(i) For every compact subset Y of X and every positive measure $\mu$ on Y, the endomorphism of $L^2(Y, \mu)$ defined by the kernel $f|(Y\times Y)$ (Def. 1 of III, p. 29) is positive; in other words, one has

for every $h\in \mathscr{L}^2(Y, \mu)$;

(ii) For every integer $n\in \mathbf{N}$, every family $(x_i)_{0\leqslant i\leqslant n}$ in X and every family $(t_i)_{0\leqslant i\leqslant n}$ of complex numbers, one has

(iii) There exist a complex Hilbert space E and a continuous mapping $g: X\rightarrow E$ with total image such that $f(x, y) =\langle g(x)|g(y)\rangle$ for all $x$ and $y$ in X;

(iv) There exist a complex Hilbert space E and a continuous mapping $g: X\rightarrow E$ such that $f(x, y) =\langle g(x)|g(y)\rangle$ for all $x$ and $y$ in X.

It is apparent that (iii) implies (iv), and one sees that (i) implies (ii) by considering the discrete measure $\mu$ which is the image of the counting measure on $\{0, . . . , n\}$ under the mapping $i\mapsto x_i$ and the function $h$ such that

$x_i=x$

Let us prove that (iv) implies (i). Suppose that there exist a complex Hilbert space E and a continuous mapping $g: X\rightarrow E$ such that $f(x, y) =\langle g(x)|g(y)\rangle$ for every $(x, y)\in X\times X$. Let Y be a compact subset of $X,\mu$ a positive measure on Y and $h\in \mathscr{L}^2(Y, \mu)$. One has

by INT, V, p. 97, § 8, no$^o4$, Prop. 9.

Finally, let us prove that (ii) implies (iii). Let $\widetilde{E}$ be the space of complex measures with finite support on X. For $\mu_1$ and $\mu_2$ in $\widetilde{E}$, put

The sesquilinear form resulting on $\widetilde{E}$ is a positive Hermitian form. In fact, let $\mu\in \widetilde{E}$; there exist a finite family $(x_i)_{0\leqslant i\leqslant n}$ in X and complex numbers $(t_i)_{0\leqslant i\leqslant n}$ such that $\mu=\sum^n_{i=0}t_i\varepsilon_{x_i}$. One then has

by hypothesis. Define $\widetilde{g}: X\rightarrow \widetilde{E}$ by $\widetilde{g}(x) =\varepsilon_x$. The image of the mapping $\widetilde{g}$ generates $\widetilde{E}$. Moreover, for every $(x, y)\in X\times X$, one has on the one hand $f(x, y) =\langle \widetilde{g}(x)|\widetilde{g}(y)\rangle$ and on the other hand

which implies that $\widetilde{g}$ is continuous since $f$ is continuous.

Let E be the separated-completed Hilbert space of $\widetilde{E}$ (EVT, V, p. 8, cor. of the prop. 4) and $g: X\rightarrow E$ the composition of $\widetilde{g}$ and of the canonical mapping $\widetilde{E}\rightarrow E$. Then the mapping $g$ is continuous, its image is total in E, and one has $f(x, y) =\langle g(x)|g(y)\rangle$ for all $(x, y)\in X\times X$.

The method used to prove that (ii) implies (iii) is called the Gelfand–Naimark–Segal construction.

The uniqueness of $u$ results from the fact that the image of $g_1$ is total in $E_1$. Let $F =\mathbf{C}^{(X)}$ and let $(e_x)_{x\in X}$ be the canonical basis of F. For $j= 1$ and $j= 2$, denote by $u_j$ the linear mapping from F into $E_j$ determined by $u_j(e_x) =g_j(x)$, and denote by $F_j$ its image; the space $F_1$ is dense in $E_1$.

Let $t=\sum t_xe_x$ be an element of F. One has

Consequently, there exists an isometric linear mapping $v$ from $F_1$ into $F_2$ such that $u_2=v\circ u_1$, and in particular $g_2(x) =v(g_1(x))$ for all $x\in X$. Since the image of $g_1$ is total in $E_1$, this mapping extends to an isometric linear mapping $u$ from $E_1$ into $E_2$ such that $g_2=u\circ g_1$.

