Let us prove that $|\chi (x)|\leqslant \|x\|$ for every $x\in$ A. Replacing A, if necessary, by the Banach algebra generated by $x$, we may suppose that A is commutative; then, $\chi \in \mathsf{X}'(A)$. For every $x\in A$, one has $\chi (x)\in$ Sp$'_A(x)$ (I, p. 9, No.$^o7$), hence $|\chi (x)|\leqslant \varrho (x)\leqslant \|x\|($I, p. 28, cor. 5), whence the first assertion.

If, moreover, A is unital, the equality $\chi (1) = 1$ implies that $\|\chi \|\geqslant 1$, whence the required equality.

Let $A'$ be the dual of the Banach space A. Its unit ball $A'_1$ is weakly compact (EVT, III, p. 17, cor. 2). By Theorem 1, one has $\mathsf{X}'(A)\subset A'_1$. Moreover, $\mathsf{X}'(A)$ is closed in $A'$ for the weak topology, because it is the intersection of the weakly closed sets

$X_{x,y}=\{f\in A'|f(xy)-f(x)f(y) = 0\}$ (for $x,y\in A$).

It follows that $\mathsf{X}'(A)$ is compact and that $\mathsf{X}(A) =\mathsf{X}'(A)-\{0\}$ is locally compact.

If A admits a unit element 1, then $\mathsf{X}(A)$ is the set of $\chi \in$ $\mathsf{X}'(A)$ such that $\chi (1) = 1$, and is therefore a closed subset, and hence compact, of $\mathsf{X}'(A)$.

Every compact space is homeomorphic to $\mathsf{X}(A)$ for a suitable commutative unital Banach algebra A (cf. I, p. 32, cor. 2).

Let I be a regular maximal ideal of A. It is closed (I, p. 22, cor. 2), hence $A/I$ is a Banach algebra. Since it is a field (A, VIII, p. 426, prop. 2), the Gelfand-Mazur theorem (I, p. 26, cor. 2) implies that $A/I$ is of dimension 1 over $\mathbf{C}$. Hence I is of codimension 1 in A. The theorem then follows from Lemma 3 of I, p. 10.

It may happen that a nonzero commutative Banach algebra A, without a unit element, has no maximal ideal; then $\mathsf{X}'(A)$ is reduced to $\{0\}($cf. I, p. 186, exercise 31).

This theorem shows that the sets J(A) and $\widehat{A}$ of I, p. 11, which are identified since A is commutative, may also be identified with the set $\mathsf{X}(A)$. There is therefore reason to consider on $\mathsf{X}(A)$ the weak topology, and the Jacobson topology which is less fine (I, p. 15, prop. 5). The closed sets V(M) of the Jacobson topology, for $M\subset A$, are the sets

of $\mathsf{X}(A)$, which will sometimes still be denoted by V(M). Similarly, for every subset M of $\mathsf{X}(A)$, we shall denote by Υ(M) the ideal intersection of the kernels of the $\chi \in M$; it is equal to the set $\Upsilon (M')$ defined in I, p. 13 for the subset $M'\subset J(A)$ corresponding to M when J(A) and $\mathsf{X}(A)$ are identified.

When a topological notion is used in $\mathsf{X}(A)$ without specifying which topology is meant, it will always be the weak topology.

The set $I_{\Phi}$ is a closed ideal of $\mathscr{C}_0(X)$.

Let $\Phi \not= \Phi '$ be closed subsets of X. Exchanging Φ and $\Phi '$ if necessary, we may suppose that there exists $x\in \Phi '$ such that $x\notin \Phi$, and then there exists a function $f\in \mathscr{C}_0(X)$ which is zero on Φ and non-zero at $x$ (TG, IX, p. 43, prop. 1). We have $f\in I_{\Phi}$ and $f\notin I_{\Phi'}$, so that the mapping $\Phi \mapsto I_{\Phi}$ is injective.

Let I be a closed ideal of $\mathscr{C}_0(X)$. Let Φ be the set of $x\in X$ such that $f(x) = 0$ for every $f\in I$; this is a closed subset of X, and one has $I\subset I_{\Phi}$. Let us prove that $I_{\Phi}\subset I$, which will imply that $I = I_{\Phi}$ and complete the proof of the proposition.

