
> 
> 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 $\Delta\_1$.
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> 
> Furthermore, let $a \in A$ with $\text{Sp}(a)=\{0,1\}$. Here $\text{Sp}(a)$ denotes the spectrum of $a$ in $A$.
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> 
> 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?



Help trying to show that $p_0a_1 =0$
Apply [holomorphic functional calculus](https://en.wikipedia.org/wiki/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 


$$a\_0 = \frac{1}{2\pi i} \int\_{\Gamma\_0}\lambda(\lambda -a)^{-1}d\lambda.$$


