If a bounded operator has finitely many eigenvalues above zero, is it compact?

Question

In order to give you some context: In the paper "Magnetic Lieb-Thirring estimates" by Laszlo Erdös he claims that since the number of eigenvalues of a bounded operator $K$ acting on $L^2(\mathbb{R}^3)$ which are above some $\rho>0$ is finite the operator has to be compact. I would be grateful if someone could provide a source on why this is true or if this is remotely true in some cases and if yes in which.

Answer 1

The exact context is not clear. The identity operator on an in finite dimensional space is not com pact but its only eigen value is $1$ so there are no eigen values greater than $1$.

Answer 2

Now me assume that $T$ is normal. Then we can use the spectral theorem to get $$ Tx = U^*( f\cdot (Ux)) $$ where $U:L^2(\mathbb R^3)\to L^2(\mu)$ is unitary, $f\in L^\infty(\mu)$, and $f\cdot $ denotes multiplication.

Let $k$ be given. Then define $$ \chi_k(s) = \begin{cases} 1 & \text{ if } |f(s)|\ge 1/k,\\ 0 & \text{else,}\end{cases} $$ the truncation of $f$ by $$ f_k(s) =\chi_k(s)\cdot f(s), $$ and $$ T_k:=U^*(f_k\cdot U). $$ The assumption on the eigenvalues I would like to translate to $$ \dim R ( U^*(\chi_k\cdot U)) <+\infty \quad\forall k, $$ i.e., the dimension of the range of the spectral projection to large eigenvalues is finite.

Then $T_k\to T$ in the operator norm. In addition, $R(T_k)$ is closed for all $k$. Hence by the closed range theorem and normality of $T_k$, $R(T_k)=N(T_k^*)^\perp = N(T_k)^\perp$ follows. Since $N(T_k)^\perp=N(U^*(\chi_k\cdot U))^\perp = R(U^*(\chi_k\cdot U))$, the operators $T_k$ have finite-dimensional range. It follows $T$ is compact.