Let $f\in I_{\Phi}$. For every real number $\varepsilon >0$, let $C_{\varepsilon}$ denote the set of $x\in X$ such that $|f(x)|\geqslant \varepsilon$. Since $f$ tends to 0 at infinity, the set $C_{\varepsilon}$ is compact. Let $x\in C_{\varepsilon}$; since $f(x)\not= 0$ and $f\in I_{\Phi}$, one has $x\notin \Phi$; by definition of Φ, there then exists a function $\varphi_x\in I$ such that $|\varphi_x(x)|>1$, hence such that $|\varphi_x(y)|>1$ for every $y$ belonging to a neighbourhood $V_x$ of $x$. The open sets $V_x\cap C_{\varepsilon}$ cover $C_{\varepsilon}$. Since the set $C_{\varepsilon}$ is compact, there exists a finite subset $T_{\varepsilon}\subset X$ such that

Then the element

1

$x\in T_{\varepsilon}$

of $\mathscr{C}_0(X)$ belongs to I, and one has $g_{\varepsilon}\geqslant \varepsilon^{-1}$ on $C_{\varepsilon}$. The function

belongs to I. For $x \notin C_{\varepsilon}$, one has

and for $x\in C_{\varepsilon}$, one has

Thus $f_{\varepsilon}$ converges uniformly to $f$ on X when $\varepsilon$ tends to 0. Hence $f\in \overline{I}$, whence $f\in I$ since I is closed.

This follows at once from Prop. 1.

Let us denote by $X'$ the Alexandroff compactification of X, that is, the compact space deduced from X by adjunction of a point at infinity $\omega_X$ (TG, I, p. 67 and 68). The algebra $\mathscr{C}_0(X)$ is identified with the Banach algebra of continuous complex-valued functions on $X'$ vanishing at $\omega_X$.

For every $x\in X'$, we denote by ev$_x$ the character of $\mathscr{C}_0(X)$ defined by ev$_x(f) =f(x)$ for every $f\in \mathscr{C}_0(X)$.

The mapping ev$:x\mapsto$ ev$_x$ of $X'$ into $\mathsf{X}'(\mathscr{C}_0(X))$ is injective. It is surjective by Corollary 1 and Theorem 2 of I, p. 30. It is continuous, for for every function $f\in \mathscr{C}_0(X)$ and every open subset U of $\mathbf{R}$, we have

ev$^{-1}(\{\chi \in \mathsf{X}'(\mathscr{C}_0(X))|\chi (f)\in U\}) =\overset{-1}{f}(U)$

which is open in X. The mapping ev is therefore a homeomorphism since $X'$ is compact. The restriction of ev to X is then a homeomorphism onto $\mathsf{X}(\mathscr{C}_0(X))$.

If F is a weakly closed subset of $\mathsf{X}(\mathscr{C}_0(X))$, it corresponds under the homeomorphism ev to a closed subset Φ of X; precisely, by Prop. 1, we have $F =\{\chi \in \mathsf{X}(\mathscr{C}_0(X))|I_{\Phi}\subset$ Ker $\chi \}$, which is closed for the Jacobson topology.

The mapping $f\mapsto \widetilde{f}$ is a bijection between the set of proper partial mappings $f$ of X into Y and the set of continuous mappings $g$ of $X'$ into $Y'$ such that $g(\omega_X) =\omega_Y$.

Let $f$ be a proper partial mapping of X into Y, and let U be its domain. We shall prove that the mapping $\widetilde{f}$ is continuous. It is continuous at every point of U, since U is open in $X'$. We shall prove that it is also continuous at every point $x$ of $X'-$ U; then $\widetilde{f}(x) =\omega_Y$. Let V be an open neighbourhood of $\omega_Y$ in $Y'$; we shall prove that $\widetilde{f}^{-1}(V)$ is a neighbourhood of $x$. By definition of the topological space $Y'$, we may suppose that V is of the form $Y'-$ K, where K is a compact subset of Y. Since $f$ defines a proper mapping of U into Y, the set $f^{-1}(K)$ is compact in U (TG, I, p. 77, prop. 6), hence in $X'$. In particular it is a closed subset of $X'$, and $\widetilde{f}^{-1}(V) =$ $X'-f^{-1}(K)$ is an open subset of $X'$, and is therefore a neighbourhood of $x$.

