dots)$$
Then $C$ is compact and injective, but $(1, 0, 0, 0, 0, \dots) $ is not in the closure of the range of $C$.

However, if we assume that $C$ is self-adjoint, then the statement is true.
In fact, if $\mathcal H$ is a Hilbert space (but not necessarily separable) and if $C: \mathcal H \to \mathcal H$ is self-adjoint, injective and continuous (but not necessarily compact), then the closure of the range of $C$ is the whole of $\mathcal H$.
For it's generally the case that
$$ \overline{{\rm ran}(C)} = ({\rm ker}(C^\star))^{\perp}.$$
If $C$ is self-adjoint, then $C^\star$ = $C$, and if $C$ is injective, then ${\rm ker}(C) = {0}$ and hence $({\rm ker}(C))^{\perp}$ is the whole of $\mathcal H$.

In my original example at the top, my $C$ was not self-adjoint. Its adjoint, $C^\star$, maps
$$ (y_1, y_2, y_3, y_4, ...) \mapsto (y_2, \tfrac 1 2 y_3, \tfrac 1 3 y_4,  \dots)$$
So ${\rm ker}(C^\star)$ is the one-dimensional subspace spanned by $(1, 0, 0, 0, \dots)$. Hence $\overline{{\rm ran}(C)}$ is the subspace of elements where the first component is zero.
