Unbounded operator $K\to H$ that is Hilbert--Schmidt from $Q^{1/2}K\to H$

Question

Regarding a part from Gawarecki and Mandrekar (2010), Stochastic Differential Equations in Infinite Dimensions, Springer

$K,H$: separable Hilbert spaces

$Q\colon K\to K$, either a symmetric positive definite trace class operator, or $Q=I_K$, identity on $K$.

Let $K_Q=Q^{1/2}K$, and let $L_2(K_Q,H)$ be the space of Hilbert--Schmidt operators from $K_Q$ to $H$.

Questions

In p. 25 they say

The space $L_2(K_Q,H)$ consists of linear operators $L\colon K\to L$ not necessarily bounded, with domain $D(L) \supset Q^{1/2}K$, and such that $\mathrm{tr}(LQ^{1/2})(LQ^{1/2})^*$ is finite. If $Q = I_K$, then $K_Q = K$. We note that the space $L_2(K_Q,H)$ contains genuinely unbounded linear operators from $K$ to $H$.

Exercise 2.7 Give an example of an unbounded linear operator from $K\to H$, which is an element of $L_2(K_Q,H)$.

1. (Seems this first question is resolved)When they say "the space $L_2(K_Q,H)$ contains genuinely unbounded linear operators from $K$ to $H$", (even though this is written right after the remark regarding the case $Q=I_K$) they mean when $Q$ is not the identity, am I correct?

Reasoning: If $Q=I_K$, then $L_2(K_Q,H)=L_2(K,H)$. Thus, $T\in L_2(K_Q,H)$ is compact as an operator from $K$ to $H$, in particular bounded?

2. What would be an example of an unbounded linear operator from $K\to H$, which is an element of $L_2(K_Q,H)$? If the reasoning above is correct, $Q$ cannot be identity?

Answer 1

Using Gawarecki-Mandrekar's notation, suppose the eivenvalues of $Q$ satisfies $\sum_{j=1}^{\infty}\lambda_j^{\frac12}<\infty$. This is stronger than $Q$ being simply of trace class. Let $$ Tk:=\sum_{j=1}^{\infty} \frac1{\lambda_j^\frac14}\langle k,f_j\rangle_{K}f_j, $$ for $k$ such that $Tk\in K$. Then, $T\colon K\to K$ is unbounded, as choosing $\kappa\in K$ such that $|\langle \kappa,f_j\rangle_{K}|=\lambda_j^\frac14$ we have $\kappa\in K$ but $\|T\kappa\|_{K}=\infty$.

Now, $$ \|T\|_{\mathscr{L}_2(K_Q,K)}^2 = \sum_{j=1}^{\infty}\|T(\lambda_j^\frac12f_j)\|^2 = \sum_{j=1}^{\infty}\lambda_j^{\frac12}<\infty. $$