Let $\mathscr{S}(E, F)$ be the space of all continuous sesquilinear forms on $E \times F$, endowed with the norm

We define the space $\mathscr{S}(F, E)$ similarly. We defined (V, p. 16, cor. 2) a Banach space isomorphism from $\mathscr{L}(E; F)$ onto $\mathscr{S}(F, E)$, denoted by $u \mapsto \Phi_u$ and characterized by

In an analogous way we define an isomorphism from $\mathscr{L}(F, E)$ onto $\mathscr{S}(E, F)$. Finally we define a mapping $\Phi \mapsto \Phi^*$ from $\mathscr{S}(F, E)$ onto $\mathscr{S}(E, F)$ by

This mapping is bijective, semi-linear and isometric. But formula (1) translates as $\Phi_{u^*} = (\Phi_u)^*$, hence the proposition.

(11)

By prop. 1, $\|u^*\| = \|u\|$, hence $\|u^*u\| \leq \|u^*\|\cdot\|u\| \leq \|u\|^2$. On the other hand,

hence $\|u^*u\| = \|u\|^2$. Replacing $u$ by $u^*$, we get $\|uu^*\| = \|u^*\|^2$, hence (11) follows since $\|u\| = \|u^*\|$.

Let $E_1, ..., E_n$ and $F_1, ..., F_n$ be hilbertian spaces, and for every integer $i$ between 1 and $n$, let $u_i$ be a continuous linear mapping from $E_i$ into $F_i$. Then

(12)

Let $v$ be the continuous linear mapping $u_1 \hat{\otimes}_2 ... \hat{\otimes}_2 u_n$ from

and $w$ the continuous linear mapping $u_1^* \hat{\otimes}_2 ... \hat{\otimes}_2 u_n^*$ from $F$ into $E$. It is enough to prove the equality $\langle y|v(x)\rangle = \langle w(y)|x\rangle$ for $x \in E$ and $y \in F$. By linearity and continuity, we reduce to the case when $x$ and $y$ have the following form

with $x_i \in E_i$ and $y_i \in F_i$ for $1 \leq i \leq n$. From the definition of scalar product in a tensor product (V, p. 27, formula (6)), we then get

This proves our assertion.

Let $E$ and $F$ be two hilbertian spaces, $u \in \mathscr{L}(E ; F)$ and $n$ a positive integer. If we put $u_1 = ... = u_n = u$ in formula (12) we obtain the result that the continuous linear mapping $\hat{T}^n(u^*)$ from $\hat{T}^n(F)$ into $\hat{T}^n(E)$ is the adjoint of the continuous linear mapping $\hat{T}^n(u)$ from $\hat{T}^n(E)$ into $\hat{T}^n(F)$. The formulas

can be established in the same way as formula (12), on account of the definition of the scalar product in $\hat{S}^n(E)$ (V, p. 30, formula (15)) and in $\hat{\Lambda}^n(E)$ (V, p. 33, formula (26)).

(i) $u$ is a topological vector space isomorphism, with an inverse equal to $u^*$; (ii) $u$ is surjective and $u^*u = 1_E$; (iii) $u$ is injective and $uu^* = 1_F$; (iv) $u$ is an isomorphism of normed spaces; (v) $u$ is a hilbertian space isomorphism.

Condition (1) means that we have $u^*u = 1_E$ and $uu^* = 1_F$. Hence the equivalence of (i), (ii) and (iii) follows from E, II, § 3, No. 8, prop. 8. We have already seen the equivalence of (iv) and (v) (V, p. 5). Finally, the relation $u^*u = 1_E$ is equivalent to

$^1$ See P. A. M. Dirac, Quantum Mechanics, Oxford University Press, New York, 1935.

$\langle x|u^*u(y)\rangle = \langle x|y\rangle$, that is, to $\langle u(x)|u(y)\rangle = \langle x|y\rangle$ for all $x, y$ in $E$, and evidently implies that $u$ is injective; this proves the equivalence of (ii) and (v).

An automorphism of the hilbertian space $E$ is also called a unitary operator, that is, an operator $u \in \mathscr{L}(E)$ satisfying $uu^* = u^*u = 1_E$.

(ii) Suppose that $E = F$. Let $M$ be a closed vector subspace of $E$ and $M^\circ$ its orthogonal. The relations $u(M) \subset M$ and $u^*(M^\circ) \subset M^\circ$ are equivalent.

Let $Q = \overline{u(E)}$ be the final subspace of $u$. The orthogonal $Q^\circ$ of $Q$ in $F$ consists of all vectors $y$ such that $\langle u(x)|y\rangle = 0$ for all $x \in E$; this is equivalent to : $\langle x|u^*(y)\rangle = 0$ for all $x \in E$, or to $u^*(y) = 0$. Hence we have $Q^\circ = \mathrm{Ker}\, u^*$, and $Q$ is the initial subspace of $u^*$. Since $u$ is the adjoint of $u^*$, the final subspace of $u^*$ is also the initial subspace of $u$. This proves (i).

