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I'm trying to do a problem for my Hilbert spaces class and am a bit stuck, it'd be nice to get a hint for the following question.

An operator $T\in B(H)$ is positive and we write $T \geq 0$ if $T$ is self-adjoint and $\langle Tx,x\rangle \geq 0$ $\forall x \in H$. Let $S,T$ be self-adjoint operators in $B(H)$. We say that $S\leq T$ if $T-S \geq 0$. Prove that if $0 \leq S \leq T$ then $\|S\| \leq \|T\|$. Suggestion: prove that $$|\langle Sx,y\rangle |^2 \leq \langle Sx,x\rangle \langle Sy,y\rangle \leq \langle Tx,x\rangle \langle Ty,y\rangle $$.

I have been unable to neither prove the suggestion nor the statement itself.

Many Thanks

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  • $\begingroup$ Use the definition: $||S|| = \sup_{||x|| = 1} <Sx,x>$ and the fact that $T-S\ge 0$ $\endgroup$
    – Ben
    Commented Mar 14, 2020 at 21:19
  • $\begingroup$ right... im currently trying to prove this definition is equivalent to the standard one but i suspect this holds true for $S$ both self adjoint and positive. $\endgroup$
    – THIG
    Commented Mar 14, 2020 at 21:32
  • $\begingroup$ $T-S\ge 0$ so $<Tx,x> \ge <Sx,x>$ $\endgroup$
    – Ben
    Commented Mar 14, 2020 at 21:34
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    $\begingroup$ Sorry i meant that im trying to show that $\sup_{||x||=1} ||Tx||$ is equivalent to your definition of the operator norm $\endgroup$
    – THIG
    Commented Mar 14, 2020 at 21:36
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    $\begingroup$ @qbert oh you are right. That's a special property of self-adjoint operator... $\endgroup$
    – Ben
    Commented Mar 14, 2020 at 21:38

2 Answers 2

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Hints:

  • $\ |\langle Sx, y\rangle|^2 \le \langle Sx, x\rangle \langle Sy, y\rangle\ $ is a Cauchy-Schwarz-like inequality for the "inner product" $\ \langle x, y\rangle_S\stackrel{\text{def}}{=} \langle Sx, y\rangle\ $ When $\ S\ $ is positive definite, $\ \langle.,. \rangle_S\ $ is an inner product (i.e. has all the properties required of an inner product). When $\ S\ $ is merely positive, $\ \langle.,. \rangle_S\ $ is not quite an inner product, since there may be non-zero vectors $\ x\ $ for which $\ \langle x, x\rangle_S=0\ $, but the Cauchy-Schwarz-like inequality can still be proved in the same way as the standard Cauchy-Schwarz inequality.
  • $\ \langle Sx, x\rangle \le \langle Tx, x\rangle\ $ is an immediate consequence of $\ 0\le\langle (T-S)x, x\rangle\ $.
  • $\ \sup_\limits{\|x\|\le1\\\|y\|\le1} |\langle Sx, y\rangle|=\|S\|\ $ and $\ \sup_\limits{\|x\|\le1} \langle Tx, x\rangle\le\|T\|\ $
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  • $\begingroup$ okay this makes sense! but doesn't the first equality in your last line lead us to try and prove something similar to @Ben first comment? $\endgroup$
    – THIG
    Commented Mar 14, 2020 at 23:14
  • $\begingroup$ Similar, yes, but not the same. The definition of $\ \|S\|\ $ is $\ \sup_\limits{\|x\|\le1}\|Sx\|\ $, not $\ \sup_\limits{\|x\|\le1} \langle Sx, x\rangle \ $. If $\ \sup_\limits{\|x\|\le1} \langle Sx, x\rangle =\|S\|\ $ is one of the theorems you've already met, then presumably you would be able to use it to complete your exercise. As far as I can see, however, proving that theorem from scratch will take quite a bit more work than proving $\ \sup_\limits{\|x\|\le1\\\|y\|\le1} |\langle Sx, y\rangle|=\|S\|\ $. $\endgroup$
    – lonza leggiera
    Commented Mar 14, 2020 at 23:42
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It's not so difficult to prove that $$ |\langle Sx,y\rangle|^2 \le \langle Sx,x\rangle\langle Sy,y\rangle \le \langle Tx,x\rangle\langle Ty,y\rangle. $$ The first inequality is the Cauchy-Schwarz inequality for a pseudo inner product. From this it follows that $$ |\langle Sx,y\rangle|^2 \le \|Tx\|\|x\|\|Ty\|\|y\| \le \|T\|^2\|x\|^2\|y\|^2 \\ \|S\|^2=\sup_{\|x\|=\|y\|=1}|\langle Sx,y\rangle|^2 \le \|T\|^2 \\ \|S\| \le \|T\|. $$

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