Let $\Omega$ be a probability space and $\mathbb{E} \colon L^\infty(\Omega) \to L^\infty(\Omega)$ be a conditional expectation such that $\mathbb{E}(|f|^2)$ implies $f=0$. Suppose $1<p<\infty$. If $f \in L^\infty(\Omega)$, we let $$ \|f\|=\big\|(\mathbb{E}(|f|^2)^{\frac{1}{2}}\big\|_{L^p(\Omega)}. $$ How show that $\|\cdot\|$ is a norm ? It suffices to prove the triangle inequality.
I am mainly interested by the case $1<p<2$.
It is possible to prove an equivalent of the Cauchy-Schwarz inequality, that is, for $f$ and $g$ non-negative and bounded, $$ E\left(fg\right)\leqslant \sqrt{E\left(f^2\right)}\sqrt{E\left(g^2\right)}\mbox{ a.s.}. $$ One uses when $f$ and $g$ are positive the inequality $ab\leqslant a^2/2+b^2/2$, monotonicity of $E$ and the fact that $E(f)$ and $E(g)$ are positive so that we can normalise and assume that $E(f)=1$. To be reduced to this case, work with $f_n:=f \mathbf 1_{f\geqslant 1/n}$ instead of $f$ (and the same for $g$) and use monotone convergence.
Then we deduce that almost surely,
$$ E\left(\left(f+g\right)^2\right)\leqslant \left( \sqrt{E\left(f^2\right)}+\sqrt{E\left(g^2\right)}\right)^2.$$