Exercise 11, § 8

Let X be a metric space, whose distance is denoted by $d$. A function $f: X\rightarrow \mathbf{C}$ is said to be Lipschitz if there exists a real number $c\geqslant 0$ such that, for all $x$ and $y$ in X, one has

(cf. FVR, IV, p. 7, déf. 1). We denote by Lip$_b(X)$ the set of bounded Lipschitz functions on X.

a) The set Lip$_b(X)$ is a subalgebra of $\mathscr{C}(X)$.

b) Endowed with the norm

$\|f\|=$ sup$_{x\in X}|f(x)|+$ sup$_{x,y\in X}\frac{|f(x) -f(y)|}{d(x, y)}$,

$x\not=y$

and with the involution $f\mapsto \overline{f}$, the algebra Lip$_b(X)$ is a commutative involutive Banach algebra. It is not a star algebra in general.

c) For every commutative Banach algebra A, we denote by $\mathsf{X}_m(A)$ the metric space $\mathsf{X}(A)$ endowed with the distance induced by the norm of the dual $A'$ of A. The metric space $\mathsf{X}_m(A)$ is complete and has diameter $\leqslant 2$.

d) Let A be a commutative unital Banach algebra. The Gelfand transform maps A into Lip$_b(\mathsf{X}_m(A))$. If A is without radical, then it induces a homeomorphism of A onto its image.

In what follows, we suppose that X is a compact metric space. We denote by A = Lip$_b(X)$.

e) The mapping $\varphi$ which sends $x\in X$ to the character $f\mapsto f(x)$ is a homeomorphism of X onto $\mathsf{X}(A)$.

f) Let $d_1$ be the distance on X induced by the homeomorphism $\varphi$. Then $d_1$ is equivalent to $d$.