$L^p$ norm inequality between random variable and conditional expectation (hypercontractivity)

Question

Suppose we have two random variable $X$ and $Y$ and $1\leq q\leq p$ such that for any $Y$-measurable set $A$, \begin{align*} &\| \mathbb E[1_A(Y)|X] \|_p \leq \| 1_A(Y) \|_q\\ \Leftrightarrow&\mathbb E[\mathbb P(A|X)^p]^{\frac{q}{p}}\leq \mathbb P(A) \end{align*} Where $\|W\|_a = \left( \mathbb E[|W|^p]\right)^{\frac{1}{p}}$ is the $a$-norm of $W$. I want to prove that for two $Y$-measurable sets $A$ and $B$ we have \begin{align*} \| \mathbb E[1_A(Y)+1_B(Y)|X] \|_p \leq \| 1_A(Y)+1_B(Y) \|_q \end{align*} I went as far as proving that this is equivalent to (by writting $A\cup B = (A\setminus B) \cup (B\setminus A) \cup (A\cap B)$) \begin{align*} \mathbb E[(\mathbb P(A\cup B|X)+\mathbb P(A\cap B|X))^p]^{\frac{q}{p}}\leq \mathbb P(A\cup B)+(2^q-1)\mathbb P(A\cap B) \end{align*} And it feels doable to justify that \begin{align*} \mathbb E[(\mathbb P(A\cup B|X)+\mathbb P(A\cap B|X))^p]^{\frac{q}{p}} \leq \mathbb E[\mathbb P(A\cup B|X)^p]^{\frac{q}{p}}+(2^q-1) \mathbb E[\mathbb P(A\cap B|X)^p]^{\frac{q}{p}} \end{align*} which would finish the proof. I don't know how to finish the proof from there, mostly because $q/p \leq 1$ hence $x^{\frac{q}{p}}$ is concave. Maybe writing instead the equation as the following would help \begin{align*} \mathbb E[(\mathbb P(A\setminus B \cup B\setminus A|X)+2\mathbb P(A\cap B|X))^p]^{\frac{q}{p}} \leq \mathbb E[\mathbb P(A\setminus B \cup B\setminus A|X)^p]^{\frac{q}{p}}+2^q \mathbb E[\mathbb P(A\cap B|X)^p]^{\frac{q}{p}} \end{align*}

Answer 1

Here is an attempt of solution, any comment is welcome.

If we take the last equation and write $x=\frac{q}{p}$ and $q=px$, $C=A\setminus B\cup \setminus A$ and $D=a\cap B$, then we get \begin{align*} \mathbb E[(\mathbb P(C|X)+2\mathbb P(D|X))^p]^{x} \leq \mathbb E[\mathbb P(C|X)^p]^{x}+ \mathbb E[(2\mathbb P(D|X))^p]^{x} \end{align*} We can take the $\log$ and subtract the RHS to the LHS to get \begin{align*} f(x) = x \log \mathbb E[(\mathbb P(C|X)+2\mathbb P(D|X))^p] - \log\left( \mathbb E[\mathbb P(C|X)^p]^{x}+ \mathbb E[(2\mathbb P(D|X))^p]^{x} \right)\leq 0 \end{align*} Now this is a increasing function on $[0,1]$, indeed it's derivative is \begin{align*} &\log \mathbb E[(\mathbb P(C|X)+2\mathbb P(D|X))^p] - \frac{\log(\mathbb E[\mathbb P(C|X)^p])\mathbb E[\mathbb P(C|X)^p]^{x}+ \log(\mathbb E[(2\mathbb P(D|X))^p])\mathbb E[(2\mathbb P(D|X))^p]^{x}}{\mathbb E[\mathbb P(C|X)^p]^{x}+ \mathbb E[(2\mathbb P(D|X))^p]^{x}}>0\\ &\Leftrightarrow \left(\frac{\mathbb E[(\mathbb P(C|X))^p]}{\mathbb E[(2\mathbb P(D|X))^p]}\right)^x > -\frac{\log \mathbb E[(\mathbb P(C|X)+2\mathbb P(D|X))^p]-\log\mathbb E[(2\mathbb P(D|X))^p]}{\log \mathbb E[(\mathbb P(C|X)+2\mathbb P(D|X))^p]-\log \mathbb E[(\mathbb P(C|X))^p]} \end{align*} The LHS is positive, and the RHS is negative since $p>1$ and so $(a+b)^p\geq a^p$ hence both the numerator and denominator are positives. So $f$ is increasing so it is sufficient to prove that $f(q/p)\leq f(1)\leq 0$. Now $f(1)\leq 0$ is equivalent to \begin{align*} \mathbb E[(\mathbb P(C|X)+2\mathbb P(D|X))^p] \leq \mathbb E[\mathbb P(C|X)^p]+ \mathbb E[(2\mathbb P(D|X))^p] \end{align*} and is true since for $a$ and $b$ positive $(a+b)^p\leq a^p+b^p$ (I haven't proved that one but I think it should be doable).

Does this answer seem correct ? Can this be simplified ? It seems a bit too heavy for a statement that may be easy to prove using some known inequalities.