Riesz idempotent (projection) of singleton in $\ell^p$

Question

Edit: After I posted my question, I realised that in this post the same question was asked (and answered). In the meantime useful comments were made under my current post. I've written an answer complementing the answer in the other post. If it's better that I delete my current post, as it is a duplicate, I'll gladly do so!

I'm going through examlple VII.6.10 in Conway's Functional Analysis:

Let $(\alpha_n)\in \ell^\infty$, $1\leq p\leq \infty$, and define $A:\ell^p\to\ell^p$ by $(Ax)_n=\alpha_n x_n$. For each $k$, define $N_k=\{n\in\mathbb N: \alpha_n=\alpha_k\}$ and define $P_k:\ell^p\to\ell^p$ by $P_kx=\chi_{N_k}x$. If $\alpha_k$ is an isolated point of $\sigma(A)$, then $$ E(\{\alpha_k\};A)=P_k, $$ where $E(\{\alpha_k\};A)$ is the Riesz idempotent corresponding to $\{\alpha_k\}$, i.e., $$ E(\{\alpha_k\};A)=\frac 1{2\pi i}\int_\Gamma (z-A)^{-1}dz $$ where $\Gamma$ is positively oriented Jordan system such that $\alpha_k\in\operatorname{ins}\Gamma$ ans $\sigma(A)-\{\alpha_k\}\subset\operatorname{out}\Gamma$.

I want to show that $$ E(\{\alpha_k\};A)=P_k. $$

I don't know how to compute the Riesz idempotent explicitly, so I cannot show that $$ E({\alpha_k};A)=P_k. $$ Let's write $E=E({\alpha_k};A)$. Using Riesz functional calculus, I can prove several properties of E (Proposition VII.4.11 in Conway). But even if I'm able to show that $P_j$ satisfies these properties, I would then need some unicity result... so do the properties of Proposition 4.11 uniquely determine $E$?

Another way to go about this is to determine $\ker E$ and $\operatorname{ran}E$, since these uniquely determine E (because $E$ is an idempotent). Could we somehow determine the order of the singularity αk

I've seen some results about the range of the Riesz projection, however, then we need to know what kind of singularity $\alpha_k$ is.

Answer 1

In this answer, I will add some details to supplement the comments by jd27.

The little lemma's we need are:

Let $G\subset\mathbb C$ be open, let $\gamma$ a rectifiable curve in $G$, and let $L:G\to X$ be continuous (where $X$ is a Banach space).

• If $T:X\to Y$ is a bounded linear map (where $Y$ is also a Banach space), then $$ \int TL \,d\gamma=T\left(\int L d\gamma\right). $$ This can be checked using the definition of the Riemann-Stieltjes integral (and using continuity of $T$).

• If $x\in X$, then $$ \int L\, d\gamma x=\int Lx\, d\gamma, $$ which can be checked in the same way (using continuity of $L$).

Therefore, we may evaluate $\int_\Gamma (z-A)^{-1} dz$ at $x\in\ell^p$ by computing $\int_\Gamma(z-A)^{-1}x dz$. Also, since the projection $\pi_n:x\to x_n$ is continuous, we may obtain $\int_\Gamma(z-A)^{-1}x dz$ by $$ \pi_n\left(\int_\Gamma (z-A)^{-1}x dz\right)=\int_\Gamma ((z-A)^{-1}x)_n dz. $$ Since $((z-A)^{-1}x)_n=(z-\alpha_n)x_n$, we are done by basic complex analysis.