Conditional cauchy schwarz inequality

By [F1], XY is integrable. Fix finite versions $a=E[X^2\mid \mathcal{G}]$, $b=E[XY\mid \mathcal{G}]$, and $c=E[Y^2\mid \mathcal{G}]$. Positivity gives $a,c\ge 0$ almost surely. For every rational t, $(X+tY{)}^2$ is integrable and nonnegative, and linearity and positivity give $a+2tb+t^{2}c\ge 0$ almost surely. By [F5] one null union removes every rational-parameter exception.

At a remaining point, the polynomial $q(t)=a+2tb+t^{2}c$ is continuous: $q(t)-q(s)=(t-s)(2b+c(t+s))$, which tends to zero as $t\to s$. If q were negative at any real s, it would stay negative on an interval around s, containing a rational by [F4], contrary to step 1.1. Thus q is nonnegative for all real t.

If $c=0$ and $b\mathrm{\ne }0$, the choice $t=-(a+1)\mathrm{/}(2b)$ gives $q(t)=-1$, impossible; hence $b=0$ and $b^2\le ac$. If $c>0$, put $t=-b\mathrm{/}c$ to get $0\le a-b^2\mathrm{/}c$, so again $b^2\le ac$. The cases cover every remaining point and prove the conditional inequality.