Nuclear Operator with Finite Dimensional Range

Question

I have a question regarding nuclear operators. Let me first give the definition: Let $\Xi$ and $H$ be real Hilbert spaces, and let $T\in L(\Xi,H)$. $T$ is a nuclear operator if there exists a sequence $(\alpha_k)\subset H$ and a sequence $(\xi_k)\subset \Xi$ such that \begin{equation}\tag{1} T\xi = \sum_{k=1}^{\infty} \alpha_k \langle \xi, \xi_k \rangle \end{equation} for all $\xi\in\Xi$, and \begin{equation} \sum_{k=1}^{\infty} \| \alpha_k\| \| \xi_k\| <\infty. \end{equation} Now my question: Let $H=\mathbb{R}$, and let $(\xi_k)$ be an orthonormal basis of $\Xi$. For $\xi = \sum_{k=1}^{\infty} \alpha_k \xi_k \in \Xi$, define \begin{equation} T\xi := \sum_{k=1}^{\infty} \frac{1}{k} \langle \xi,\xi_k\rangle \end{equation} Then \begin{equation} \sum_{k=1}^{\infty} \frac{1}{k} \| \xi_k\| = \infty. \end{equation} However, $T$ should be nuclear, since $T$ has finite dimensional range. Is it possible to find the representation as in (1) for $T$?

Answer 1

You have $$ T=\frac\pi{\sqrt6}\,\langle \xi,\eta\rangle, $$ where $\eta=\frac{\sqrt6}\pi\,\sum_k\frac1k\,\xi_k\in\Xi$. That is, you take $\alpha_1=\pi/\sqrt6$, $\alpha_k=0$ for $k\geq 2$, and any orthonormal basis such that its first element is $\eta$.