Bounding the Probability of the L^2-Norm

Question

Let $Y$ be an random Variable, and $f(x,Y)$ be a measurable function. I have for every $x\in[0,1]$

$$ \mathbb{P}\left(|f(x,Y)|\geq \varepsilon\right)\leq \delta $$ Is there any chance to bound $\|f(.,Y)\|_{L^2_{[0,1]}}$ in probability by $C\varepsilon$ or something?

Answer 1

Suppose $m \le |f(\cdot, \cdot)| \le M$. By the reverse Cauchy-Schwarz inequality, \begin{align*} \|f(\cdot, Y)\|_{L^2[0, 1]} \le C\|f(\cdot, Y)\|_{L^1[0, 1]} \end{align*} where $C = \frac{1}{4}\frac{(m+M)^2}{mM}$. Therefore, \begin{align*} \mathbb{P}(\|f(\cdot, Y)\|_{L^2[0, 1]} \ge C\epsilon) \le \mathbb{P}(C\|f(\cdot, Y)\|_{L^1[0, 1]} \ge C\epsilon) \le \delta \end{align*}