A inequality about pointwise absolute value vectors

Question

Let $\Gamma$ be a discrete group and $\xi\in l^{2}(\Gamma)$ be a unit vector. If $|\xi|$ be the pointwise absolute value of $\xi$, then how to verify: ($S$ is a linear bounded operator on $l^{2}(\Gamma)$) $$|\langle S\xi, \xi\rangle|\leq\langle S|\xi|, |\xi|\rangle?$$ I suppose the pointwise absolute value of $\xi$ means: if $\xi=\{(x_{i})_{i\in \Gamma}: \sum\limits_{i\in \Gamma}|x_{i}|^{2}<\infty\}$, then $|\xi|=\{(|x_{i}|)_{i\in \Gamma}: \sum\limits_{i\in \Gamma}|x_{i}|^{2}<\infty\}$.

Answer 1

That inequality is not true. Let $\Gamma=\mathbb Z_2$, so $\ell^2(\Gamma)=\mathbb C^2$. Take $$ \xi=\begin{bmatrix}1\\-1\end{bmatrix},\ \ S=\begin{bmatrix}1&-1\\0&0\end{bmatrix}. $$ Then $S|\xi|=0$ and $\langle S|\xi|,|\xi|\rangle=0$, while $$ \langle S\xi,\xi\rangle=\left\langle\begin{bmatrix}2\\0\end{bmatrix},\begin{bmatrix}1\\-1\end{bmatrix}\right\rangle=2 $$

Edit: even when $S$ is positive the inequality fails. You have $$ \langle S\xi,\xi\rangle=\left\langle\begin{bmatrix}1&-1\\-1&1\end{bmatrix}\,\begin{bmatrix}1\\-1\end{bmatrix},\begin{bmatrix}1\\-1\end{bmatrix}\right\rangle=\left\langle\begin{bmatrix}2\\-2\end{bmatrix},\begin{bmatrix}1\\-1\end{bmatrix}\right\rangle=4, $$ $$ \langle S|\xi|,|\xi|\rangle=\left\langle\begin{bmatrix}1&-1\\-1&1\end{bmatrix}\,\begin{bmatrix}1\\1\end{bmatrix},\begin{bmatrix}1\\1\end{bmatrix}\right\rangle=\left\langle\begin{bmatrix}0\\0\end{bmatrix},\begin{bmatrix}1\\-1\end{bmatrix}\right\rangle=0. $$