TITLE: Is there an injective operator with a dense nonclosed range?
QUESTION [5 upvotes]: Let $H$ be an infinite dimensional separable Hilbert space.   

Is there an operators $A \in B(H)$ such that  $Im(A) \subsetneq \overline{Im(A)} = H$ and $Ker(A) = \{0\}$ ?

Bonus : We can build such operators by using some compact or shift operators (see the answers).
Is there others possibilities ?
How classify this phenomenon ?

REPLY [3 votes]: By definition of the continuous spectrum $\sigma_{c}$ (possibly empty) :
$\forall \lambda \in \sigma_{c}(T)$ , $T-\lambda I$ is injective with nonclosed dense range.  
Example: the shift  
Let $H=l^{2}(\mathbb{Z})$ and $S \in B(H)$ defined by $S.e_{n} = e_{n+1}$, called the shift.
It's well-known that $\sigma(S) = \sigma_{c}(S) = \mathbb{S}^{1}$ (see here).   So $\forall \lambda \in \mathbb{S}^{1}$, $S-\lambda I$ is such an operator.