Bounded increasing positive nets have strong suprema

For $\xi \in H$, the real numbers $q_i(\xi ):=⟨x_i\xi ,\xi ⟩$ increase and satisfy $0\le q_i(\xi )\le C\mathrm{\parallel }\xi {\mathrm{\parallel }}^2$. Define $q(\xi ):={\sup }_{i\in I}{q}_i(\xi )$. For each $i$, the complex polarization identity gives $⟨x_i\xi ,\eta ⟩=\frac{1}{4}{\sum }_{k=0}^3{i}^k{q}_i(\xi +i^k\eta )\mathrm{.}$ Each term on the right converges as $i$ increases, so the left-hand side converges for every $\xi ,\eta$. Let $b(\xi ,\eta )$ be its limit. The forms $⟨x_i\xi ,\eta ⟩$ are sesquilinear and uniformly bounded by $C\mathrm{\parallel }\xi \mathrm{\parallel }\mathrm{\parallel }\eta \mathrm{\parallel }$, so their pointwise limit $b$ is sesquilinear with the same bound. For fixed $\xi$, Riesz representation applied to $\eta \mapsto \overline{b(\xi ,\eta )}$ gives a unique vector $x\xi$ with $b(\xi ,\eta )=⟨x\xi ,\eta ⟩$. Sesquilinearity and uniqueness make $\xi \mapsto x\xi$ linear, and the bound gives $\mathrm{\parallel }x\xi \mathrm{\parallel }\le C\mathrm{\parallel }\xi \mathrm{\parallel }$; thus $x\in \mathcal{B}(H)$ and $\mathrm{\parallel }x\mathrm{\parallel }\le C$. Since each $x_i$ is self-adjoint, $b(\xi ,\eta )=\overline{b(\eta ,\xi )}$, hence $⟨x\xi ,\eta ⟩=⟨\xi ,x\eta ⟩$ and $x=x^{\ast }$. Also $⟨x\xi ,\xi ⟩=q(\xi )\ge 0$, so $x$ is positive. By [F2], $x_i\to x$ weakly. Since $M$ is weakly closed by [F1], $x\in M$.

For every $\xi$, $q_i(\xi )\le q(\xi )=⟨x\xi ,\xi ⟩$, so $x_i\le x$. If $y\in M_{+}$ is any other upper bound, then $q_i(\xi )\le ⟨y\xi ,\xi ⟩$ for every $i,\xi$; taking the supremum over $i$ gives $⟨x\xi ,\xi ⟩\le ⟨y\xi ,\xi ⟩$ for every $\xi$, hence $x\le y$. Thus $x={\sup }_i{x}_i$ in $M_{+}$.

If $H=\{0\}$, strong convergence is immediate. Otherwise fix $\xi \in H$ and put $D_i=x-x_i$. Then $0\le D_i\le x\le CI$ and $⟨D_i\xi ,\xi ⟩=q(\xi )-q_i(\xi )\to 0$. Positivity gives the Cauchy–Schwarz inequality for the form $(\alpha ,\beta )\mapsto ⟨D_i\alpha ,\beta ⟩$: applying positivity to $\xi +z\eta$ for arbitrary $z\in \mathbb{C}$ yields the nonnegative quantity $u+2\mathrm{Re}(\bar{z}a)+\mathrm{\mid }z{\mathrm{\mid }}^{2}v,$ where $u=⟨D_i\xi ,\xi ⟩$, $a=⟨D_i\xi ,\eta ⟩$, and $v=⟨D_i\eta ,\eta ⟩$. If $v>0$, taking $z=-a\mathrm{/}v$ gives $\mathrm{\mid }a{\mathrm{\mid }}^2\le uv\mathrm{.}$ If $v=0$ and $a\mathrm{\ne }0$, taking $z=-ta$ for sufficiently large real $t>0$ makes the quantity negative; hence $a=0$ and the same inequality holds. Therefore $\mathrm{\mid }⟨D_i\xi ,\eta ⟩{\mathrm{\mid }}^2\le ⟨D_i\xi ,\xi ⟩⟨D_i\eta ,\eta ⟩\le C\mathrm{\parallel }\eta {\mathrm{\parallel }}^2⟨D_i\xi ,\xi ⟩\mathrm{.}$ Taking the supremum over unit $\eta$ gives $\mathrm{\parallel }{D}_i\xi {\mathrm{\parallel }}^2\le C⟨D_i\xi ,\xi ⟩\to 0$. Therefore $\mathrm{\parallel }(x_i-x)\xi \mathrm{\parallel }\to 0$ for every $\xi$, which is strong-operator convergence.