Consider two real-valued random variables $X,Y$ on a probability space $(\Omega,\mathcal{A}, \mathbb{P})$, where $Y$ is standard normal distributed and let $q >2$. Further let $\Vert X\Vert_q := (\mathbb{E}(\vert X \vert^q))^{1/q}$ denote the $L^q$ norm. What I'm looking for is an estimation of the kind $$\tag{$*$}\Vert (XY)^q \Vert_1 \le c_q \Vert X^q \Vert_1.$$
If both were independent, then $$\Vert (XY)^q \Vert_1 = \Vert Y^q \Vert_1 \Vert X^q \Vert_1$$ would hold and $c_q = \Vert Y^q \Vert_1 < \infty$. If they are not independent then my idea was to apply Hoelder's inequality and obtain $$\Vert (XY)^q \Vert_1 \le \Vert X^q \Vert_1 \Vert Y^q \Vert_\infty,$$ but $\Vert Y^q \Vert_\infty < \infty$ does not hold. I tried different variations of Hoelder and tried to use the fact that $\Vert Z\Vert_p \le \Vert Z \Vert_q$ whenever $p \le q$, but nothing worked.
I assume that something like $(*)$ is not possible without the independence $Y$ and $X$, has someone a counterexample in mind?
This is indeed not possible without the independence of $X$ and $Y$. Let $p \in (\frac{1}{q},1+\frac 1q)$ and $$X = \frac{e^{\frac{Y^2}{2q}}}{|Y|^p}1_{|Y| > 1}.$$ Then \begin{align*} \mathbb{E}[|X|^q] &= \frac{2}{\sqrt{2 \pi}} \int_1^\infty \frac{e^{\frac{y^2}{2}}}{|y|^{pq}} e^{-y^2/2}dy \\ &= \frac{2}{\sqrt{2 \pi}} \int_1^\infty \frac{1}{|y|^{pq}}dy < \infty \end{align*} because $pq > 1$, but \begin{align*} \mathbb{E}[|YX|^q] &= \frac{2}{\sqrt{2 \pi}} \int_1^\infty |y|^q\frac{e^{\frac{y^2}{2}}}{|y|^{pq}} e^{-y^2/2}dy \\ &= \frac{2}{\sqrt{2 \pi}} \int_1^\infty |y|^{q(1-p)}dy = \infty \end{align*} because $q (1-p) > -1$.