The relation $u(M) \subset M$ implies that $u(M)$ is orthogonal to $M^\circ$, and the relation $u^*(M^\circ) \subset M^\circ$ implies that $u^*(M^\circ)$ is orthogonal to $M$. But we have $\langle u(x)|y\rangle = \langle u^*(y)|x\rangle$ for all $x \in M$ and $y \in M^\circ$; hence (ii) follows.

We remark that prop. 4 can be deduced from the general properties of the transpose (II, p. 51, cor. 2) in view of remark 3, V, p. 41.

(i) The mapping $u^* \in \mathscr{L}(F; E)$ is partially isometric, with initial subspace $Q$ and final subspace $P$. The isomorphism from $P$ onto $Q$ induced by $u$ is then the inverse of the isomorphism from $Q$ onto $P$ induced by $u^*$.

(ii) We have $u^*u = p$ and $uu^* = q$.

On account of prop. 4 (i), assertion (i) is a consequence of (ii).

We now prove (ii). Since $P$ contains the image of $u^*$, the mapping $u^*u$ maps $E$ into $P$. Let $x \in E$ and $y \in P$, then

If $x$ belongs to $P$, then $\langle u(x)|u(y) \rangle = \langle x|y \rangle$ by the definition of a partially isometric mapping; if $x$ belongs to the kernel $N$ of $u$, then $u(x) = 0$, hence $\langle u(x)|u(y) \rangle = 0$ and $\langle x|y \rangle = 0$ since $N$ and $P$ are orthogonal. Since $E = P \oplus N$, we have $\langle u^*u(x) - x|y \rangle = 0$ in all the cases, and so $u^*u$ is the orthoprojector $p$ from $E$ onto $P$. That $uu^* = q$ follows by interchanging $u$ and $u^*$ in the above.

(i) $u$ is partially isometric ; (ii) $u^*$ is partially isometric ; (iii) $u^*u$ is an orthoprojector ; (iv) $uu^*$ is an orthoprojector ; (v) *$uu^*u = u ;*$ (vi) *$u^*uu^* = u^* .*$

By prop. 5, (i) is equivalent to (ii).

(i) $\Rightarrow$ (v) : Suppose $u$ is partially isometric. Then $u^*u$ is the initial orthoprojector of $u$ by prop. 5. Hence for every $x \in E$, $u^*u(x) - x$ belongs to the kernel of $u$, that is, $uu^*u(x) = u(x)$.

(v) $\Rightarrow$ (iii) : Suppose that $uu^*u = u$ and let $p = u^*u$; then $p = p^*$ and $p^2 = p$. Let $M$ (resp. $N$) be the image (resp. the kernel) of $p$. For $x \in M$ and $y \in N$, we have $\langle x|y \rangle = \langle p(x)|y \rangle = \langle x|p^*(y) \rangle = \langle x|p(y) \rangle = 0$. Since $M$ and $N$ are orthogonal, $p$ is the orthoprojector from $E$ onto $M$.

(iii) $\Rightarrow$ (i) : Suppose $p = u^*u$ is an orthoprojector with image $M$ and kernel $N$.

For all $x \in E$, we have

Hence $u(x) = 0$ for $x \in N$ and $\|u(x)\| = \|x\|$ for $x \in M$, and so $u$ is partially isometric with kernel $N$ and initial subspace $M$.

We have proved the equivalence of (i), (iii) and (v). Replacing $u$ by $u^*$, we can deduce the equivalence of (ii), (iv) and (vi). This proves prop. 6.

We define a hermitian form $\Phi$ on $E$ by

For $u$ to be normal, it is necessary and sufficient that $\Phi = 0$. By the polarization formulas (V, p. 2), this is equivalent to $\Phi(x, x) = 0$ for all $x \in E$. The proposition now follows since

Prop. 7 shows that $u$ and $u^*$ have the same kernel $N$. By prop. 4, (ii) of V, p. 41, the subspace $M$ of $E$ is stable under $u$ and $u^*$ since this is so for $N = M^\circ$, since $M \cap N = \{0\}$, the endomorphisms of $M$ induced by $u$ and $u^*$ are injective. Let $v = u^m(u^*)^n$; the preceding argument shows that the restriction of $v$ to $M$ (resp. $N$) is injective (resp. null), hence $N$ is the kernel of $v$. Consequently, $M = N^\circ$ is the initial subspace of $v$. By prop. 4, (i) of V, p. 41, the final subspace of $v$ is equal to the initial subspace of $v^*$. But $v^* = u^n(u^*)^m$ and so the initial subspace of $v^*$ is equal to $M$ by the preceding.

c) the primary subspace of u relative to $\lambda$ (in other words, by LIE, VII, § 1, No. 1, the set of all vectors $x$ of E for which there exists an integer $n \geqslant 0$ such that $(u - \lambda.1_E)^n(x) = 0$); d) the primary subspace of $u^*$ relative to $\overline{\lambda}$.

