Let $X$ and $Y$ be two random variable. Let $\| W \|_p = \mathbb E[|W|^p]^{\frac{1}{p}}$. Let $f$ and $g$ be two positive $Y$-measurable functions such that $f\cdot g = 0$, $Y$-almost surely.
Suppose that for some $1\leq q\leq p$ we have that both $\| \mathbb E[f(Y)|X] \|_p\leq \| f(Y) \|_q$ and $\| \mathbb E[g(Y)|X] \|_p\leq \| g(Y) \|_q$. Prove that $\| \mathbb E[f(Y)+g(Y)|X] \|_p\leq \| f(Y)+g(Y) \|_q$.
I can't prove this statement or disprove it. The closest I got to proving it was by making a mistake but here is the attempt proof :
Since $f\cdot g = 0$, then \begin{align} \| f(Y)+g(Y) \|_q &= \left( \mathbb E[f(Y)^q] + \mathbb E[g(Y)^q] \right)^\frac{1}{q}\\ &=\left(\| f(Y) \|_q^q + \| g(Y) \|_q^q \right)^{\frac{1}{q}}\\ &\geq \left(\| \mathbb E[f(Y)|X] \|_p^q + \| \mathbb E[g(Y)|X] \|_p^q \right)^{\frac{1}{q}} \end{align} Now this is the $q$-norm of the vector $\left(\| \mathbb E[f(Y)|X] \|_p, \| \mathbb E[g(Y)|X] \|_p \right)$ and since $p\geq q$, it is lower bounded by its $p$-norm. So \begin{align*} \| f(Y)+g(Y) \|_q &\geq \left(\| \mathbb E[f(Y)|X] \|_p^p + \| \mathbb E[g(Y)|X] \|_p^p \right)^{\frac{1}{p}}\\ &= \left( \mathbb E[\mathbb E[f(Y)|X]^p + \mathbb E[g(Y)|X]^p]\right)^{\frac{1}{p}} \end{align*} And then concluded by saying that for $a$ and $b$ positive and $p\geq 1$, $(a+b)^p \leq a^p+b^p$ but the inequality is in the other direction.
I think that the second inequality is messing things up, the following example shows it :
Let $X$ and $Y$ be two independent Bernouli randomvariable with parameter $1/2$, and $f(y) = y$ and $g(y)=1-y$. Then $f\cdot g=0$ and in order to satisfy the requirement we need to have (after rewriting $f$ and $g$ as indicator and expectation of indicator is probability and conditional expectation is expectation) $1/2\geq (1/2)^q$ for any $p$, hence $q=1$ and any $p\geq 1$ is fine. Now $f+g=1$ and so the desired inequality holds with equality (both sides are trivially one), But the part after the first inequality is $(1/2^q+1/2^q)^{1/q} =1$ since $q=1$. Now when we use the second inequality we get $(1/2^p+1/2^p)^{1/p} =2^{\frac{1-p}{p}}<1$ for $p>1$.
Does any one have an idea on how to fix that proof ? In other words can someone prove that
\begin{align*} \| \mathbb E[f(Y)|X] \|_p^q + \| \mathbb E[g(Y)|X] \|_p^q \geq \| \mathbb E[f(Y)+g(Y)|X] \|_p^q \end{align*} Observe that $f\cdot g$ is still necessary, otherwise we can use $f=g$ to get that $2\| \mathbb E[f(Y)|X] \|_p^q \geq \| \mathbb E[2f(Y)|X] \|_p^q$ hence $2\geq 2^q$ which is false.