# Norm defined by a conditional expectation

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 $$1. If $$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.

## 1 Answer

1. 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.

2. Then we deduce that almost surely,

$$E\left(\left(f+g\right)^2\right)\leqslant \left( \sqrt{E\left(f^2\right)}+\sqrt{E\left(g^2\right)}\right)^2.$$