Help trying to show that $p_0a_1 =0$ [closed]

Question

Let $A$ be a Banach algebra and let $\Gamma_0, \Gamma_1$ be circles of centres 0 and 1 respectively, each of radius less that $\frac{1}{2}$, which bound the two open disks $\Delta_0$ and $\Delta_1$.

Furthermore, let $a \in A$ with $\text{Sp}(a)=\{0,1\}$. Here $\text{Sp}(a)$ denotes the spectrum of $a$ in $A$.

Define $$p_0 = \frac{1}{2\pi i} \int_{\Gamma_0}(\lambda-a)^{-1}d\lambda$$ and $$a_1 = \frac{1}{2\pi i} \int_{\Gamma_1}\lambda(\lambda -a)^{-1}d\lambda.$$

I am now trying to show that $p_0a_1=0$.

I have the following so far:

Let $\lambda\neq \mu$ not be in $\text{Sp}(a)$, then: \begin{align*} p_0a_1 &= \bigg[\frac{1}{2\pi i} \int_{\Gamma_0}(\lambda-a)^{-1}d\lambda\bigg]\bigg[ \frac{1}{2\pi i} \int_{\Gamma_1}\mu(\mu -a)^{-1}d\mu \bigg] \\ &= -\frac{1}{4\pi^2}\int_{\Gamma_0}(\lambda-a)^{-1}d\lambda \cdot \int_{\Gamma_1}\mu(\mu -a)^{-1}d\mu \\ &= -\frac{1}{4\pi^2}\int_{\Gamma_0}\int_{\Gamma_1}\mu(\lambda -a)^{-1}(\mu-a)^{-1}d\mu d\lambda \\ &= -\frac{1}{4\pi^2}\int_{\Gamma_0}\int_{\Gamma_1}\mu\bigg[\frac{(\lambda-a)^{-1} - (\mu-a)^{-1}}{\mu - \lambda} \bigg]d \mu \lambda \\ &= -\frac{1}{4\pi^2} \bigg[ \int_{\Gamma_0}\int_{\Gamma_1}\mu\frac{(\lambda-a)^{-1}}{\mu - \lambda}d\mu d\lambda - \int_{\Gamma_0}\int_{\Gamma_1}\mu\frac{(\mu-a)^{-1}}{\mu-\lambda}d\mu d\lambda \bigg] \end{align*}

However, this is where I am stuck... Can anyone please help guide me on the right direction to showing that the above equates to zero?

Answer 1

Apply holomorphic functional calculus to the following functions which are defined on a disconnected open set in the plane containing $\Gamma_0 , \Gamma_1$

$f(z)=\begin{cases} 1& \text{Around 0}\\0& \text{around 1}\end{cases}$

$g(z)=\begin{cases} 0& \text{Around 0}\\z& \text{around 1}\end{cases}$

Then $p_0=f(a), a_1=g(a)$ but $fg=0$.

By the same argument one can shows that $p_0a=a_0$ where

$$a_0 = \frac{1}{2\pi i} \int_{\Gamma_0}\lambda(\lambda -a)^{-1}d\lambda.$$