Let $E$ be a complex Hilbert space, $\mathcal{L}(E)$ be the algebra of all bounded linear operators on $E$.
For a positive semidefinite operator $S$, we consider the following two subalgebras of $\mathcal{L}(E)$: $$\mathcal{L}_1(E)=\left\{A\in \mathcal{L}(E);\;\exists \,c_1> 0;\;\|SAy\| \leq c_1\|Sy\|,\;\forall\,y\in E\right\},$$ and $$\mathcal{L}_2(E)=\left\{A\in \mathcal{L}(E);\;\exists \,c_2> 0;\;\|S^{1/2}Ay\| \leq c_2\|S^{1/2}y\|,\;\forall\,y\in E\right\}.$$ We have $\mathcal{L}_1(E)\subseteq \mathcal{L}_2(E)$ and the equality hold if $S$ has a closed range.
I aim to prove that $\mathcal{L}_1(E)\subsetneq\mathcal{L}_2(E)$.
I try with the following example: Consider $$ A= \left[\begin{matrix}0 & 1 \\ &0 & 1 \\ &&\ddots&\ddots\end{matrix}\right], $$ And $S=\mbox{diag}(0,\frac{1}{2},\frac{1}{3},\frac{1}{4},\cdots)$.
By the answer of @mechanodroid in this question, we deduce that $S$ is neither injective nor has a closed range.
Could you please help me to show that $A\in \mathcal{L}_2(E)$ but $A\notin \mathcal{L}_1(E)$? Or to find $A$ such that $A\in \mathcal{L}_2(E)$ but $A\notin \mathcal{L}_1(E)$.
We need to modify your example a bit. Let $S$ be the diagonal operator on $\ell^2(\Bbb Z)$ given by $Se_n= S(n) e_n$. Let $A$ be the operator given by $A e_n=\sqrt{A(n)}e_{-n}$. $S(n)$ and $A(n)$ are some positive numbers (bounded from above). Then we have
$$\|S^{1/2}Ax\|^2=\sum_{n\in\Bbb Z} S(-n)A(n)\, |x_n|^2, \qquad \|S^{1/2}x\|=\sum_{n\in\Bbb Z} S(n)\,|x_n|^2,\\ \|SAx\|^2=\sum_{n\in\Bbb Z} S(-n)^2A(n)\, |x_n|^2,\qquad \|Sx\|^2=\sum_{n\in\Bbb Z} S(n)^2\,|x_n|^2.$$
Take $$S(n)=\begin{cases} \frac1{n^2}& n>0\\ \frac1{|n|}& n<0\\ 0&n=0\end{cases}\qquad A(n)= \begin{cases}\frac1n&n>0\\0&\text{else}\end{cases}.$$
Then the top left is equal to $$\|S^{1/2}Ax\|^2=\sum_{n>0} \frac1{n^2}\,|x_n|^2≤\sum_{n>0}\frac1{n^2}\,|x_n|^2+\sum_{n<0}\frac1{|n|}\,|x_n|^2=\|S^{1/2}x\|^2.$$ Thus $A\in \mathcal L_2$. However We have $\|SA e_n\|^2=1/{n^3}$ for $n>0$ whereas $\|Se_n\|^2=1/n^4$. Thus a constant $C$ so that $$\|SAe_n\|^2≤C\,\|Se_n\|^2$$ is not possible. This shows that $A$ cannot lie in $\mathcal L_1$.