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Let $X$ be a Banach space, $A\in\mathbb{B} (x)$ and let $B\in\mathbb{K} (x)$ be a compact operator on $X$. show that $$\sigma(A+B)\subset\sigma(A)\cup\sigma_p(A+B)$$ (where $\sigma_p,\sigma_c,\sigma_r$ are point spectrum, continuous spectrum and residual spectrum respectivelyenter image description here)

I have tried using the existence of $(T-\lambda I)^{-1}$ and it is suggested that we should prove $\sigma_c(A+B)\cup \sigma_r(A+B)\subset\sigma (A)$ instead, why?

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    $\begingroup$ How to ask a homework question $\endgroup$ – Nate Eldredge Mar 15 '15 at 15:45
  • $\begingroup$ What are $\sigma_p,\ \sigma_c$ and $\sigma_r$? $\endgroup$ – Berci Mar 15 '15 at 17:28
  • $\begingroup$ $\sigma_p,\sigma_c,\sigma_r$ are point spectrum, continuous spectrum and residual spectrum respectively. $\endgroup$ – Daniel Mar 17 '15 at 6:42
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Suppose $\lambda \notin \sigma(A)\cup \sigma_p(A+B)$, then $\exists D \in B(X)$ such that $$ (A-\lambda I)D = D(A-\lambda I) = I $$ and $\nexists x\in X$ such that $(A+B)x = \lambda x$. We want to show that $(A+B-\lambda)$ is invertible. By the bounded inverse theorem, it suffices to prove that it is bijective.

By hypothesis, $(A+B-\lambda)$ is injective.


Claim: $(I+DB)$ is invertible.

Proof: Since $DB$ is compact, by the Fredholm alternative, it suffices to prove that $(I+DB)$ is injective. However, if $(I+DB)y = 0$, then $$ y = -DBy \Rightarrow (A-\lambda)y = -By \Rightarrow (A+B-\lambda) y = 0 $$ Since $(A+B-\lambda)$ is injective, this implies that $y=0$, proving the claim.


Now to show that $(A+B-\lambda)$ is surjective, consider $y \in X$, then set $$ x := (I+DB)^{-1}Dy $$ then $$ x + DBx = Dy \Rightarrow x = D(y-Bx) \Rightarrow (A-\lambda)x = y - Bx $$ and so $(A+B-\lambda)x = y$.

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