By Lemma 8 of I, p. 107, the image of $u$ is closed in $E_2$. If $(E_2, g_2)$ is a cyclic Hilbertian realization, the image of $u$ is also dense in $E_2$ and $u$ is therefore an isomorphism.

real numberIt is elementary that if$t\geqslant 0$, and that $\frac{f}{f}\in \in$ NoyNoy$_{++}(X)$.(X), then $tf\in$ Noy$_+(X)$ for every

If $(E_1, g_1)$ and $(E_2, g_2)$ are Hilbertian realizations of universally positive kernels $f_1$ and $f_2$ on X, then the pair $(E_1\oplus E_2, g_1+g_2)$ (resp. the pair $(E_1\widehat{\otimes}_2E_2, g_1\otimes g_2)$) is a Hilbertian realization of $f_1+f_2$ (resp. of $f_1f_2$); these are therefore universally positive kernels.

Characterization (ii) of Noy$_+(X)$ (Theorem 1) implies that this set is closed in $\mathscr{C}(X\times X)$ endowed with the topology of simple convergence.

The first assertion follows from I, p. 97. From what precedes, the mapping $\varphi$ of $\mathscr{O}$(Sp$_A(a)$) into A defined by $f\mapsto (f^*(a^*))^*$ is a continuous unital morphism of $\mathscr{O}$(Sp$_A(a)$) into A such that the image of the germ in a neighbourhood of Sp$_A(a)$ of the identity function of $\mathbf{C}$ is equal to $a$. Consequently, $\varphi$ is the mapping $f\mapsto f(a)$ of the holomorphic functional calculus (I, p. 74, th. 5).

The radius of convergence of the power series

$n=0$

is equal to 1 and its sum $f$ defines a holomorphic function on D (VAR,

R1, p. 27, 3.2.9) taking the value 1 at 0. It satisfies $f(x) =1-x$ for every $x\in D\cap \mathbf{R}$ (FVR, III, p. 19), hence $f(z)^2= 1-z$ for $z\in D$ since the difference $f(z)^2-(1-z)$ is a holomorphic function all of whose successive derivatives vanish at 0 (VAR, R1, p. 27, 3.2.5). By definition, one verifies that $f^*=f$.

Let $g$ be a holomorphic function on D such that $g(z)^2= 1-z$ for every $z\in D$ and such that $g(0) = 1$. The function $g$ does not vanish and the continuous function $f /g$ on D takes its values in $\{-1,1\}$; since D is connected and $f(0) =g$(0), one has $f=g$.

a) For all $a$ and $b$ in A, one has $\lambda (a^*b) =\overline{\lambda(b^*a)}$ and

b) If A is unital, then the linear form $\lambda$ is continuous and its norm is equal to $\lambda (1)$.

The mapping $(a, b)\mapsto \lambda (a^*b)$ is a positive Hermitian form on A; it therefore satisfies $\lambda (a^*b) =\lambda (b^*a)$ and $|\lambda (a^*b)|^2\leqslant \lambda (a^*a)\lambda (b^*b)$ for all $a$ and $b$ in A (EVT, V, p. 2, remark and p. 3, prop. 2).

Let us prove b). Let $a\in A$ be a hermitian element of norm $<1$. The spectral radius of $a$ is less than $\|a\|$, hence the spectrum of $a$ is contained in the open unit disc D of $\mathbf{C}$ (Theorem 1 of I, p. 24). Let $f$ be a holomorphic function on D such that $f(z)^2= 1-z$ for all $z\in D$ (Lemma 1 of V, p. 435). Applying the holomorphic functional calculus to the element $a$ and to the function $f$, the element $b=f(a)$ satisfies $b^2= 1-a($I, p. 74, Theorem 5). By Prop. 3 of V, p. 435, one moreover has $f(a)^*=f^*(a^*) =f(a)$, hence $b$ is hermitian. We then have $\lambda (1-a) =\lambda (b^*b)\geqslant 0$, whence $\lambda (a)\leqslant \lambda (1)$.

