I am currently reading a paper that claims the following fact:
Let $X$, $Y$ be $L^2$ random variables on some probability space. The $L^2$ norm is denoted by $\| \cdot \|_{2}$. Then there exists $C>0$ such that \begin{eqnarray} && \mathbb{E} \bigg[ \mathbf{1}_{\{ |Y| \geq \|Y\|^{1/2}_2 \}} (1+ |X| + |Y|)|Y| \bigg] \\ & \leq & C \|Y\|_2 \bigg( \|Y\|_2 + \sqrt{\|Y\|_2} + \sup_{A: \mathbb{P}(A) \leq \|Y\|_2} \mathbb{E}[|X|^2 \mathbf{1}_A]^{1/2} \bigg). \end{eqnarray}
It seems to involve the Cauchy-Schwarz inequality, applying to the terms $ \mathbf{1}_{\{|Y| \geq \|Y\|^{1/2}_2\}}(1+|X|+|Y|)$ and $|Y|$. But I don't see how the $L^2$ norm of the first term can be bounded.
We can estimate each term separately. Observe the following: $$ \Bbb E\left[1_{\{ |Y| \geq \|Y\|^{1/2}_2 \}}|Y|\right]\le \Bbb E\left[1_{\{ |Y| \geq \|Y\|^{1/2}_2 \}}\frac{|Y|^2}{\|Y\|_2^{1/2}}\right]\le\|Y\|_2^{3/2}, $$ $$ \Bbb E\left[1_{\{ |Y| \geq \|Y\|^{1/2}_2 \}}|Y|^2\right]\le \|Y\|_2^2, $$
$$ \Bbb P(|Y| \geq \|Y\|^{1/2}_2)=\Bbb P(|Y|^2 \geq \|Y\|_2)\le \frac{\Bbb E|Y|^2}{\|Y\|_2}=\|Y\|_2, $$ and $$\begin{eqnarray} \Bbb E\left[1_{\{ |Y| \geq \|Y\|^{1/2}_2 \}}|X||Y|\right]^2&\le& \Bbb E\left[1_{\{ |Y| \geq \|Y\|^{1/2}_2 \}}|X|^2\right]\Bbb E\left[1_{\{ |Y| \geq \|Y\|^{1/2}_2 \}}|Y|^2\right]\\ &\le&\sup_{A:\Bbb P(A)\le \|Y\|_2}\Bbb E[X^2 1_A]\cdot \|Y\|^2_2 \end{eqnarray}$$ by Cauchy-Schwarz. This gives $$ \begin{eqnarray} \mathbb{E} \left[ 1_{\{ |Y| \geq \|Y\|^{1/2}_2 \}} (1+ |X| + |Y|)|Y| \right] \le \|Y\|_2\left(\|Y\|_2+\|Y\|_2^{1/2}+\left(\sup_{A:\Bbb P(A)\le \|Y\|_2}\Bbb E[X^2 1_A]\right)^{1/2}\right) \end{eqnarray} $$ as desired.