Disreteness of spectra of Dirichlet laplacians I have fundamental questions on Dirichlet Laplacians.
Let $D \subset \mathbb{R}^d$ be an open subset and $\mathcal{L}$ be the (non positive) Dirichlet laplacian on $D$.
We denote by $T_t=e^{t\mathcal {L}}$ the semigroup on $L^{2}(D,m)$.
Here $m$ is the Lebesgue measure on $D$.
I am concerned with when $T_t$ becomes a compact operator on $L^{2}(D,m)$.
Each $T_t$ has a nonnegative kernel $p_t(x,y)$: 
$$T_tf=\int_{D}p_t(x,y)f(y)dm(y),\quad f \in L^{2}(D,m).$$
Thus, if it holds that $\int_{D}\int_{D}p_t(x,y)^2\,dm(x)\,dm(y)<\infty,$
$T_t$ becomes a Hilbert-Schimit operator and also becomes a compact operator on $L^{2}(D,m)$. 
For any $t>0$, there exists a positive constant $C_t$ such that $p_t(x,y) \le C_t \exp(-|x-y|^2/2t)$ for any $x,y \in D$. Thus, if $m(D)<\infty$, $T_t$ becomes a compact operator on $L^{2}(D,m)$.  
My question
Is there an unbounded open subset $D$ which satisfies the following conditions?:


*

*For any  $t>0$, $\int_{D}\int_{D}p_t(x,y)^2\,dm(x)\,dm(y)=\infty,$

*For any $t>0$, $T_t$ becomes a compact operator on $L^{2}(D,m)$.

 A: Yes. Take $D$ to be the union of disjoint intervals $(n, n + a_n)$, where $a_n$ takes values in $(0, 1)$ slowly converges to zero. (Or, in higher dimensions, take $D$ to be the union of disjoint balls with slowly decreasing radii. One can even make $D$ connected by joining the balls using sufficiently narrow rods).
On one hand, $T_t$ are all compact operators. I think this is a very classical result, but I do not know a good reference. The result is proved in Lemma 1 in this paper of mine, but let me stress it: this seems to be something very standard.
On the other hand, by choosing $a_n$ which converges to zero sufficiently slow, one can make the integral
$$
 I_t = \int_D \int_D (p_t(x,y))^2 dx dy
$$
infinite for every $t > 0$. Indeed: we have
$$
 \int_{(n, n+a_n)} \int_{(n, n+a_n)} (p_t(x,y))^2 dx dy = \sum_{k = 1}^\infty e^{-2 t \pi^2 k^2 / a_n^2} ,
$$
and therefore
$$
 I_t = \sum_{n = 1}^\infty \sum_{k = 1}^\infty e^{-2 t \pi^2 k^2 / a_n^2} .
$$
Set, for example, $a_n = (\log \log (n + e))^{-1/2}$. Then
$$
 I_t \geqslant \sum_{n = 1}^\infty e^{-2 t \pi^2 / a_n^2} = \sum_{n = 1}^\infty (\log(n + e))^{-2 t \pi^2} = \infty .
$$
