Exercise 10, § 4

Let E be a hilbertian space, u a positive endomorphism of E. a) Show that for all $x \in E$, we have

(observe that $\langle u(x)|u(x) \rangle^2 \leq \langle u(x)|x \rangle \langle u^2(x)|u(x) \rangle$, by V, p. 3, prop. 2). b) Let M be a closed vector subspace of E, $M^\circ$ its orthogonal complement. Let $x \in M$, and let $f(x)$ be the lower bound of $\langle u(x+y)|x+y \rangle$ as $y$ ranges over $M^\circ$. For every $\varepsilon > 0$, let $E(x, \varepsilon)$ be the set of all $y \in M^\circ$ such that $\langle u(x+y)|x+y \rangle \leq f(x) + \varepsilon$. Show that $E(x, \varepsilon)$ is convex and that for all $z \in M^\circ$, we have (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 $g(0) \leq f(x) + \varepsilon$. c) For every integer $n \geq 1$, let $y_n \in E(x, 1/n)$; show that the sequence $(u(x+y_n))$ tends to a limit $x_1'$ belonging to M, and that the sequence $(\langle u(y_n)|y_n \rangle)$ is bounded (find a bound for 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 $z \in M^\circ$ using inequality (*)). d) Let $(y'_n)$ be a sequence of points of $M^\circ$ such that $\langle u(y'_n-y'_m)|y'_n-y'_m \rangle$ is arbitrarily small when $m$ and $n$ are large enough, and such that the sequence $(u(x+y'_n))$ has a limit $x_1' \in M$;

show that $x'_1 = x_1$. (Let $Q(z) = \langle u(z)|z \rangle$; first show that the number $Q((y_p - y'_p) - (y_q - y'_q))$ is arbitrarily small as soon as $p$ and $q$ are large enough, 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 all $p$, show 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 we put $u_1(x) = x_1$, then $u_1$ is a linear mapping from $M$ into itself. Show that $0 \leq \langle u_1(x)|x \rangle \leq \langle u(x)|x \rangle$ for all $x \in M$ and consequently that $u_1$ is continuous and is an endomorphism of $M$ which is $\geq 0$. (Observe that $\langle u(x + y_n)|y_n \rangle$ tends to 0 and $\langle u(x + y_n)|x + y_n \rangle$ tends to $f(x)$.) f) Let $p_M$ be the orthoprojector with image $M$, and let $u_0 = u_1 \circ p_M$. We have $0 \leq u_0 \leq u$, and $M$ is stable under $u_0$, and the restriction of $u_0$ to $M^\circ$ is null. Show that $u_0$ is the largest element in the family of all endomorphisms $v \geq 0$ such that $v \leq u$, that $M$ is stable under $v$ and is such that the restriction of $v$ to $M^\circ$ is null.