Is there a name for this class of operators, $T(x) = \sum_{j\geq 1} \lambda_j \langle x, u_j \rangle u_j$?

Question

Let $\lambda_j$ be an increasing sequence of positive numbers such that $\lambda_j \to \infty$ as $j \to \infty$.

Let us consider the map which takes $x \in \ell^2(\mathbb{N})$ and sends it to $$ Tx = \sum_{j =1}^\infty \lambda_j \langle x, u_j \rangle u_j, $$ where $\{u_j\}_{j \geq 1}$ is an orthonormal basis of $\ell^2(\mathbb{N})$.

I can see certain properties of this operator:

• in general it seems possible that $Tx \not \in \ell^2(\mathbb{N})$ even though $x \in \ell^2(\mathbb{N})$. For instance, one can take $u_j = e_j$, and consider $x = (1/j)$ and $\lambda= (j)$, then $Tx \equiv 1$, which is evidently not in $\ell^2(\mathbb{N})$.
• for $x$ such that $Tx \in \ell^2(\mathbb{N})$ we have $\langle Tx, x \rangle \geq 0$.
• it is an unbounded operator since $Tu_j = \lambda_j u_j$ and $\lambda_j \to \infty$.
• it is self-adjoint: if $x, y \in \ell^2(\mathbb{N})$ are such that $Tx, Ty \in \ell^2(\mathbb{N})$ then $\langle y, Tx \rangle = \langle x, Ty \rangle$.

Hence, I can see the type of operator I am considering is an unbounded, positive, self-adjoint operators when considered as an operator from $D \to \ell^2$, where $D = \{x : Tx \in \ell^2\}$.

Are all unbounded positive self-adjoint operators on a subspace of $\ell^2$ of this form?

Answer 1

No, not every self-adjoint operator in $\ell^2$ is of this form. Note that for $z\in\mathbb C\setminus \mathbb R$, the resolvent $(T-z)^{-1}$ of the operator you defined is given by $$ (T-z)^{-1}x=\sum_{j=1}^\infty (\lambda_j-z)^{-1}\langle x,u_j\rangle u_j. $$ Since $\lambda_j\to\infty$, the operator $(T-z)^{-1}$ is compact.

Not every positive self-adjoint operator has compact resolvent. For example, take the Laplacian with domain $H^2(\mathbb R)\subset L^2(\mathbb R)$. [If you insist on $\ell^2$ as your Hilbert spaces, you can transplant this example using a unitary from $\ell^2$ onto $L^2(\mathbb R)$.]

If you add to the list of your properties that the resolvent $(T-z)^{-1}$ is compact for some (equivalently for all) $z\in\mathbb C\setminus\mathbb R$, then every operator is indeed of the form you described: By the spectral theorem for compact operators, $(T-z)^{-1}$ must be of the form $$ (T-z)^{-1}x=\sum_{j=1}^\infty\mu_j \langle x,u_j\rangle u_j $$ for a (complex) null sequence $(\mu_j)$. A direct computation shows that $$ Tx=\sum_{j=1}^\infty (\mu_j^{-1}+z)\langle x,u_j\rangle u_j. $$ You can check that if $T$ is positive, $\mu_j^{-1}+z$ must be positive numbers. Moreover, since $\mu_j\to 0$, $\mu_j^{-1}+z\to\infty$.