Lemma 5

Let $E_1$ be a closed vector subspace of E, stable under $u^*u$. If $E_1 \neq \{0\}$, then there exists a vector of norm 1 in $E_1 \cap P$.

Since B is weakly compact, so is the weakly closed subspace $B \cap E_1$ of B. Hence there exists (GT, IV, § 6, No. 1, th. 1) a point $x_0$ in $B \cap E_1$ such that $Q(x_0) \geq Q(x)$ for all $x \in B \cap E_1$. If $Q(x_0) = 0$, we have $Q(x) = 0$ and so $u(x) = 0$ for all $x \in B \cap E_1$. Thus $E_1 \subset P$ and lemma 5 follows in this case.

Suppose now that $Q(x_0) > 0$, then $x_0 \neq 0$. Since the vector $\|x_0\|^{-1} \cdot x_0$ belongs to $B \cap E_1$, we have

i.e. $\|x_0\| = 1$. We shall prove that $x_0$ belongs to P; let $y \in E$ be orthogonal to $x_0$. It is enough to prove that $u(y)$ is orthogonal to $u(x_0)$. But since $y$ is the sum of a vector of $E_1$ and a vector orthogonal to $E_1$, and both orthogonal to $x_0$ (since $x_0 \in E_1$), it is enough to consider the following two cases :

a) $y$ is orthogonal to $E_1$: since $E_1$ is stable under $u^*u$, $u^*u(x_0) \in E_1$, hence $0 = \langle y|u^*u(x_0)\rangle = \langle u(y)|u(x_0)\rangle$.

b) $y$ belongs to $E_1$: for all $t \in \mathbf{R}$, the vector $x(t) = (x_0 + ty)/\|x_0 + ty\|$ belongs to $B \cap E_1$. We have $Q(xt) = f(t)/g(t)$ with

In view of the definition of $x_0$, we have $Q(x(0)) \geq Q(x(t))$ for all real $t$, hence $\frac{d}{dt} Q(x(t))$ is zero for $t = 0$. But $f(0) = \|u(x_0)\|^2$, $g(0) = 1$, $f'(0) = 2 \Re \langle u(x_0)|u(y)\rangle$, $g'(0) = 0$. Since we conclude that $f'(0) = 0$, that is, $\Re \langle u(x_0)|u(y)\rangle = 0$. When $K = \mathbf{R}$, $u(x_0)$ is orthogonal to $u(y)$, when $K = \mathbf{C}$, the vector $iy$ belongs to $E_1$ and is orthogonal to $x_0$, hence $\mathscr{I} \langle u(x_0)|u(y)\rangle = -\Re \langle u(x_0)|u(iy)\rangle = 0$, and finally $u(x_0)$ is orthogonal to $u(y)$. This proves lemma 5.

Now we prove th. 2. Applying th. 1 of S, III, § 4, No. 5 we see, as in V, p. 23, that there exists a set $S$ which is maximal among the orthonormal subsets of $E$ contained in $P$. Let $E_1$ be the set of all vectors orthogonal to $S$. Let $y \in E_1$; if $x \in S$, the vectors $x$ and $y$ are orthogonal, and since $S \subset P$, we conclude that $u(x)$ and $u(y)$ are orthogonal; then

and $u^*u(y)$ is orthogonal to $S$. Hence $E_1$ is stable under $u^*u$. If we had $E_1 \neq \{0\}$, there would exist a vector $x$ of norm 1 in $E_1 \cap P$ (lemma 5) and $S \cup \{x\}$ would be an orthonormal subset of $E$ contained in $P$. This would contradict the maximal character of $S$. Hence $E_1 = \{0\}$ and $S$ is an orthonormal basis of $E$. Q.E.D.