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I am reading the book Perturbation theory for linear operators from Kato.

He defines (§5 Section 3) for an operator $T : X\to X$ on a finite Banach Space the resolvent as

$$ R(x) = (T- x)^{-1}.$$

He then consideres the Laurent Series of the resolven on a singularity $\lambda$ of $R$ and finds a decomposition of $R$ as $$ R(x) = -(x-\lambda)^{-1}P - \sum_{n=1}^\infty (x- \lambda)^{-n-1}D^n + \sum_{n=0}^\infty (x-\lambda)^n S^{n+1}$$ on some set $$\Gamma = \{ x\in \mathbb{C} \mid r < |x-\lambda | <r' \} $$where $P^2 = P$ and $PD=DP=D, \, PS=SP=0$. Since $P$ is a projection one can write $$ X= U\oplus V$$ with $U=PX$ and $V=(1-P)X$.

The author makes then the following two conclusions:

  1. He claims that $Range(D) = U$. I see that $Range(D) \subset U$ but how can one explain the other inclusion?

  2. He states:

    As the principal part of a Laurent Series at an isolated singularity $$ -(x-\lambda)^{-1}P -\sum_{n=1}^\infty (x- \lambda)^{-n-1}D^n$$ is convergent for $x \neq \lambda$, so that the part of $R(x)$ in $U$ has only the one singularity $x=\lambda$, and the spectral radius of $D$ must be zero.

    I do not understand how he concludes that this part of the Laurent Series has no further singularites (in $\mathbb{C}$ ?)

Adam
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  • When you say "finite Banach Space" do you mean finite-dimensional Banach Space? The reason I ask is that it is rather straightforward to show in the context of finite-dimensional spaces that the resolvent has poles at its singularities. Actually, I did that for you in another problem. – Disintegrating By Parts Feb 12 '15 at 17:38
  • @T.A.E. it did not see the connection between these questions. Now I do. I will post it as an answer. Thank you. – Adam Feb 16 '15 at 15:08

1 Answers1

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To 1)

The operator $$ [R(x) + (x-\lambda)^{-1} P ] : U \to U$$ is bijective. Therefore, for any $u \in U$ we find a $v\in U$ with $$ [R(x) + (x-\lambda)^{-1} P ]v = u$$ and the following sequence converges to $u$ $$Range(D) \supset \sum_{i=1}^n (x-\lambda)^{-i-1}D^n v \to u.$$ Since $Range(D)$ is a closed subspace we have $u\in Range(D)$ and therefore $Range(D)=U$.

To 2) As explained here Laurent Series of operator-value function the singularites of $R$ are isolated. This means that

$$ R(x) = -(x-\lambda)^{-1}P - \sum_{n=1}^\infty (x- \lambda)^{-n-1}D^n + \sum_{n=0}^\infty (x-\lambda)^n S^{n+1}$$ converges in a ball $$ B = \{ x \in \mathbb{C} \mid 0 < |x - \lambda |< R \}.$$ For any $x\in\mathbb{C}\backslash\{\lambda\}$ we find a $x_0 \in \mathbb{C}\backslash\{\lambda\}$ with $$ | x - \lambda| > | x_0 - \lambda|.$$ If you think of $D$ as a matrix then the convergence of $$\sum_{n=1}^\infty (x- \lambda)^{-n-1}D^n$$ is equivalent to the convergence of $$\sum_{n=1}^\infty (x- \lambda)^{-n-1}[D^n]_{ij}$$ which converges for $x_0$ and therefore also for $x$ (see comment from T.A.E.).

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Adam
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    If $F$ is holomorphic in an annulus $0 \le r_{1} < |z-z_{0}| < r_{2}$, then you get $F(z) = \sum_{n=0}^{\infty}a_{n}(z-z_{0})^{n}+\sum_{n=1}^{\infty}b_{n}(z-z_{0})^{-n}$. The power series converges in $|z-z_{0}| < r_{2}$ and the singular series converges in $r_{1} < |z-z_{0}|$. Hence, the sum of the two converges on the annulus. This is a property of radius of convergence. – Disintegrating By Parts Feb 16 '15 at 15:54
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    Another way to understand this: If $\sum_{n=0}^{\infty} a_{n} z^{n}$ converges (conditionally or absolutely) for some $z$, then it converges absolutely for $w$ for which $|w| < |z|$. If $\sum_{n=1}^{\infty}a_{n}z^{-n}$ converges for some $z$, then it converges absolutely for all $w$ for which $\frac{1}{|w|} < \frac{1}{|z|}$ or for $|w| > |z|$. – Disintegrating By Parts Feb 16 '15 at 16:00
  • @T.A.E. thank you! I edited my answer. – Adam Feb 16 '15 at 17:46