# If $S$ and $T$ are distinct self-adjoint operators on $H$ such that $S\leq T$, can we conclude that $\langle Sx,x\rangle<\langle Tx,x\rangle$?

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?

• Certainly false for $x=0$. May 16, 2020 at 12:51
• Your second edit already tells you, that your first question has a negative answer. E.g. in dimension 2, let $e_1, e_2$ be the standard basis vectors and set $S=Id$ and $T(e_1)=e_1, T(e_2)=2e_1$. Then $$\langle x, S x \rangle = x_1^2 x_2^2\leq x_1^2 +2x_2^2 = \langle x, Tx \rangle.$$ But $S(e_2)= e_2 \neq 2 e_2 = T(e_2).$ May 16, 2020 at 13:04
• @SeverinSchraven Thanks for your reply! I don't immediately see that $x_{1}^{2}x_{2}^{2}=x_{1}^{2}+2x_{2}^{2}$ for some $x\in\mathbb{C}^{2}$. And what if we restrict to unit vectors $x$? May 16, 2020 at 13:08
• Sorry for the typos and the confusion. I assumed that you are working over the reals. You would get in general $$\langle x, Sx \rangle = \vert x_1 \vert^2 + \vert x_2 \vert^2 \leq \vert x_1 \vert^2 + 2 \vert x_2 \vert^2 = \langle x, Tx \rangle.$$ May 16, 2020 at 13:11

Let $$S$$ and $$T$$ be ortohogonal projectors onto subspaces $$V_S$$, $$V_T$$ respectively, with $$V_S$$ a proper subspace of $$V_T$$, so that $$S\neq T$$.
Then take $$x\in V_S$$: clearly $$(Sx,x)=(Tx,x)=\|x\|^2$$.