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 $L^{2}(D,m)$.



Disreteness of spectra of Dirichlet laplacians
**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](http://dx.doi.org/10.1007/s11118-009-9125-9) 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
$$
 I\_t \geqslant \sum\_{n = 1}^\infty e^{-2 t \pi^2 / a\_n^2} = \sum\_{n = 1}^\infty (\log(n + e))^{-2 t \pi^2} = \infty .
$$


