For $A\in L(X)$, with $A$ being a closed operator and $X$ is a Banach space (with bounded $\sigma(A)$), define $$P=\frac{1}{2\pi i} \int_\gamma R(\lambda ,A)d\lambda$$ to be the spectral projection that commutes with $A$. Since this is a projection onto $X$, why is it that $P=I$ if and only if $A\in L(X)$?
If in the case $X$ was a Hilbert space, could we not show that since $A$ commutes with $P$, $A$ is diagonalizable? However, since $X$ is a Banach space, I am stuck.