Finding operator with given sequence of eigenvalues

Question

I am currently dealing with the spectral theorem for compact, self-adjoint operators. From this theorem, we know that for a linear, bounded, compact and self-adjoint operator the seuqence of eigenvalues is real and the only possible accumulation point is $0$. Hence, we can analyze e.g. the convergence of the sum of all eigenvalues of such operators.

I was wondering, if we there are statements for 'the other way around'. Assume, we have a sequence $(\lambda_n)_{n \in\mathbb{N}}$ of real values with accumulation point $0$. Let's assume that $(\lambda_n)_{n \in\mathbb{N}} \in \ell^1(\mathbb{R})$. Is there any linear, bounded, compact, self-adjoint operator $T$, such that this operator has the eigenvalues $(\lambda_n)_{\mathbb{N}}$. In other words, is there an one-to-one correspondence between the set of $\ell^1$ sequences and such operators $T$, or at least some relation?

I haven't seen anything in my lecture notes or in books on this topic. Is this, because it is only an 'unconventional' question or is it because there is no such statement?

Thank you in advance for your help!

EDIT: Of course, we need an operator, acting on an infinite-dimensional space

Answer 1

Let $\sigma = (\lambda_n)_{n=0}^\infty\subset \mathbb{R}$ be a sequence of real numbers accumulating at zero, such that $\sum|\lambda_n| < \infty$. (The summability condition implies that $|\lambda_n|$ has finite multiplicity and is bounded above.)

Let $H$ be a separable Hilbert space with inner product $\langle \cdot,\cdot\rangle$ and let $\{e_n\}_{n=0}^\infty$ be an orthonormal basis of $H$. Define an operator $T$ by setting $Te_n = \lambda_ne_n$, with domain $$D(T) = \bigg\{ v \in H\ \bigg|\ \sum_n (\lambda_n\langle v, e_n\rangle)^2 < \infty \bigg\}$$ Then:

1. Because $\sigma$ is absolutely summable, the operator $T$ is bounded and has domain $D(T) = H$.
2. As $\langle e_j, T^*e_k \rangle = \lambda_k$, the operator $T$ is manifestly self-adjoint.
3. For each $k \geq 0$, define the finite rank operator $T_k$ by setting $$T_ke_n = \begin{cases}\lambda_ne_n, & n \leq k \\ 0, & n > k\end{cases}$$ Denoting by $\|\cdot \|$ the operator norm, observe that $\|T_k - T\| \leq \sup_{i > k} |\lambda_k| \to 0$, so $T_k\to T$ in the operator norm. As $T$ is the operator norm limit of finite rank operators, it is compact.