Now let $b\in A$ be of norm $<1$. The element $b^*b$ is hermitian and $\|b^*b\|<1$, hence applying a) with $a= 1$, we find

with equality if $b= 1$. The linear form $\lambda$ is therefore continuous, and its norm is equal to $\lambda (1)$. Assertion b) is proved.

a) For every $a$ in A, one has $\lambda (a^*) =\overline{\lambda(a)}$ and $|\lambda (a)|^2\leqslant \|\lambda \|\lambda (a^*a)$;

b) Let $\widetilde{A}$ be the involutive algebra obtained from A by adjunction of a unit element, and let $e$ be its unit element. There exists a continuous positive linear form $\widetilde{\lambda}$ on $\widetilde{A}$ extending $\lambda$ and such that $\widetilde{\lambda}(e) =\|\lambda \|$;

c) For all $a$ and $b$ in A, one has $|\lambda (b^*ab)|\leqslant \|a\|\lambda (b^*b)$.

Let us prove a). Let $\mathfrak{F}$ be an approximate unit of A. Let $a\in A$. Using Lemma 2, a) and the definition of an approximate unit, one finds

$\lambda (a^*) =$ lim$_{f,\mathfrak{F}}\lambda (f a^*) =$ lim$_{f,\mathfrak{F}}\lambda (af^*) =$ lim$_{f,\mathfrak{F}}\lambda ((f a^*)^*) =\overline{\lambda(a)}$,

whence (loc. cit.)

$|\lambda (a)|^2=$ lim$_{f,\mathfrak{F}}|\lambda (f a)|^2\leqslant \lambda (a^*a)$ lim sup$_{f,\mathfrak{F}}\lambda (f f^*)\leqslant \|\lambda \|\lambda (a^*a)$,

since $\mathfrak{F}$ is a filter on the unit ball of A.

To prove b), one may suppose that $\lambda$ is not zero, then that $\|\lambda \|= 1$. For $a\in A$ and $z\in \mathbf{C}$, let us put $\widetilde{\lambda}(a+z\cdot e) =\lambda (a) +z$. The mapping $\widetilde{\lambda}$ is a continuous linear form on $\widetilde{A}$ which extends $\lambda$ and satisfies $\widetilde{\lambda}(e) = 1$. It is positive: for all $a\in A$ and $z\in \mathbf{C}$, by a), we calculate

Finally, let $b$ be an element of A. The mapping $a\mapsto \widetilde{\lambda}(b^*ab)$ is a positive linear form on the unital involutive Banach algebra $\widetilde{A}$. It is therefore continuous of norm equal to its value at $e$ (Lemma 2, b)), which is equal to $\lambda (b^*b)$, whence assertion c).

a) For every Hilbert space E, every morphism of involutive algebras $\varphi : A\rightarrow \mathscr{L}(E)$ and every vector $x\in E$, the linear form defined on A by $\lambda (a) =\langle x|\varphi (a)x\rangle$ is a continuous positive linear form;

b) Let $\lambda \in A'_+$. The mapping $f$ of $A\times A$ into $\mathbf{C}$ defined by $f(a, b) =$ $\lambda (a^*b)$ for every $(a, b)\in A\times A$ is a universally positive kernel on A.

Let us prove a). For every $a\in A$, one has

therefore $\lambda$ is a positive linear form on A; it is continuous since $\varphi$ is continuous (Prop. 2 of I, p. 104).

Let us prove b). The function $f$ is continuous. For every $n\in \mathbf{N}$, every family $(a_i)_{0\leqslant i\leqslant n}$ in A and every family $(t_i)_{0\leqslant i\leqslant n}$ of complex numbers, one has

whence the result (Def. 1 of V, p. 433).

One may suppose that A is a unital algebra (Lemma 3, b)).

Let $\lambda$ be a continuous positive linear form on A. Let $(E, g)$ be a cyclic Hilbertian realization of the universally positive kernel $f$ on A defined by $f(a, b) =\lambda (a^*b)$ (Prop. 4 and Theorem 1). The mapping $g$ is continuous, its image is total in E, and one has $\lambda (a^*b) =\langle g(a)|g(b)\rangle$ for every $(a, b)\in A^2$. Put $x=g(1)\in E$.

