Assume $A\subseteq B(H)$ is a non-unital C$^*$-subalgebra. Is the following statement true?
$$\forall T\in A,\ \sigma(T)'\subseteq\{0\}\ \ \ \ \iff\ \ \ \ A\subseteq K(H)$$.
Here, $\sigma(T)'$ represents the accumulation points of $\sigma(T)$.
If the right hand is true then the left hand is true. But for the other direction is there a counterexample? If $A$ was unital, $\{ \lambda 1: \lambda \in \mathbb{C}\} $ was a counterexample.
If I can prove that every eigenvalue of $T$ has finite multiplicity, then the other direction will be true. Can someone help me or guide me?
This is not true. In essence, this is because the spectrum of any $a\in A$ doesn't depend on how $A$ is embedded in $B(H)$, so we can take an infinite sum of the identity representation an destroy any chance of elements of $A$ being compact. I'll include an example to (hopefully) elucidate my claims.
Consider $A=C_0(\mathbb N)$, and consider the canonical representation of $A$ on $\ell^2(\mathbb N)$. Now let $H_0$ be any separable infinite-dimensional Hilbert space, let $H=\ell^2(\mathbb N,H_0)$ (i.e., $H$ is the collection of all functions $\xi:\mathbb N\to H_0$ such that $\sum_{n\in\mathbb N}\|\xi(n)\|^2<\infty$), and define a faithful representation $\pi:A\to B(H)$ by $$(\pi(f)\xi)(n)=f(n)\xi(n).$$ (Another way of thinking about this is by considering $H=\ell^2(\mathbb N)\otimes H_0$ and $\pi(f)=f\otimes 1$.)
If $f\in A$ then $\sigma(\pi(f))=\sigma(f)=f(\mathbb N)$, and thus $\sigma(\pi(f))'\subset\{0\}$. However, if $f\in A$ is non-zero, then $\pi(f)\notin K(H)$ (if $f\neq 0$, its easy to construct a bounded sequence $(\xi_n)$ in $H$ such that $(\pi(f)\xi_n)$ has no convergent subsquence).