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?

  • 2
    $\begingroup$ Certainly false for $x=0$. $\endgroup$ May 16, 2020 at 12:51
  • $\begingroup$ 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 $S(e_2)= e_2 \neq 2 e_2 = T(e_2).$ $\endgroup$ May 16, 2020 at 13:04
  • $\begingroup$ @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 $x$? $\endgroup$
    – Calculix
    May 16, 2020 at 13:08
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    $\begingroup$ Sorry for the typos and the confusion. I assumed that you are working over the reals. You would get in general $$ \langle x, Sx \rangle = \vert x_1 \vert^2 + \vert x_2 \vert^2 \leq \vert x_1 \vert^2 + 2 \vert x_2 \vert^2 = \langle x, Tx \rangle. $$ $\endgroup$ May 16, 2020 at 13:11

1 Answer 1


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 $(Sx,x)=(Tx,x)=\|x\|^2$.

Edit: This won't work if your Hilbert space is one-dimensional. In that case, I guess your result does hold :)

  • $\begingroup$ Very convincing answer! Thank you! $\endgroup$
    – Calculix
    May 16, 2020 at 13:13

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