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Let $G$ be normal operator with compact resolvent such that $\ker G$ is different from $\{0\}$.

Now Let $P$ be the orthogonal projection onto $\ker G$ and consider $G' = G + P$.

Please, I want an explication to the following question:

How $0$ belongs to resolvent set of $G'$?

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What do you mean by "compact resolvent"? –  Martin Argerami Jun 20 '12 at 14:00
    
operator with compact resolvent means that its spectrum is disrete –  salma Jun 20 '12 at 21:10
    
So your operator is unbounded, then? –  Martin Argerami Jun 21 '12 at 3:56
1  
@Martin: The usual meaning is the literal one: an (unbounded) operator is said to have compact resolvent if its resolvent (at some, equivalently at all points of the resolvent set) is a compact operator. As you correctly point out, such an operator must be unbounded (unless you work in finite dimensions). Typical examples are differential operators whose resolvents are integral operators (which tend to be compact). Many operators arising in mathematical physics do have compact resolvents. The first theorem one proves about operators with compact resolvent is that the spectrum is discrete. –  t.b. Jun 21 '12 at 14:35
    
Now, I have another question: –  salma Dec 22 '13 at 7:20

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