If $S$ and $T$ are distinct self-adjoint operators a (separable) Hilbert space $H$ such that $S\leq T$, can we conclude that $\langle Sx,x\rangle<\langle Tx,x\rangle$ for all $x\in H$?
Here $S\leq T$ means that $T-S$ is a positive operator. I know that we must have $\langle Sx,x\rangle\leq\langle Tx,x\rangle$, but I'm not sure whether this inequality becomes strict when $S\neq T$.
I was experimenting with diagonal self-adjoint operators (w.r.t. some orthonormal basis). For such operators I think that it is indeed true.
Any suggestions are greatly appreciated!
Edit 1: As @KaviRamaMurthy pointed out, this is not true for $x=0$. So what can we say for non-zero $x$? Or if necessary, for unit vectors $x$.
Edit 2: For example, if $S(e_{n})=\alpha_{n}e_{n}$ and $T(e_{n})=\beta_{n}e_{n}$ with $\alpha_{n},\beta_{n}\in\mathbb{R}$ and $\alpha_{n}<\beta_{n}$ (for some ONB $(e_{n})$), then $S$ and $T$ are self-adjoint, distinct and $S\leq T$. In this case, I think we indeed have $\langle Sx,x\rangle<\langle Tx,x\rangle$ for all non-zero $x$, since infinite series preserve strict inequalities. Is this right?