Operators bounded each other on eigenvalues

Question

Let $A$ and $B$ two self-ajoint compact positive operators defined on a Hilbert space $H$ to itself. Let $g_n$ the sequence of eigenfunctions of $A$, if $$||Bg_n||\leq C||Ag_n||\,\forall n$$ It is possible to prove that $$||Bx||\leq K||Ax||\,\forall x\in H$$?

Answer 1

Assume $\lambda_n>0$ and $\sum \lambda_n^2<\infty.$ Let $A$ and $B$ be defined on $\ell^2$ by $$Ax=\sum_{n=1}^\infty\lambda_n\langle x,e_n\rangle e_n,\quad Bx=\langle x,v\rangle v$$ where $e_n$ denote the elements of the standard basis in $\ell^2$ and $v=\sum_{n=1}^\infty \lambda_ne_n.$ Both operators are positive and compact. Moreover $A$ is strictly positive and $B$ is one dimensional. The elements $e_n$ are the eigenvectors of $A.$ We have $$\|Be_n\|=\langle e_n,v\rangle \|v\|=\lambda_n\|v\|= \|v\|\|Ae_n\|$$ On the other hand for $x^{(N)}=\sum_{n=1}^N\lambda_n^{-1}e_n$ we get $$\|Ax^{(N)}\|^2=\left \|\sum_{n=1}^N e_n\right \|^2=N$$ and $$\|Bx^{(N)}\|^2= \left ( \sum_{n=1}^N\lambda_n^{-1}\langle e_n,v\rangle \right )^2\|v\|^2=\left ( \sum_{n=1}^N 1\right )^2\|v\|^2=N^2\|v\|^2$$ Hence $${\|Bx^{(N)}\|\over \|Ax^{(N)}\|}\underset{N\to \infty}{\longrightarrow}\infty $$ Therefore the constant $K$ does not exist.

Remark If we want $B$ to be strictly positive we can replace $B$ with $A+B.$