The mapping $g$ of A into E is linear: in fact, for $(a, b, c)\in A^3$ and $(s, t)\in \mathbf{C}^2$, we have

whence the assertion, since the image of $g$ is total in E.

In particular, the image F of $g$ is a dense vector subspace of E. The kernel of $g$ is a left ideal of A: if $g(b) = 0$, then for all $a$ and $c$ in A, we have

therefore $g(ab) = 0$ since F is dense in E.

Let $a\in A$. Since the kernel of $g$ is a left ideal of A, there exists a linear mapping $\widetilde{\varphi}(a) : F\rightarrow F$ such that $\widetilde{\varphi}(a)(g(b)) =g(ab)$ for every $b\in A$. Moreover, for every $b\in A$, we have

(lemma 3, c)), therefore $\widetilde{\varphi}(a)$ is continuous and of norm $\leqslant \|a\|$. Consequently, there exists a unique endomorphism $\varphi (a)\in \mathscr{L}(E)$ which induces $\widetilde{\varphi}(a)$ by passing to the subspaces.

Let $a$ and $b$ be in A. We have

for every $c$ in A, therefore $\varphi (ab) =\varphi (a)\circ \varphi (b)$ since F is dense in E.

The mapping $\varphi$ of A into $\mathscr{L}(E)$ is linear: in fact, for every $(a, b)\in A^2$ and every $(s, t)\in \mathbf{C}^2$, we have

for every $c\in A$, whence $\varphi (sa+tb) =s\varphi (a) +t\varphi (b)$ since F is dense in E. Since $\|\varphi (b)\|\leqslant \|b\|$ for every $b\in A$, the mapping $\varphi$ is continuous. We also have $\varphi (1) = 1_E$ since $\varphi (1)(g(a)) =g(a)$ for every $a\in A$ and F is dense in E.

Lastly, let $a,b,c$ be in A. We have

whence $\varphi (a^*) =\varphi (a)^*$ since F is dense in E.

In conclusion, the mapping $\varphi$ is a continuous morphism of involutive algebras of A into $\mathscr{L}(E)$ and $g(a) =\varphi (a)x$ for every $a\in A$; consequently, the triplet $(E, x, \varphi )$ is a cyclic Hilbertian realization of $\lambda$.

a) There exists a unique continuous linear mapping $u$ from $E_1$ into $E_2$ which is a morphism of representations of $\varphi_1$ into $\varphi_2($I, p. 11) and which satisfies $u(\varphi_1(a)x_1) =\varphi_2(a)x_2$ for every $a\in A$;

b) The linear mapping $u$ is isometric;

c) If $(E_2, x_2, \varphi_2)$ is cyclic, then $u$ is an isomorphism;

d) If $(E_2, x_2, \varphi_2)$ is cyclic and if A is unital, then $u(x_1) =x_2$.

For $j\in \{1,2\}$, define $\gamma_j: A\rightarrow E_j$ by $\gamma_j(a) =\varphi_j(a)x_j$ for every $a\in A$. By definition, the pairs $(E_j, \gamma_j)$ are Hilbertian realizations of the universally positive kernel $f$ on A defined by $(a, b)\mapsto \lambda (a^*b)$. The Hilbertian realization $(E_1, \gamma_1)$ is cyclic; by prop. 1 of V, p. 434, there exists therefore a unique continuous linear mapping $u: E_1\rightarrow E_2$ such that $\gamma_2=u\circ \gamma_1$, and this mapping is isometric. Moreover, if $(E_2, x_2, \varphi_2)$ is also cyclic, then $u$ is an isomorphism. To prove a), b) and c), it is enough to prove that $u$ is a morphism of representations of $\varphi_1$ into $\varphi_2$.

Let $a\in A$. For every $b\in A$, we have

therefore the continuous linear mappings $u\circ \varphi_1(a)$ and $\varphi_2(a)\circ u$ coincide on the subspace of $E_1$ generated by the image of $\gamma_1$; this subspace is dense in $E_1$ by hypothesis, whence $u\circ \varphi_1(a) =\varphi_2(a)\circ u$.

