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}, ...) $?

Question

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?

Answer 1

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.

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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).