It is clear that $w = u - \lambda.1_E$ is a normal endomorphism of E, hence the endomorphisms $w, w^* = u^* - \overline{\lambda}.1_E, w^n$ and $(w^*)^n$ of E have the same kernel by prop. 8.

Put $\Phi = \Phi_u$ and $c = \sup_{\|x\| \leqslant 1} |\Phi(x, x)|$, then evidently $c \leqslant \|u\|$. Let $x, y$ be in E such that $\|x\| \leqslant 1, \|y\| \leqslant 1$. Then

hence but $|\Phi(t, t)| \leqslant c\|t\|^2$ for all $t \in E$, thus Let $a = \Phi(x, y)$; there exists a complex number $\lambda$ with absolute value 1 such that

λa = |a|. Replacing y by λy in the preceding inequality, we get $|\Phi(x, y)| \leq c$. By (15), $\|u\| \leq c$ and the proposition follows. Q.E.D.

Evidently every hermitian endomorphism is normal. Conversely :

For, the relation « $u = h_1 + ih_2,\ h_1^* = h_1,\ h_2^* = h_2$ » is equivalent to

In addition, we have $h_1 h_2 - h_2 h_1 = \frac{i}{2}(uu^* - u^*u)$. This proves prop. 10.

Suppose $p^2 = p$. Let M be the image of p and N its kernel. E is the topological direct sum of M and N. In order that p is an orthoprojector, it is necessary and sufficient that M is orthogonal to N, that is to say that we have $\langle p(x)|y - p(y) \rangle = 0$ for all x, y in E. This latter relation is equivalent to $p = p^*p$, and implies that $p^* = (p^*p)^* = p^*p = p$; conversely if $p^* = p$, we have $p = p^2 = p^*p$.

For, we have $(vuv^*)^* = v^{**}u^*v^* = vuv^*$. On the other hand, if $u \geq 0$, we have

for all $y \in F$, hence $vuv^* \geq 0$.

Prop. 12 shows, in particular, that $vv^*$ is positive for all $v \in \mathscr{L}(E; F)$. Since, in particular, an orthoprojector $p$ satisfies $p = p^2 = pp^*$, it is positive.

(i) The set $\mathscr{F}$ has an upper bound $u_0$ in $\mathscr{H}(E)$; we have

(ii) The filter of sections of $\mathscr{F}$ converges to $u_0$ in the space $\mathscr{L}(E)$ endowed with the topology of simple convergence.

Let $\Sigma$ be the filter of sections of $\mathscr{F}$; for every $u \in \mathscr{H}(E)$, let $\Phi_u$ be the continuous hermitian form on $E$ defined by

Let for $u \in \mathscr{H}(E)$ and $x \in E$. By the polarization formulas (V, p. 2), we have

For every $x \in E$, the mapping $u \mapsto \Psi_u(x)$ from into $\mathbf{R}$ is increasing and bounded, hence has a limit with respect to $\Sigma$. By the preceding formulas, the limit

exists for every pair $(x, y)$ of elements of $E$. It is clear that $\Phi$ is a hermitian form on $E$. If $v_1 \in \mathscr{F}$ and $v_2$ is a bound of $\mathscr{F}$, the hermitian forms $f_1 = \Phi - \Phi_{v_1}$ and $f_2 = \Phi_{v_2} - \Phi$ are positive; there exists a real number $M \geq 0$ such that hence consequently the semi-norms $x \mapsto f_i(x, x)^{1/2}$ are continuous on $E$. Since we conclude that $x \mapsto \Phi(x, x)$ is a continuous function on $E$, and by formulas (19) and (20), that $\Phi$ is continuous on $E \times E$. Therefore there exists (V, p. 16, cor. 2) an element $u_0$ of $\mathscr{H}(E)$ such that $\Phi = \Phi_{u_0}$. Formula (18) is evidently satisfied, hence $u_0$ is the upper bound of $\mathscr{F}$ in $\mathscr{H}(E)$. This proves (i).

We have, by construction

Let $v_1 \in \mathscr{F}$; given a $u \in \mathscr{F}$ such that $u \geq v_1$, let $v = u_0 - u$. If we apply the Cauchy-Schwarz inequality to the positive hermitian form $\Phi_v$ on $E$, we get since $\|v\| \leq \|u_0 - v_1\|$ by V, p. 44, prop. 9. Then by (21) we get $\lim_{u,\Sigma} \|(u_0 - u)(x)\| = 0$ for all $x \in E$; which proves assertion (ii).

In particular, prop. 13 can be applied to the case of an increasing and bounded sequence $(u_n)_{n \in \mathbf{N}}$ of elements of $\mathscr{H}(E)$. Then there exists an element $v$ of $\mathscr{H}(E)$ characterized by

and we have $v(x) = \lim_{n \to \infty} u_n(x)$ for all $x \in E$. Moreover, $v$ is the upper bound of the set of the $u_n$ in $\mathscr{H}(E)$.