Spectrum of Banach algebra with coordinate multiplication

Question

Consider $X=l^p$ , $p \in [1, \infty )$. I proved that $X$ with coordinate multiplication is commutative Banach algebra without unit. I have got a problem to find the spectrum of general element of this Banach algebra.

Answer 1

This elaborates on my comment. If we add a unit, we get an extra dimension. Hence we want $X^+=X\times\mathbb{C}$. By definition of a unit, we should get $(x,\lambda)(0,1)=(0,1)(x,\lambda)=(x,\lambda)$ and $\|(0,1)\|=1$. We also want $x\mapsto(x,0)$ to be an isometric isomorphic embedding, so $(x,0)(y,0)=(xy,0)$ and $\|(x,0)\|=\|x\|$. By the axioms of a Banach algebra, we can conclude that our multiplication should be defined as $(x,\lambda)(y,\mu)=(xy+\lambda y+\mu x,\lambda\mu)$. In order to define a norm for $X^+$ however, we have multiple options. For general Banach algebras the choice turns out to be rather unimportant, so it is usually most convenient to define $\|(x,\lambda)\|=\|x\|+|\lambda|$. Do keep in mind though that this is not always the case. For $C^*$-algebras, there is another definition with better properties, but that is more complicated and not necessary here.

Now let's try to calculate the spectrum of, for example, $x=\{x_n=n^{-2}\}\in\ell^1$. We want to know for which $\lambda\in\mathbb{C}$ the element $(x,-\lambda)\in X^+$ is invertible. So we want to know if there exists $(y,\mu)\in X^+$ with $(x,-\lambda)(y,\mu)=(0,1)$. So we want $xy-\lambda y+\mu x=0$ and $-\lambda\mu=1$. It already follows that $\lambda\neq0$ and $\mu=-\lambda^{-1}$, so $0\in\sigma(x)$. We are left with $xy-\lambda y-\lambda^{-1}x=0$. So for all $n\in\mathbb{N}$ we need $n^{-2}y_n-\lambda y_n-\lambda^{-1}n^{-2}=0$, so $(n^{-2}-\lambda)y_n=\lambda^{-1}n^{-2}$. It now follows that $\lambda\neq n^{-2}$ and $y_n=\frac{\lambda^{-1}n^{-2}}{n^{-2}-\lambda}$, so $\{n^{-2}:n\in\mathbb{N}\}\subseteq\sigma(x)$.

Claim: $\sigma(x)=\{n^{-2}:n\in\mathbb{N}\}\cup\{0\}$

Note that the only thing left to show is $y\in\ell^1$. Let $C=\frac12|\lambda|>0$ and $N=\lceil C^{1/2}\rceil$ such that for all $n>N$ we have $|n^{-2}-\lambda|\geq|\lambda|-n^{-2}\geq|\lambda|-C=C$. We get \begin{align} |\lambda|\sum_{n=1}^\infty|y_n| &=|\lambda|\sum_{n=1}^\infty\left|\frac{\lambda^{-1}n^{-2}}{n^{-2}-\lambda}\right| \\&=\sum_{n=1}^\infty\frac{n^{-2}}{|n^{-2}-\lambda|} \\&\leq\sum_{n=1}^{N-1}\frac{n^{-2}}{|n^{-2}-\lambda|}+C^{-1}\sum_{n=N}^\infty n^{-2}<\infty. \end{align}