Properties of the Riesz projector We have a hilbert space $X$, a continuous and linear function $A:X\to X$ and an eigenvalue $\lambda_0$ of $A$. Furthermore, we have an arbitrary positive oriented closed jordan curve $\Gamma$ in $\mathbb{C}$. Let the operator $A-\lambda\, \text{Id}$ be continuously invertible inside the curve for all $\lambda$ except at $\lambda=\lambda_0$ and $\textbf{assume}$ that $\lambda_0$ lies in the inside of $\Gamma$. Then, we want to show that the operator $P_{\lambda_0} :X\to X$ defined by
$$
P_{\lambda_0} := -\frac{1}{2\pi i} \int_\Gamma (A-\lambda \, \text{Id})^{-1}\, d\lambda
$$
is continuous and linear with closed range (range means $P_{\lambda_0}(X)$) and independent of the choice of $\Gamma$ (if it fulfills the same properties). We should also show, that it is a projection and that the eigenspace of $\lambda_0$ is in the range of $P_{\lambda_0}$.
 A: $\textit{Proof:}~$ Linearity is clear since $A$ and therefore $(A-\lambda \, \text{Id})^{-1}$ are linear.
Since the integrand is holomorphic in respect to $\lambda$ everywhere except on the spectrum, we get the independency from $\Gamma$ by the redidue theorem (one has to apply a function $\phi\in \big(L_b(X)\big)^*$ from the dual space first on the integral, where we can pull it inside, and see that $\phi\circ (A-\lambda\, \text{Id})^{-1}$ is holomorphic too and then use the theorem; since the dual space is a separating set, we get that the difference between two integrals with different $\Gamma$ is zero) and
$$
\|P_{\lambda_0}x\|_X \leq \frac{1}{2\pi} |\Gamma| \max_{\lambda\in\Gamma}\|(A-\lambda\, \text{Id})^{-1}\| \cdot \|x\|_X,
$$
where we used that the integrand is continuous in respect to $x$ for every $\lambda$. The maximum exists since the integrand is holomorphic and therefore continuous in respect to $\lambda$ (and since $\Gamma$ is compact).
It is a projection, since we can take another curve $\tilde{\Gamma}$ that runs inside of $\Gamma$ and get
$$\begin{align*} P_{\lambda_0}^2 &= \frac{1}{(2\pi i)^2} \int_{\tilde{\Gamma}} R_{\mu}(A) \, d\mu \int_{\Gamma} R_{\lambda}(A) \, d\lambda \\
&= \frac{1}{(2\pi i)^2} \int_{\tilde{\Gamma}}\int_{\Gamma} R_{\mu}(A) R_{\lambda}(A) \, d\lambda\, d\mu \\
&= \frac{1}{(2\pi i)^2}\int_{\tilde{\Gamma}}\int_{\Gamma}\frac{R_{\mu}(A)-R_{\lambda}(A)}{\mu-\lambda} \, d\lambda \, d\mu \\
&= \frac{1}{(2\pi i)^2}\int_{\tilde{\Gamma}}\int_{\Gamma}\frac{R_{\mu}(A)}{\mu-\lambda} \, d\lambda \, d\mu - \int_{\tilde{\Gamma}}\int_{\Gamma}\frac{R_{\lambda}(A)}{\mu-\lambda} \, d\lambda \, d\mu \\
&= \frac{1}{(2\pi i)^2}\int_{\tilde{\Gamma}}\int_{\Gamma}\frac{R_{\mu}(A)}{\mu-\lambda} \, d\lambda \, d\mu - \int_{\Gamma}\int_{\tilde{\Gamma}}\frac{R_{\lambda}(A)}{\mu-\lambda} \, d\mu \, d\lambda \\
&= \frac{1}{(2\pi i)^2}\int_{\tilde{\Gamma}} R_{\mu}(A) \int_{\Gamma}\frac{1}{\mu-\lambda} \, d\lambda \, d\mu - \int_{\Gamma} R_{\lambda}(A)\int_{\tilde{\Gamma}}\frac{1}{\mu-\lambda} \, d\mu \, d\lambda \\
&= -\frac{1}{2\pi i}\int_{\tilde{\Gamma}} R_{\mu}(A) \,d\mu = P_{\lambda_0},\end{align*}$$
where we used that by Cauchy's integral theorem
$$\int_{\Gamma}\frac{1}{\mu-\lambda} \, d\lambda = -2\pi i$$
holds and that the function in the second integral is holomorphic and therefore there exists a function $F$ such that $F'(\mu)= \frac{1}{\mu-\lambda}$ and therefore (since $\Gamma$ is closed)
$$\int_{\tilde{\Gamma}}\frac{1}{\mu-\lambda} \, d\mu = F\big(\Gamma(1)\big) - F\big(\Gamma(0)\big) = 0.$$
We also used the first resolvent identity, that is
$$R_{\mu}(A)-R_{\lambda}(A) = (\mu-\lambda)R_{\mu}(A)R_{\lambda}(A) \quad\text{or}\quad R_{\mu}(A)R_{\lambda}(A) = \frac{R_{\mu}(A)-R_{\lambda}(A)}{\mu-\lambda}.$$
The range is closed since $P_{\lambda_0}$ is a continuous projection and therefore $\text{Id}-P_{\lambda_0}$ is also continuous and $\text{range}(P_{\lambda_0})=\text{ker}(\text{Id}-P_{\lambda_0})$.
For an eigenvector $x$ with $Ax = \lambda_0 x$, it holds that $(A-\lambda\,\text{Id})^{-1}x = \frac{1}{\lambda_0-\lambda} x$ and therefore
\begin{align*}
     P_{\lambda_0}x = -\frac{1}{2\pi i} \int_{\Gamma} (A-\lambda\, \text{Id})^{-1} x \, d\lambda = -\frac{1}{2\pi i} \int_{\Gamma} \frac{1}{\lambda_0-\lambda} \,d\lambda \,x = x,
 \end{align*}
where we used Cauchy's integral theorem again. So we have $x\in \text{range}(P_{\lambda_0})$ and therefore $E(\lambda_0) \subseteq \text{range}(P_{\lambda_0})$.
