Prove that $(\ell^2, \left||.\right\|_2)$ is a commutative and semisimple Banach algebra

Question

Let $(\ell^2, \left||.\right\|_2)$ with the coordinatewise product.Prove that $(\ell^2, \left||.\right\|_2)$ is a commutative and semisimple Banach algebra

this is my attempt:

Since $(\ell^2, \left||.\right\|_2)$ is a $C$*-$ algebra, Hence The Gelfand–Naimark Theorem states

$\ell^2$ is isometrically *-isomorphic to $ C(\Phi _{\ell^2}) $ .

this allows one to say that $(\ell^2, \left||.\right\|_2)$ is a semisimple

Any help will be appreciated! Thanks

Answer 1

The space $\ell^2$ is not isometrically isomorphic to any $C^\ast$-algebra. It is reflexive, and any reflexive $C^\ast$-algebra is finite-dimensional.

Clearly $\ell^2$ is a commutative Banach algebra. To prove that $\ell^2$ is semi-simple, we must show that the character space $\Gamma$ of $\ell^2$ separates the points of $\ell^2$. This is easy enough: For every $k\in\mathbb{N}$ the map $\delta_k\colon \ell^2\to\mathbb{C}$ is linear, multiplicative and non-zero. Moreover, $\delta_k(f)=0$ for all $k\in\mathbb{N}$ implies $f=0$.