Let $\ell^2(\mathbb{N}^*)$ be the Hilbert space with the inner product $$\langle x\mid y\rangle_2:=\sum_{i=1}^{+\infty}x_i\overline{y_i},\;\,\forall\,x, y \in \ell^2(\mathbb{N}^*).$$
Consider the following operator on $\ell^2(\mathbb{N}^*)$ $$S=\begin{bmatrix}1\\&\dfrac1{2!}\\&&\dfrac1{3!}\\&&&\ddots\end{bmatrix}.$$ I want to construct an operator $A$ on $\ell^2(\mathbb{N}^*)$ which is a non scalar mutiple of the identity such that: for some positive constant $M$ we have $$\tag1 \langle SAx\mid x\rangle_2\leq M\langle Sx\mid x\rangle_2,\;\forall\,x\in \ell^2(\mathbb{N}^*).$$
I try with \begin{align*} A\colon \ell^2(\mathbb{N}^*) & \rightarrow \ell^2(\mathbb{N}^*) \\ (x_1,x_2,\cdots)&\mapsto (x_2,x_3\cdots). \end{align*}
However I find $$\langle SAe_n\mid e_n\rangle_2=\frac{1}{(n-1)!}=n!\langle Se_n\mid e_n\rangle_2,\;\forall\,n\in \mathbb{N}^*,$$ where $e_n$ is the canonical basis of $\ell^2(\mathbb{N}^*)$. So it is impossible to find a positive constant $M$ satisfying $(1)$.
Consider $A$ defined by $Ax = x_n e_n$ for some fixed $n$. Then $SAx = \frac{x_n}{n!} e_n$. Hence $\langle SAx, x \rangle = \frac{|x_n|^2}{n!}$. Now $\langle Sx, x \rangle = \sum_j \frac{|x_j|^2}{j!} \geq \frac{|x_n|^2}{n!}$ which gives the desired construction.