Is there an injective operator with a dense nonclosed range?

Question

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 ?

score 11 · Accepted Answer · answered Jul 15 '13 at 18:15

11

You can check that for $$ T:\ell_2\to\ell_2: (x_1,x_2,x_3,\ldots)\mapsto(1^{-1} x_1,2^{-1}x_2, 3^{-1}x_3,\ldots) $$ we have $\operatorname{Ker}(T)=\{0\}$, $\overline{\operatorname{Im}(T)}=\ell_2$, but $T$ is not surjective because $(1,2^{-1},3^{-1},\ldots)\notin\operatorname{Im}(T)$.

answered Jul 15 '13 at 18:15

Norbert

56,803

thank you for this canonical example. – Sebastien Palcoux Jul 15 '13 at 18:22
1

More generally, any compact operator $A:\ell_2\to \ell_2$ with ${\rm ker}(A)={ 0}$ does the job. The range of $A$ is infinite-dimensional because $A$ is $1$-$1$, and so ${\rm Im}(A)$ cannot be closed because it would then be isomorphic to $\ell_2$ and this would give a compact isomorphism from $\ell_2$ onto itself. – Etienne Jul 15 '13 at 18:28
1

No. If you take $T=Id\oplus A$ acting on $\ell_2\oplus\ell_2$ with $A$ as above, then $T$ has the required properties but it is not compact. – Etienne Jul 15 '13 at 18:40
Thank you Etienne for this generalization. What is the set of all such operators ? – Sebastien Palcoux Jul 16 '13 at 23:41
1

@SébastienPalcoux We can say it is the set of all operators such that $0$ belongs to their continuous spectrum. Other than that... – Julien Jul 17 '13 at 00:04
Thank you @julien. I had forgotten that we can decompose the spectrum into a point, continuous and residual spectrum. – Sebastien Palcoux Jul 17 '13 at 08:30

score 3 · Answer 2 · edited Apr 13 '17 at 12:21

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.

Is there an injective operator with a dense nonclosed range?

2 Answers2