Let $X$ be a hilbert space and $T\in L(X)$

Show that:

(i) $\sigma_c(T^*)=(\sigma(T))^*$

(ii) $\sigma_r(T)=((\sigma_p(T^*))^*)$\ $\sigma_p(T)$

(i): $"\subset"$

Let $\lambda\in\sigma_c(T^*)$ then $T^*-\lambda:X\rightarrow R(T^*-\lambda)$ is injective with $R(T^*-\lambda)\neq X$ and $\overline{R(T^*-\lambda)}=X$

We have $T^*-\lambda=(T-\overline{\lambda})^*$ and we need that $T-\overline{\lambda}$ is injective too with dense in X but not equal to X image.

Can someone help me? Also with the others inclusions.

• nobody can help?:( – Duke Dec 10 '14 at 12:46
• Check what you have written for (i). To see my concern, what do you know about the relation between $\sigma(T)$ and $\sigma(T^{\star})$? – DisintegratingByParts Dec 10 '14 at 16:50
• Hm..sorry, I don't know. maybe they are equal? – Duke Dec 10 '14 at 17:08
• If you have $(T-\lambda I)B=I=B(T-\lambda I)$, then what do you get if you take adjoints? This will tell you how $\rho(T)$ and $\rho(T^{\star})$ are related. Once you know that, then look at your problem (i) more carefully. – DisintegratingByParts Dec 10 '14 at 17:20
• I will get: $B^*(T-\lambda I)^*=I=(T-\lambda)^*B^*$ which is equivalent to $B^*(T^*-\overline{\lambda})=I=(T^*-\overline{\lambda})B^*$. But I can't see how this can help :/ Sorry :( – Duke Dec 10 '14 at 17:32