Comparison eigenvalues of three compact operators Let $R_1$, $R_2$ and $R_3$ three compact, self-adjoint, positive definite operators.
If $R_1\leqslant R_2 \leqslant R_3$, in the sense that
$$
\langle f, R_1(f)\rangle\leq\langle f, R_2(f)\rangle\quad \forall f
$$
It is possible to conclude that the eigenvalues $\lambda_k$ of $R_2$ are squeezed between the ones of $R_1$ and $R_3$, for $k\geq k_0$?
 A: I assume these are operators on a Hilbert space $H$.
Let's note $(\lambda_k)_{k \geq 0}$ the positive eigenvalues of $R_1$ in decreasing order, and $(e_k)_{k \geq 0}$ its associated eigenvectors, so the the family $(e_k)_{k \geq 0}$ forms a "Hilbert basis" of $H$. Similarly, let's note $(\mu_k)_{k \geq 0}$ the positive eigenvalues of $R_2$ in decreasing order, and $(f_k)_{k \geq 0}$ its associated eigenvectors, so the the family $(f_k)_{k \geq 0}$ forms a "Hilbert basis" of $H$. All of this is possible as $R_1$ and $R_2$ are compact, self-adjoint and positive definite.
We now want to show that for all $k \in \mathbb{N}$, $\lambda_k \leq \mu_k$. Consider the vector space $F = \mathrm{Span}(e_1,\cdots,e_k)$. On this vector space, as the eigenvalues of $R_1$ are in decreasing order, we have :
$$\forall x \in F, \quad \langle x , R_1x \rangle \geqslant \lambda_k \|x\|^2.$$
Consider the vector space $G = \left(\mathrm{Span}(f_1,\cdots,f_{k-1})\right)^{\bot} = \left\{ \displaystyle\sum_{i=k}^{+\infty} a_i.f_i \mid (a_i)_{i \geq k} \in \ell^2(\mathbb{N}) \right\}$. As the eigenvalues of $R_2$ are in decreasing order, we have
$$\forall x \in G, \quad \langle x , R_2 x \rangle \leqslant \mu_k \|x\|^2.$$
Now, $F$ has dimension $k$ and $G$ has comdimension $k-1$. Therefore $F\cap G \neq\{0\}$, so there is an $x \neq 0$ in $F \cap G$. For such an $x$, we get :
$$\lambda_k \|x\|^2 \leqslant \langle x , R_1x \rangle \leqslant \langle x , R_2 x \rangle \leqslant \mu_k \|x\|^2.$$
As $x$ is not $0$, this gives $\lambda_k \leqslant \mu_k$.
$\rhd$ Similarly, if $(\gamma_k)_{k \geq 0}$ are the eigenvalues of $R_3$ in decreasing order, we get $\mu_k \leqslant \gamma_k$, and therefore we have the desired "squeezing" :
$$\forall k \in \mathbb{N},\quad\lambda_k \leqslant\mu_k \leqslant \gamma_k.$$
