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Let $T$ be a self-adjoint bounded operator on a separable Hilbert space $H$ and $ T=S-K$, where $S$ and $K$ are bounded positive operators with $||S||\leq ||T||$ and $||K||\leq ||T||$. If $$sup \{ \langle Tx,x \rangle: x\in H, ||x||=1\} = M>0,$$ Can we say that $sup \{ \langle Sx,x \rangle: x\in H, ||x||=1\} \leq M$ ?. If this is not possible, can we have any other conditions on $S$ and $K$ on which this satisfies?

I could prove this when $S= T^{+}, K=T^{-}$.

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  • $\begingroup$ By "positive operators", do you mean "positive self-adjoint operators"? $\endgroup$ Commented Sep 16 at 14:03
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    $\begingroup$ The answer to the question is yes. The first displayed supremum is simply the norm of $T$. $\endgroup$ Commented Sep 16 at 15:02
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    $\begingroup$ @NikWeaver The supremum is of $\langle Tx, x \rangle$, not the absolute value of that, so it’s the maximum element of the spectrum of $T$ instead of the norm of $T$. (And in fact the answer is no. See the answer I posted.) $\endgroup$ Commented Sep 16 at 15:19
  • $\begingroup$ Oh, you're right! $\endgroup$ Commented Sep 16 at 19:56

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No. For a counterexample, let $S = \begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix}$ and $K = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$. Let $T = S - K$. Then $\|S\| = \|K\| = \|T\| = 2$, but $\sup_{\|x\| = 1} \langle Tx, x \rangle = 1 < 2 = \sup_{\|x\| = 1} \langle Sx, x \rangle$.

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  • $\begingroup$ Sorry, I forgot to mention that the sup of the set is a non zero positive value . In your counterexample, you have supremum value of <Tx,x> =0. I still feel there can be counterexample. $\endgroup$ Commented Sep 16 at 15:47
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    $\begingroup$ @KNSRIDHARANNAMBOODIRI A very minor change can generate a counterexample with positive supremum. I’ve made the needed change to the answer. $\endgroup$ Commented Sep 16 at 18:37

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