Proposition 1

Let B be a unital associative algebra with a real filtration $(B_\alpha)_{\alpha \in \mathbf{R}}$ such that B is Hausdorff and complete (§ 4, nos. 1 and 2). Let f be a mapping of X into B such that there exists $\lambda > 0$ for which $f(X) \subset B_\lambda$. Then f can be extended in one and only one way to a continuous unital homomorphism $\hat{f}$ of $\hat{A}(X)$ into B.

Let $f'$ be the unique unital algebra homomorphism of $A(X)$ into B extending $f$ (Algebra, Chapter III, § 2, no. 7, Proposition 7). We show that $f'$ is continuous: $f'(A^n(X)) \subset B_{n\lambda}$ whence $f'(\hat{A}_n(X) \cap A(X)) \subset B_{n\lambda}$. Therefore $f'$ can be extended in one and only one way by continuity to a homomorphism $\hat{f} : \hat{A}(X) \to B$.

We preserve the hypotheses and notation of Proposition 1 and let $u \in \hat{A}(X)$. The element $\hat{f}(u)$ is denoted by $u((f(x))_{x \in X})$ and called the result of substituting the f(x) for the x in u. In particular, $u((x)_{x \in X}) = u$. Now let $u = (u_y)_{y \in Y}$ be a family of elements of $\hat{A}_1(X)$ and let $v \in \hat{A}(Y)$. The above allows us to define the element $v((u_y)_{y \in Y}) \in \hat{A}(X)$. It is denoted by $v \circ u$. As $u_y((f(x))) \in B_\lambda$, the elements $u_y((f(x)))$ can be substituted for the y in v. The mappings $v \mapsto (v \circ u)((f(x)))$ and $v \mapsto v((u_y((f(x))))))$ are then two continuous homomorphisms of unital algebras of $\hat{A}(X)$ into B taking the same value $u_y((f(x)))$ at the element $y \in Y$. Therefore (Proposition 1)

for all $v \in \hat{A}(Y)$.