# closed operator, projection

Let $A: D \subset X \to X$ be a closed linear operator. X is a Banach space. Furthermore we have $\gamma: [0,1] \to \mathbb{C}$, $\gamma$ is a $C^1$ curve and $\gamma \subset \rho(A)$, where $\rho(A)$ is the resolvent set of A.

Define $P = \frac{1}{2 \pi i} \int_{\gamma} R(z) dz$.

Now, I'm trying to show that $PX$ is a subset of $D$ and $AP = \frac{1}{2 \pi i} \int_{\gamma} zR(z) dz$.

I think for the second part, one has to use the identity $AR(z) = zR(z) - id$.

Thanks for any help!

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$R: \rho(A) \to L(X)$ is the resolvent operator, so $R(z) = (z-A)^{-1}$. – thomas.en Oct 29 '12 at 18:40
Once you know that $P$ is a projection, both of the desired properties follow from the closedness of $A$ as well as the fact that the considered integral is a limit (in norm) of Riemann sums. You can find details in Vol. IV of Reed and Simon's Methods of Modern Mathematical Physics (Section XII.2, Theorem XII.5 (c)).