Conversely, let $g: X'\rightarrow Y'$ be a continuous mapping such that $g(\omega_X) =\omega_Y$, and let $\Gamma_g\subset X'\times Y'$ be its graph. The set U = X $-\overset{-1}{g}(\omega_Y)$ is open in X. The correspondence $f= (\Gamma_g\cap (U\times Y),X,Y)$ is a proper partial mapping of X into Y (TG, I, p. 77, prop. 7) such that $\widetilde{f}=g$, and it is the only one.

We shall identify the proper partial mappings of X into Y with the continuous mappings of $X'$ into $Y'$ which map $\omega_X$ onto $\omega_Y$. In particular, the proper mappings of X into Y are the proper partial mappings with domain X; they are identified with the continuous mappings $f$ of $X'$ into $Y'$ such that that $\overset{-1}{f}(\omega_Y) =\{\omega_X\}$. If X is compact, these are simply the continuous mappings of X into Y.

Let A be a commutative complex Banach algebra. Recall that $\mathsf{X}'(A)$ is identified with the compact space obtained from $\mathsf{X}(A)$ by adjoining a point at infinity (I, p. 29, corollaire).

In fact, $\mathsf{X}'(\pi )$ is a continuous mapping of $\mathsf{X}'(B)$ into $\mathsf{X}'(A)$ (I, p. 10). The point at infinity of $\mathsf{X}'(B)$ (resp. of $\mathsf{X}'(A)$) is the null character, and one has $\mathsf{X}'(\pi )(0) = 0$.

b) The mapping $\varphi \mapsto \varphi^*$ is a bijection of the set of proper partial mappings of X into Y onto the set of algebra morphisms of $\mathscr{C}_0(Y)$ into $\mathscr{C}_0(X)$. Its inverse bijection is the mapping $\pi \mapsto \mathsf{X}'(\pi )$.

Let us prove a). Let $\varphi$ be a proper partial mapping of X into Y, identified with a continuous mapping of $X'$ into $Y'$ such that $\varphi (\omega_X) =\omega_Y$. For $f\in \mathscr{C}_0(Y)$, one has $(f\circ \varphi )(\omega_X) =f(\omega_Y) = 0$, hence the mapping $\varphi^*$ is well-defined. It is an algebra morphism.

Let us prove that $\mathsf{X}'(\varphi^*)$ identifies with $\varphi$. Let $x\in X$. For every function $f\in \mathscr{C}_0(Y)$, the character $\mathsf{X}'(\varphi^*$)(ev$_x$) associates to $f$ the complex number

(ev$_x\circ \varphi^*$)$(f) =$ ev$_x(f\circ \varphi ) =f(\varphi (x))$,

hence $\mathsf{X}'(\varphi^*$)(ev$_x$) $=$ ev$_{\varphi(x)}$. This proves the assertion.

Conversely, let $\pi :\mathscr{C}_0(Y)\rightarrow \mathscr{C}_0(X)$ be an algebra morphism. Let us prove that $\mathsf{X}'(\pi )^*=\pi$. Let $f\in \mathscr{C}_0(Y)$, and let us write $g=\mathsf{X}'(\pi )^*(f)\in$ $\mathscr{C}_0(X)$. For every $x\in X$, one has

$g(x) = (f\circ \mathsf{X}'(\pi ))(x) =$ ev$_{\mathsf{X}'(\pi)(x)}(f) =$ (ev$_x\circ \pi$ )$(f) =\pi (f)(x)$,

since $\mathsf{X}'(\pi )$ satisfies ev$_{\mathsf{X}'(\pi)(x)}=$ ev$_x\circ \pi$. Hence $g=\pi (f)$, which permits us to conclude that $\mathsf{X}'(\pi )^*=\pi$.

In an entirely similar manner, one has:

a) For every continuous mapping $\varphi : X\rightarrow$ Y, the mapping $\varphi^*:f\mapsto f\circ \varphi$ is an algebra morphism of $\mathscr{C}(Y)$ into $\mathscr{C}(X)$;

b) The mappings $\varphi \mapsto \varphi^*$ and $\pi \mapsto \mathsf{X}(\pi )$ are inverse bijections between the set of continuous mappings of X into Y and the set of algebra morphisms of $\mathscr{C}(Y)$ into $\mathscr{C}(X)$.

