Conditions for the Hamiltonian's spectrum to be discrete I came across this article [1], in which the author studies some Hamiltonian that have a discrete spectrum even though they do not go to infinity at infinity.
In there, the author makes several claims, that I don't really get :

If $H_1 \geqslant H_2$ and $H_2$ has a discrete spectrum, then so does $H_1$.


If $\operatorname{Tr}e^{-tH}< \infty$ for any $t$, then $H$ has a discrete spectrum.

I would like to understand the intuition behind those results, as well as their formal proofs.
[1] Some quantum operators with discrete spectrum but classically continuous spectrum; B. Simon (Caltech); Published in: Annals Phys. 146 (1983), 209-220
 A: (I propose a second answer because the first was affected by a trivial but devastating   mistake).
First of all
Definition. A selfadjoint operator $A: D(A) \to H$, where $H$ is a complex Hilbert space and $D(A)\subset H$  a dense subspace,   has discrete spectrum if its spectrum $\sigma(A)$ is made of isolated points, thus eigenvalues, and the associated eigenspaces are all finite dimensional.
A first fact is true.
Proposition. Let $H$ be infinite dimensional and let $A =A^*$ be bounded below.
(a) If $A$  has discrete spectrum then the resolvent operator $(A -\lambda I)^{-1} : H \to D(A)$ is compact for every real  $\lambda\not \in \sigma(A)$.
(b) if the resolvent operator is compact for some real  $\lambda\not \in \sigma(A)$ then $A$ has discrete spectrum (and thus the resolvent is compact for all reals as above for (a)).
Remark. The resolvent operators of selfadjoint operators are always defined in the whole Hilbert space when the parameter is not part of the spectrum. This is the case for closed operators as the selfadjoint ones. Furthermore, essentially by definition, $D(A-aI) =D(A)$.
Proof.(a) As the Hilbert space is infinite dimensional and $A$ must admit a Hilbert basis of eigenvectors, and finally every eigenspace is finite dimensional, the spectrum must contain infinite isolated points. Since the spectrum is closed, it contains its accumulation points. As the points of the spectrum are isolated, the spectrum cannot admit accumulation points at all, and thus it must be unbounded. Finally, we assumed that the spectrum is bounded below, so that  the only possibility is that it is made of a sequence of elements $$\lambda_1< \lambda_2 < ...< \lambda_n \to +\infty$$
From spectral calculus, if the real  $\lambda$ does not belong to the spectrum,
$$(A-\lambda I)^{-1}= \sum_n (\lambda_n-\lambda)^{-1}P_n$$
where $P_n$ is the orthogonal projector onto the finite dimensional eigenspace of $\lambda_n$.
The series converges in the uniform operator topology because
$$||\sum_{n=N}^{+\infty}(\lambda_n-\lambda)^{-1}P_nx||^2\leq \sum_{n=N}^{+\infty}(\lambda_{n}-\lambda)^{-2}||P_nx||^2\leq (\lambda_{N}-\lambda)^{-2}||x||^2$$
and thus, taking the sup over the norm one vectors,
$$||\sum_{n=N}^{+\infty}(\lambda_n-\lambda)^{-1}P_n||^2\leq (\lambda_{N}-\lambda)^{-2}\to 0.$$
We have established that the resolvent is the uniform limit of finite rank operators hence it is compact.
(b) immediately arises from the Hilbert-Schmidt decomposition theorem of selfadjoint compact operators and standard spectral calculus. QED
Now, and this is the best I managed to prove, we have my version of the obscure statement in Simon's paper.
Theorem. Let  $A$ and $B$ be selfadjoint operators in the infinite dimensional complex Hilbert space $H$ such that
1. $D(B)\subset D(A)$,
2. $B \geq A$, $\quad$ (i.e. $\langle x, Bx\rangle \geq \langle x,Ax\rangle$ if $x\in D(B)$)
3. $A$ is bounded below.
Then $B$ has discrete spectrum if $A$ has discrete spectrum.
Proof. Since $A\geq cI$ for some $c>-\infty$, (2) entails that $B\geq cI$ and furthermore $c$ is also a lower bound of both spectra, so that  $\lambda < c$ cannot belong to $\sigma(A)\cup \sigma(B)$. In view of (a) in the proposition above, to conclude it is sufficient to show that if $$X^{-1}:=(A-\lambda I)^{-1}$$ is compact (with $\lambda$ as above), then $$Y^{-1}:=(B-\lambda I)^{-1}$$ is compact as well. To this end, observe that
$$Y^{-1} = X^{-1} (X Y^{-1})\:.$$
Since $D(X)=D(A)$, $D(Y)=D(B)$ and (1) holds, the compositions are well posed. In particular $Ran(Y^{-1}) = D(Y)\subset D(X)$. Now observe that, since $X$ is closed it being selfadjoint, and $Y^{-1}$ is bounded, the closed graph theorem easily implies that $X Y^{-1}: H \to H$ is bounded as well. Since $X^{-1}$ is compact,
$Y^{-1} = X^{-1} (X Y^{-1})$ is compact because it is the product of a bounded operator and a compact operator. QED
