# functional analysis – Characters on \$ell^infty\$ and characteristic functions

Let $$ell^infty$$ be the set of bounded complex-valued sequences, which we regard as a commutative $$C^*$$ algebra under pointwise multiplication. Write $$Phi$$ for the set of characters of $$ell^infty$$, i.e. non-zero algebra homomorphisms $$ell^inftyto mathbb{C}$$. Note that, if $$Bsubsetmathbb{N}$$ and $$chi_B$$ is its characteristic function, then $$varphi(chi_B)in {0,1}$$ for any $$varphiinPhi$$ because
$$varphi(chi_B)=varphi(chi_B^2)=(varphi(chi_B))^2$$

Fix $$varphiin Phi$$, $$varepsilon>0$$, and $$xin ell^infty$$. Put
$$A={ninmathbb{N}:|x_n-varphi(x)|
I’d like to show that $$varphi(chi_A)=1$$.

Some context: By the Gelfand-Naimark Theorem, $$C(Phi)cong ell^infty$$, and $$nin mathbb{N}$$ can be identified with the projection $$xmapsto x_n$$, which obviously belongs to $$Phi$$. Under this identification, $$mathbb N$$ is dense in $$Phi$$ and so $$Phi$$ can be regarded as the Stone-Cech compacfication of $$mathbb N$$. If we let $$beta mathbb N$$ be the set of ultrafilters on $$mathbb N$$, then we can find explicit bijections between $$Phi$$ and $$beta mathbb N$$ by considering
$$F: Phi to betamathbb N\ varphimapsto {Asubset mathbb N:varphi(chi_A)=1}$$
and
$$G:betamathbb{N}to Phi \ mathcal{U}mapsto left(xmapsto lim_mathcal{U} xright)$$
The identity I’m trying to prove arises when checking that $$(Gcirc F)(varphi)=varphi$$ for every $$varphiinPhi$$. Of course, one can also verify that they are continuous and then check that the compositions are the identity on $$mathbb{N}$$, which as a by product yields what we want. However, I suspect there’s a more direct, elementary argument that I’m not seeing.