By definition (cf. n$^o7$ of I, p. 9), the function $\mathscr{G}'_A(x) :\chi \mapsto \chi (x)$ is continuous on $\mathsf{X}'(A)$ and zero at 0. Since $\mathsf{X}'(A)$ is identified with the Alexandroff compactification of $\mathsf{X}(A)$ by cor. 1 of I, p. 29, the proposition follows.

a) The union of the set of values of $\mathscr{G}(x)$ and of $\{0\}$ is equal to Sp$'_A(x)$ ;

b) If A has a unit element, the set of values of $\mathscr{G}(x)$ is Sp$_A(x)$. In particular, in order that $x$ be invertible, it is necessary and sufficient that $\mathscr{G}(x)$ should not vanish.

Suppose that A admits a unit element. It is known that, for every $\chi \in \mathsf{X}(A)$, one has $\chi (x)\in$ Sp$_A(x)$. Conversely, let $\lambda \in$ Sp$_A(x)$. Then $x-\lambda$ is not invertible, hence belongs to a maximal ideal of A. There then exists $\chi \in \mathsf{X}(A)$ such that $\chi (x-\lambda ) = 0$ (th. 2 of I, p. 30), whence b).

Let us pass to the general case. Let $\widetilde{A}$ be the Banach algebra obtained from A by adjoining a unit element; it is commutative. The set Sp$'_A(x)$ is equal to Sp$_{\widetilde{A}}(x)$, that is to say to the set of values of $\mathscr{G}_{\widetilde{A}}(x)$ on $\mathsf{X}(\widetilde{A}) =\mathsf{X}'(A)$. Whence a).

a) The Gelfand transform $\mathscr{G}$ defines a morphism of A into $\mathscr{C}_0(\mathsf{X}(A))$ such that $\|\mathscr{G}(x)\|=\varrho (x)\leqslant \|x\|$ for every $x\in A$;

b) In order that the Gelfand transform $\mathscr{G}$ be isometric, it is necessary and sufficient that $\|x^2\|=\|x\|^2$ for every $x\in A$.

The mapping $\mathscr{G}$ is a morphism of A into $\mathscr{C}_0(\mathsf{X}(A))$ by No. $^o7$ of I, p. 9 and Prop. 5 of I, p. 37, and satisfies $\|\mathscr{G}(x)\|=\varrho (x)$ by Prop. 6 and Cor. 5 of I, p. 28. Assertion b) follows from a) and Remark 1 of I, p. 21.

a) One has $\varrho (xy)\leqslant \varrho (x)\varrho (y)$ and $\varrho (x+y)\leqslant \varrho (x) +\varrho (y)$;

b) If $y$ is quasi-nilpotent, then Sp$'_A(x) =$ Sp$'_A(x+y)$; if moreover A is unital, then Sp$_A(x) =$ Sp$_A(x+y)$.

By considering the Banach algebra deduced from A by adjoining a unit element, one is first reduced to the case where the algebra A is unital. Then by considering the closed full subalgebra of A generated by $x$ and $y$, one is reduced to the case where A is commutative and unital. Assertion a) is then a consequence of Prop. 7, a), and assertion b) follows from Prop. 6 of I, p. 37 and Cor. 5 of I, p. 28.

(i) The kernel of the Gelfand transform;

(ii) The set of elements $x$ of A such that Sp$'_A(x) =\{0\}$;

(iii) The set of quasi-nilpotent elements of A;

(iv) The radical of A.

Let us denote these sets by $N_1,N_2,N_3,N_4$, respectively. One has $N_1=$ $N_2$ (Prop. 6, a)), and $N_2= N_3($I, p. 28, Cor. 5). By definition, the set $N_4$ is the intersection of the regular maximal ideals of A; it is therefore the intersection of the kernels of the characters of A (th. 2 of I, p. 30), which is equal to $N_1$.

