Exercise 10, § 4

Let $E$ be a Hilbert space, and let $u$ be a positive endomorphism of E. a) Now prove that for every $x\in E$, one has

(observe that $\langle u(x)|u(x)\rangle^2\leq\langle u(x)|x\rangle\langle u^2(x)|u(x)\rangle$ by virtue of V, p.3, prop. 2). b) Let M be a closed vector subspace of E, and let $M^\circ$ be its orthogonal supplement. Let $x\in M$, and let $f(x)$ be the greatest lower bound of $\langle u(x+y)|x+y\rangle$ when $y$ runs through $M^\circ$. For every $\varepsilon>0$, let $E(x,\varepsilon)$ be the set of the $y\in M^\circ$ such that $\langle u(x+y)|x+y\rangle\leq f(x)+\varepsilon$. Now prove that $E(x,\varepsilon)$ is convex and that for every $z\in M^\circ$, one has, for $y\in E(x,\varepsilon)$, ()

$$\langle u(x+y)|z\rangle^2\leq\varepsilon\langle u(z)|z\rangle$$

(consider the function $g:t\mapsto\langle u(x+y+tz)|x+y+tz\rangle$ of the real variable $t$, which attains its minimum at a point $t_0$ and note that $g(t_0)\geq f(x)$ and $g(0)\leq f(x)+\dot{\varepsilon}$. c) For every integer $n\geq 1$, let $y_n\in E(x,1/n)$; now prove that the sequence $(u(x+y_n))$ tends to a limit $x_1$ which belongs to M, and that the sequence $(\langle u(y_n)|y_n\rangle)$ is bounded (majorize the numbers $\langle u(y_n-y_m)|y_n-y_m\rangle$ for $m\geq n$ and $|\langle u(x+y_n)|z\rangle|$ for $z\in M^\circ$ by means of the inequality ()). d) Let $(y'_n)$ be a sequence of points of $M^0$ such that $\langle u(y'_n-y'_m)|y'_n-y'_m\rangle$ is arbitrarily small as soon as $m$ and $n$ are sufficiently large, and such that the sequence $(u(x+y'_n))$ has a limit $x'_1\in M$; now prove that one has $x'_1=x_1$. (Putting, for brevity, $Q(z)=\langle u(z)|z\rangle$, prove first that the number $Q((y_p-y'_p)-(y_q-y'_q))$ is arbitrarily small as soon as $p$ and $q$ are sufficiently large, and deduce that the sequence $(Q(y_n-y'_n))$ is bounded; using the fact that $\langle x'_1-x_1|y_p-y'_p\rangle=0$ for every $p$, prove that the sequence $(Q(y_n-y'_n))$ tends to 0 and use a.).)

e) Deduce from d) that the point $x_1$ does not depend on the choice of the $y_n \in E(x, 1/n)$, and that if one sets $u_1(x) = x_1$, $u_1$ is a linear mapping of $M$ into itself. Now prove that one has $0 \leq \langle u_1(x)|x \rangle \leq \langle u(x)|x \rangle$ for all $x \in M$, and consequently that $u_1$ is continuous and is an endomorphism $\geq 0$ of $M$. (Observe that $\langle u(x + y_n)|y_n \rangle$ tends to 0 and $\langle u(x + y_n)|x + y_n \rangle$ to $f(x)$.) f) Let $p_M$ be the orthogonal projector with image $M$, and let $u_0 = u_1 \circ p_M$. One has $0 \leq u_0 \leq u$, $M$ is stable under $u_0$ and the restriction of $u_0$ to $M^\circ$ is null. Now prove that $u_0$ is the greatest element of the set of endomorphisms $v \geq 0$ such that $v \leq u$, that $M$ is stable under $v$ and that the restriction of $v$ to $M^\circ$ is null.