eigenvalue for the compostion of operators Assume we have a compact self-adjoint linear operator $A$ on $L^{2}( \Omega )$ . Spectrum of $A$ is countably infinite with finite number of negative eigenvalues (multiplicity counted). Let  $B$ be a compact linear operator on $L^{2}( \partial \Omega ) \rightarrow  L^{2}( \Omega )$ and $B^*$ be its adjoint.
We then have $N=B^*AB$ a self-adjoint compact linear operator on  $L^{2}( \partial \Omega )$. Let $C_{-} (\cdot)$ denote the number of negative eigenvalues.
Does it imply $C_{-}(N)\leq C_{-}(A)$?
This is a question related to step 4 in Theorem 5.7 proof BEH
 A: I think I have solved it.
Before proceeding with the claim, we need a preparatory lemma:
Let $A$ be a self-adjoint compact linear operator on Hilbert space $H$. Let $d \in \mathbb{N}_0$ and $r \geq 0$. Then, the following statements are equivalent:

*

*$A$ has most  $d$ eigenvalues larger than  $r$

*There exists a subspace $V \subset H$ with dim$( V ) \leq d$ such that
\begin{align*}
   \langle Av,v \rangle \leq r \|v\| \text{ for all } v \in V^{\perp}
  .\end{align*}
(see Lemma 3.2)

Let $V$ be the subspace spanned by the eigenvectors of $A$, which are associated with negative eigenvalues. Since $A$ is compact, for each negative eigenvalue $e_i$, its associated eigenspace is finite. Hence,  $V$ is a finite dimensional subspace in  $L^{2}( \Omega ).$ It is clear that
\begin{alignat*}{2}
 &\langle A v,v  \rangle < 0 &&\text{ for all } v \in V\setminus \{0\},\\
 &\langle Av,v \rangle \geq 0 && \text{ for all } v \in V^{\perp}
.\end{alignat*}
Let $f \in L^{2}( \partial \Omega )$ with $Bf \subset V^{\perp}$,
\begin{align*}
 \langle B^*ABf , f \rangle &=\langle ABf, Bf   \rangle \geq 0
.\end{align*}
By the adjoint property of $B,$
\begin{align*}
 0=\langle B f , v \rangle=\langle f,B^*v  \rangle \text{for all } v \in V
.\end{align*}
Hence, \begin{align*}
 \langle B^*ABf,f \rangle \geq 0 \text{ for all } f \in ( B^{*}( V ) )^{\perp} 
.\end{align*}
By the previous Lemma, the number of negative eigenvalue of $B^*AB$ is at most $dim(B^*(V)).$ Since $dim(B^*( V ))\leq dim( V )=C_{-}(A) <\infty $, we have $C_{-}(B^*AB)\leq dim(B^*(V))\leq dim(V)=C_{-}(A)<\infty$.