Let Υ(M) be the intersection of the kernels of the elements of M. The set M is closed for the Jacobson topology if and only if M = V(Υ(M)) (cf. I, p. 13 and I, p. 30). This condition is equivalent to saying that the elements $\chi$ of M are precisely the characters which vanish on Υ(M). Consequently M is closed if and only if for every character $\chi \notin M$, there exists $x\in \Upsilon (M)$ such that $\chi (x)\not= 0$. This translates into $\mathscr{G}(x)(\chi )\not= 0$ and $\mathscr{G}(x)|M = 0$.

Let $h: A\rightarrow B$ be an algebra morphism and let $(a, b)\in A\times B$ be a point adherent to the graph Γ of $h$. Let $\chi \in \mathsf{X}'(B)$. The function $x\mapsto \chi (h(x))$ from A into $\mathbf{C}$ is an algebra homomorphism, hence is continuous (I, p. 29, th. 1). The map from $A\times B$ into $\mathbf{C}$ given by $(x, y)\mapsto \chi (h(x))-\chi (y)$ is therefore continuous; it is zero on Γ, hence zero at $(a, b)$. Thus one has $\chi (h(a)) =\chi (b)$ for every $\chi \in \mathsf{X}'(B)$. Since B is without radical, one has $h(a) =b$. Thus the graph of $h$ is closed and therefore $h$ is continuous (EVT, I, p. 19, cor. 5).

It is enough to apply prop. 9 to the identity map of the algebra.

Let A and B be commutative Banach algebras. By No.$^o7$ of I, p. 9, if $h: A\rightarrow B$ is a surjective morphism, $\mathsf{X}'(h)$ is a homeomorphism of $\mathsf{X}'(B)$ onto a closed subspace of $\mathsf{X}'(A)$ which transforms 0 into 0. (In the case where $h$ is the injection of a subalgebra into $\mathscr{C}(X)$, the map $\mathsf{X}'(h)$ is injective under much weaker hypotheses, cf. I, p. 142, prop. 1, d).)

Now let $h: A\rightarrow B$ be an injective morphism. In general, $\mathsf{X}'(h)$ is not surjective, but the following proposition provides a necessary condition for this to be the case.

Let $x\in A$ be such that $h(x)$ is invertible in B. For every $\chi \in \mathsf{X}(A)$, there exists $\xi \in \mathsf{X}(B)$ such that $\chi =\mathsf{X}(h)(\xi )$, hence $\chi (x) =\xi (h(x))\not= 0$. Prop. 6 of I, p. 37 then shows that $x$ is invertible in A, and hence that $h(x)$ is invertible in $h(A)$.

The necessary condition of the proposition is not sufficient, even if $h$ is isometric (I, p. 168, exerc. 14). One has nevertheless the following result:

(i) $\mathsf{X}(h)$ is surjective;

(ii) $h(A)$ is a full subalgebra of B;

(iii) Sp$_A(a) =$ Sp$_B(h(a))$.

(i) $=\Rightarrow$ (ii) follows from prop. 10.

(ii) $=\Rightarrow$ (iii) results from the formula Sp$_A(a) =$ Sp$_{h(A)}(h(a)) =$ Sp$_B(h(a))$, valid since $h(A)$ is a full subalgebra of B.

(iii) $=\Rightarrow$ (i) by the formula (4) of No.$^o6$ of I, p. 6, we have the commutative diagram

Sp$_B(h(a))\rightarrow$ Sp$_A(a)$

where the vertical arrows denote surjective mappings (I, p. 37, prop. 6) and $i$ is the canonical inclusion. Assumption (iii) means that $i$ is bijective. Moreover the surjective mapping $\mathscr{G}_A(a) :\mathsf{X}(A)\rightarrow$ Sp$_A(a)$ is bijective: in fact, for every characters $\chi_1$ and $\chi_2$ of A, the set $A_{\chi_1,\chi_2}=\{x\in A|\chi_1(x) =\chi_2(x)\}$ is a closed full subalgebra of A. By hypothesis, therefore, one has $A_{\chi_1,\chi_2}$ = A if $a\in A_{\chi_1,\chi_2}$, that is to say $\chi_1=\chi_2$ if $\mathscr{G}_A(a)\chi_1=\mathscr{G}_A(a)\chi_2$. The diagram then implies that the mapping $\mathsf{X}(h)$ is surjective.