# Difference of two positive operators

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

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

• $$\ |\langle Sx, y\rangle|^2 \le \langle Sx, x\rangle \langle Sy, y\rangle\$$ is a Cauchy-Schwarz-like inequality for the "inner product" $$\ \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\|\$$
• 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\|\$. Commented Mar 14, 2020 at 23:42
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 $$|\langle Sx,y\rangle|^2 \le \|Tx\|\|x\|\|Ty\|\|y\| \le \|T\|^2\|x\|^2\|y\|^2 \\ \|S\|^2=\sup_{\|x\|=\|y\|=1}|\langle Sx,y\rangle|^2 \le \|T\|^2 \\ \|S\| \le \|T\|.$$