‎If ‎‎$‎X$ is an infinite-dimensional Banach space and ‎‎$‎‎u‎\in ‎B(X)‎$ ,then $\bigcap_{v\in K(X)}\sigma(u+v) =\cdots$

‎If ‎‎$‎X$ is an infinite-dimensional Banach space and ‎‎$‎‎u‎\in ‎B(X)‎$,why the following equality is true?

$$\bigcap_{v\in K(X)}\sigma(u+v) =\sigma(u) \setminus \{\lambda \in\mathbb{C}\mid u - \lambda \text{ is Fredholm of index zero} \}$$ ‎‎‎ The index of $‎‎u‎\in A$ is defined to be

$$\operatorname{ind}(u)=\operatorname{nul}(u)-\operatorname{def}(u)$$ where co-dimension of $u(X)$ in $Y$ be denoted by $\operatorname{def}(u)$ and $\dim(\ker(u))$ denoted by $\operatorname{nul}(u)$

$u\in B(X,Y)$ is Fredholm operator, if $\ker(u)$ is finite-dimensional and $u(X)$ is finite-co-dimensional in $Y$

$$K(X,X):=K(X)=\{u : X \longrightarrow X\mid u\text{ is compact operator}\} ‎$$

Reference: Schecter, "Principles of Functional Analysis", pg. 180. The object on the left hand side is known as the Essential Spectrum $(\sigma_e(u))$.

One direction is not so hard; if $\lambda \notin \sigma_e(u)$ then for some $K$ compact we have that $u + K - \lambda$ is invertible. This implies that $\lambda \in \Phi(u+K)$ and $i(u+K - \lambda) = 0$. Recall now that compact perturbations of Fredholm operators are fredholm, and that such perturbations preserve the index $i$. So perturb $u+K-\lambda$ by $-K$, and we see that $u-\lambda$ is fredholm with index zero.

The other direction is trickier; check the reference for the details. The idea is to construct an operator $F$ of finite rank such that $N(u-\lambda) \cap N(F) = \{0\}$ and $R(u - \lambda) \cap R(F) = \{0\}$. Equipped with this, $u-\lambda+F$ is both Fredholm of index zero and injective; thus $\lambda$ belongs to the resolvent set of $u+F$. As operators of finite rank are compact, we're done.