Looking for an example of a nuclear operator which is not a diagonal linear operator

Question

Using the following definition for Diagonal Linear operator,

Let $\mathbb K \in $ {$\mathbb R, \mathbb C$} and let $(H,<.,.>_{H},||\cdot||_{H})$ be a $\mathbb K$-Hilbert Space. Then $A$ is a diagonal linear operator on $(H,<.,.>_{H},||\cdot||_{H})$ iff, $\exists \quad \mathbb H \in \mathcal P(H)$ (-the power set of $H$) & a mapping $\lambda$ from $\mathbb H$ to $\mathbb K$ such that:

1. $\mathbb H$ is an orthonormal basis of $H$,

2. $A$ is a mapping from {$v \in H : \sum_{h \in \mathbb H}|\lambda_{h}<h,v>_{H}|^{2}$ $\lt +\infty$} to $H$,

3. $\forall v \in \mathcal D(A)$, $Av = \sum_{h \in \mathbb H}\lambda_{h}<h,v>_{H}h$

and using the following definition for Nuclear Operator,

Let $\mathbb K \in \{\mathbb R, \mathbb C\}$. Let $(V, ||\cdot||_{V})$ and $(W,||\cdot||_{W})$ be $\mathbb K$-Banach spaces. Then $A:(V, ||\cdot||_{V}) \to (W,||\cdot||_{W})$ is a $\mathbb K$-linear Nuclear operator iff $\exists \{v_{n}\}_{n \in \mathbb N} \in V^{'}$ and $\exists \{w_{n}\}_{n \in \mathbb N} \in W$ such that

i)$A$ is $\mathbb K$-linear operator from $V$ to $W$

ii) $\Sigma_{n=1}^{\infty}||v_{n}||_{V^{'}}||w_{n}||_{W} \lt \infty$

iii)$\forall x \in V$, $Ax = \Sigma_{n=1}^{\infty}v_{n}(x)w_{n}$

I want to ask the following question:

can someone provide me an example of a Nuclear operator which is not Bounded-Linear ?? Because, every Nuclear operator is bounded linear. This fact is stated and proved almost everywhere. But, I haven't seen an example regarding the failure of the converse inclusion.

Mathematically speaking of the example (counterexample may be, I am having the feeling!! (don't know why)),

I want to prove or disprove the fact:

For every $\mathbb K \in \{\mathbb R , \mathbb C\}$, every $\mathbb K$-Hilbert Space $(H, <\cdot,\cdot>_{H}, ||\cdot||_{H})$, every orthonormal basis $\mathbb H \subseteq H$ of $H$, every diagonal linear operator $A$ with $\sum_{h \in \mathbb H} \lambda_{h}\lt \infty$ is a Nuclear operator.

P.S.:- Since many counterexamples in measure theory arise due to counting measure, I am trying the condition $\sum_{h \in \mathbb H} \lambda_{h}\lt \infty$ !!

Can anyone prove or disprove my claim??

Answer 1

You need $\sum_k |\lambda_k| < \infty$ condition, not just $\sum_k \lambda_k < \infty$. For counterexample you just need any series which convergence but not absolutely convergence. Take $a_n = \frac{(-1)^n}{n}$. Then $\sum_{n=1}^\infty \frac{(-1)^n}{n} = - \ln(2)$ is finite, but operator $T=\sum_{k=1}^\infty e_k (?,e_k) a_k$ are not nuclear. Because we can represent it like $T(x) = \sum_{k=1}^\infty e_k (x,e_k) a_k$ and $\sum_{k=1}^\infty ||e_k|| ||(?,e_k)|| |a_k| = \sum_{k=1}^\infty |a_k| = \infty$.

Conversely, if you have some diagonal operator $T_a(e_n) = a_n e_n$ such that $\sum_n |a_n| < \infty$ then you can represent it like $T_a(x)=\sum_n e_n (x,e_n) a_n$, and you can check that $\sum_n ||e_n|| ||(?,e_n)|| |a_n| = \sum_n |a_n| < \infty$.