Suppose $T$ is bounded linear operator on a separable Hilbert space $H$ such that, $0\in \sigma_{e}(T)$, where $\sigma_{e}(T)$ represents the essential spectrum of $T$. Then does this implies that the operator $T$ is compact?
I am a little confused since $0\in \sigma_{e}(T)$ then $T$ is not Fredholm hence the image of $T$ under calkin algebra is not invertible. So does this make $T$ compact?
Best Answer
I don't think so.
Consider \begin{align*}T:\ell^2&\to\ell^2, \\ (a_n)_{n\in\mathbb N} &\mapsto\left(\begin{cases}0,&\text{if }m\text{ is odd}\\a_{m/2},&\text{if }m\text{ is even}\end{cases}\right)_{m\in\mathbb N},\end{align*}
i.e. $T(a_1,a_2,\dots)=(0,a_1,0,a_2,\dots)$. Then $T$ is not compact (left to the reader) but it is also not Fredholm, since the cokernel contains distinct equivalence classes for all odd unit vectors (what I mean are the equivalence classes of $(1,0,0,\dots)$ and $(0,0,1,0,\dots)$ etc. under the image of $T$) and thus cannot be finite dimensional.