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In this article http://en.wikipedia.org/wiki/Decomposition_of_spectrum_(functional_analysis) spectrum decomposition of the shift operator on $\ell_p(N)$ has been discussed.

Question: Is it possible figure out the decomposition without help of the adjoint operator?

For example, to show that the open unit disc, say, {$\lambda\in{\bf C}:|\lambda|<1$} includes in the residual spectrum of the right shift operator $R$, one needs to show that range$(R-\lambda I)$ is not dense in $\ell_p(N)$ or equivalently, there is an $x\in \ell_p(N)$ such that it is not adherent to range$(R-\lambda I)$.

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By Hahn-Banach, it is enough to show that the adjoint operator from $\ell_q$ to $\ell_q$ is not injective. Now this becomes an easy exercise –  Yemon Choi Mar 17 '11 at 3:51
The adjoint operator should be from $\ell_q$ to $\ell_p$.(typo?) –  Jack2 Mar 17 '11 at 17:23
Yu Cao: No. If T is a linear operator from a Banach space $X$ to itself, then the adjoint is a linear operator from $X^*$ to itself. All this is explained in, for instance, Rudin's Functional Analysis (Chapter 4 I think?) –  Yemon Choi Mar 17 '11 at 19:03
I have voted to close, because this really is quite a basic question that one should be able to understand after a course in linear functional analysis, or after reading the relevant parts of Rudin's book. –  Yemon Choi Mar 20 '11 at 9:53
Why on earth would you insist on doing this "without help of the adjoint operator"? –  Mark Meckes Mar 20 '11 at 18:34
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