Can I write $\mathcal H: \{c_n\} \to \{(n+\frac{1}{2})c_n\}$ as $H=\operatorname{diag}(\frac{1}{2}, 1 +\frac{1}{2}, 2+\frac{1}{2}, ...) $? We are in the Hilbert space $l^2(\mathbb{C})$ and this operator is given:
$$ \mathcal H : \{c_n\} \to \{(n+\frac{1}{2})c_n\}$$
Can I write $\mathcal H$ as $H=\operatorname{diag}(\frac{1}{2}, 1 +\frac{1}{2}, 2+\frac{1}{2}, ...) $ ?
If I can, I think that H would be:


*

*Invertible

*Not bounded

*Symmetric 

*Self-adjoint


Is it right?
 A: You can write $H=\operatorname{diag}(\frac12,\frac32,\frac52,\ldots)$ although this only holds in a formal way. As I have also stated in the comment section of your other question, $H:\ell_2\to\ell_2$, $(c_n)_n\mapsto ((n+\frac12)c_n)_n$ is not well-defined as it there are elements which leave $\ell_2$ when $H$ gets applied to them (e.g. $x=(\frac1n)_{n\in\mathbb N}$ is in $\ell_2$ but $Hx$ is not anymore as $(Hx)_n\overset{n\to\infty}\longrightarrow 1\neq 0$ so it's not even in $c_0$ anymore). So one way to define your operator is
$$
H:c_{00}\to\ell_2\qquad (c_n)_n\mapsto \Big(\Big(n+\frac12\Big)c_n\Big)_n.
$$
As $c_{00}$ (space of vectors with finitely many non-zero entries) is dense in $\ell_2$, $H$ is densely defined. Only now we can talk about properties of $H$.


*

*Obviously, $H$ is unbounded as $\Vert He_n\Vert=n+\frac12\overset{n\to\infty}\longrightarrow \infty$ where $e_n$ is the $n$-th standard vector.

*One also quickly sees that $H$ is symmetric, just check $\langle x,Hy\rangle=\langle Hx,y\rangle$ for any $x,y\in c_{00}=D(H)$. 

*$H$ as we've defined it here is essentially self-adjoint due to Nelson's Theorem but not self-adjoint for the following reason. Since $H$ is densely defined, $H^\dagger$ is closed by a well-known result. But $H$ when defined on $c_{00}$ is not closed (as can be seen easily) so $H$ and $H^\dagger$ can't coincide (basically because the domain of $H^\dagger$ is larger than the domain of $H$).

*With this definition of $H$, it is not invertible as it is not bijective (e.g. there is no $x\in c_{00}$ such that $Hx=(\frac1n)_{n\in\mathbb N}$ so $H$ is not surjective). If you modify the target space of $H$ to be $c_{00}$, then $H:c_{00}\to c_{00}$ as defined as above is bijective with inverse
$$
H^{-1}:c_{00}\to c_{00}\qquad (x_n)_{n\in\mathbb N}\mapsto \Big(\frac2{2n+1}x_n\Big)_{n\in\mathbb N}
$$
Writing $H^{-1}$ again in a formal way would yield $H^{-1}=\operatorname{diag}(2,\frac23,\frac25,\ldots)$ (but the domain of $H^{-1}$ is not all of $\ell_2$).


However, if you enlarge the domain of $H$ as follows
$$
D(H):=\lbrace x\in\ell_2\,|\,Hx\in\ell_2\rbrace\supset c_{00}
$$
then $H:D(H)\to\ell_2$ is an unbounded, densely defined operator which is self-adjoint and bijective. 
The point I'm trying to make here is that some of the properties you asked about heavily depend on the domain of your (densely defined) unbounded operator.

For more on the topic of unbounded operators I refer to "Methods of Modern Mathematical Physics. I: Functional Analysis" (1980) by Reed & Simon (Section VIII).