Finally, let us prove d). Suppose therefore that A is unital and that $(E_2, x_2, \varphi_2)$ is cyclic. There exists a cyclic Hilbertian realization $(E_3, x_3, \varphi_3)$ of $\lambda$ such that $\varphi_3$ is a unital morphism (prop. 5). Let $j\in \{1,2\}$. Applying the foregoing to $(E_j, x_j, \varphi_j)$ and $(E_3, x_3, \varphi_3)$, we see that there exists an isometric isomorphism $\widetilde{u}_j$ of $E_j$ into $E_3$ such that $\widetilde{u}_j\circ \varphi_j(a) =\varphi_3(a)\circ \widetilde{u}_j$ for every $a\in A$. Taking $a= 1_A$, we see that $\varphi_j$ is unital. We then have $x_2=\varphi_2(1_A)x_2=u(\varphi_1(1_A)x_1) =u(x_1)$.

Let us first prove that $\lambda$ is bounded on the intersection of $A_+$ and the unit ball of A. Suppose this is not so. Then there exists a sequence $(x_n)_{n\geqslant 1}$ in $A_+$ such that $\|x_n\|\leqslant 1$ and $\lambda (x_n)\geqslant n$ for every integer $n\geqslant 1$. The series with general term $n^{-2}x_n$ converges to an element $x$ of A (cf. TG, IV, p. 33, Example 3). For every integer $N\geqslant 1$, since

$+\infty$

$\sum\frac{1}{n^2}x_n\in A_+$,

$n=N+1$

(I, p. 116, Prop. 14), it follows that

N N $+\infty$

$\sum\frac{1}{n}\leqslant \sum\frac{1}{n^2}\lambda (x_n) =\lambda (x)-\lambda (\sum\frac{1}{n^2}x_n)\leqslant \lambda (x)$,

$n=1n=1n=N+1$

which is absurd (TG, IV, p. 33, Example 4).

Since every element of the unit ball of A is a linear combination with coefficients bounded by 1 of at most four positive elements of A of norm $\leqslant 1$(I, p. 96, Lemma 2 and formula (4) of I, p. 117), we conclude that $\lambda$ is continuous.

There exist involutive Banach algebras which admit positive linear forms that are not continuous (Exercise 3, (a) of V, p. 493).

Consider the real Banach space $A_h$ of hermitian elements of A. The set $A_+$ is a salient pointed closed convex cone in $A_h$ (Prop. 14 of I, p. 116). The element $a^*a$ is positive (Theorem 2 of I, p. 118) and nonzero; hence the hermitian element $-a^*a$ is therefore not positive. By EVT, II, p. 42, Cor. 5, there exists a continuous real linear form $\lambda \in A'_h$ such that $\lambda (-a^*a)<0$ and $\lambda (A_+)\subset \mathbf{R}_+$. The linear form $\lambda$ extends to a hermitian $\mathbf{C}$-linear form on A (cf. I, p. 98), which is positive and has the required property.

For every $b\in A-\{0\}$, let $\lambda_b$ be a continuous positive linear form on A such that $\lambda_b(b^*b)>0$ (Prop. 8). Since A admits an approximate unit (I, p. 121, Prop. 18), there exists a Hilbert realization $(E_b, x_b, \varphi_b)$ of $\lambda_b$ (Prop. 5). If A is unital, one may suppose that $\varphi_b$ is unital (loc. cit.). We have $\|\varphi_b(b)\|^2=\lambda_b(b^*b)\not = 0$.

Let E be the external Hilbert sum of the spaces $E_b$ for $b$ belonging to A $-\{0\}$. For every $a\in A$, let us denote by $\varphi (a)\in \mathscr{L}(E)$ the unique continuous linear mapping whose restriction to $E_b$ coincides with $\varphi_b(a)$ for every $b\in A-\{0\}$. The mapping $a\mapsto \varphi (a)$ is a morphism of involutive algebras; it is injective, hence isometric (Prop. 9 of I, p. 112), and satisfies $\varphi (1) = 1_E$ if A is unital. This completes the proof.

The category whose objects are stellar algebras and whose morphisms are morphisms of involutive algebras is therefore equivalent to the category whose objects are closed involutive subalgebras of the algebras of endomorphisms of Hilbert spaces, and whose morphisms are morphisms of involutive algebras.