Let's say we have $A \leq B$. Is it then true that $||Ax|| \leq ||Bx||$ (where $x, A, B$ all belong to the same finite-dimensional Hilbert space $H$)?

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    $\begingroup$ Uh?? How are $\,x,A,B\,$ all "in the same space"? What is $\,A\le B\,$ in a vector space? And what is $\,Ax\,$ (product of vectors) in a vector space?? $\endgroup$ – DonAntonio Mar 18 '13 at 22:10
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    $\begingroup$ There are two different ways we can view $A \leq B$. It could either mean that $\|Ax\|\leq\|Bx\|$ for all $x\in H$ OR it could mean that $\|A\|\leq \|B\|$, where $\|\cdot\|$ here is the operator norm. In the first case, the answer is obvious. In the second case, we cannot (in general) conclude anything about these two quantities. $\endgroup$ – Ian Coley Mar 18 '13 at 22:27
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    $\begingroup$ By definition, as far as I am aware, doesn't it just mean that $A$ and $B$ are positive operators such that $B - A$ is a positive operator in $H$ as well? $\endgroup$ – wemblem Mar 18 '13 at 22:41
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    $\begingroup$ @FrankMcGovern Most likely: $A\leq B$ means $(Ax,x)\leq (Bx,x)$ for all $x$. That is: the operator $B-A$ is positive (in the sense of operators on $H$, which is actually nonnegative, in a way). $\endgroup$ – Julien Mar 18 '13 at 22:45
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    $\begingroup$ @wemblem Correct. Except that $A$ and $B$ need not be positive when we write $A\leq B$. Apparently it is a further assumption here. $\endgroup$ – Julien Mar 18 '13 at 22:48

Counterexample: $$ A=\left(\matrix{1&1\\1&1}\right)\qquad B=\left(\matrix{2&1\\1&1}\right) $$

Details: Your question needs some context. In operator theory, this reads: is the function $f(t)=t^2$ operator monotone on $[0,+\infty)$, i.e does $A\leq B$ imply $f(A)\leq f(B)$ for every self-adjoint matrices with nonnegative spectrum? For instance, $g(t)=\sqrt{t}$ is operator monotone on $[0,+\infty)$. But $f(t)=t^2$ is not, as the above example shows.

Positive operators: An operator $A$ in $B(H)$ is called positive if and only if one of the following equivalent assertions holds. In this case, one writes $A\geq 0$.

1) $A$ is self-adjoint and $(Ax,x)\geq 0$ for all $x\in H$ (note that the latter implies the former in the complex case by polarization, but not in the real case).

2) $A$ is self-adjoint with nonnegative spectrum.

3) There exists $B$ self-adjoint such that $A=B^2$.

Partial order: For every self-adjoint operators $A,B$, one writes $A\leq B$ if $B-A\geq 0$. This defines a partial order. You question may be restated as follows: $$ 0\leq A\leq B\quad\Rightarrow?\quad A^2\leq B^2. $$ Indeed, $$ \|Bx\|^2-\|Ax\|^2=(Bx,Bx)-(Ax,Ax)=(B^2x,x)-(A^2x,x)=((B^2-A^2)x,x). $$

Counterexample check: It is easily seen that $A$ and $B$ are positive, as they are self-adjoint with spectra $\{0,2\}$ and $\{\frac{3\pm\sqrt{5}}{2}\}$. Now $B-A\geq 0$ as it is self-adjoint with spectrum is $\{0,1\}$. But $B^2-A^2$ has determinant $-1$, so it is not positive.

Commuting case: If we assume that $A$ and $B$ commute, the result holds. With matrices, you can prove it easily by simultaneous diagonalization in an orthonormal basis. I will now give a proof which works in any dimension.

Write $B^2-A^2=(B-A)(A+B)$. Then observe that $B-A$ and $A+B$ are two commuting positive operators. Write $B-A=C^2$ and $A+B=D^2$ with $C=\sqrt{B-A}$ and $D=\sqrt{A+B}$ the unique positive square roots given by continuous functional calculus for normal operators. Since $B-A$ and $A+B$ commute, $C$ and $D$ commute by functional calculus. Therefore $B^2-A^2=C^2D^2=(CD)^2\geq 0$.

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  • $\begingroup$ There is a weak converse. If $ x \mapsto x^{2} $ is an operator-monotone function for a $ C^{*} $-algebra $ A $, then $ A $ must be commutative. This follows from the monotone version of the Stinespring Factorization Theorem. $\endgroup$ – Leonard Huang Mar 20 '13 at 3:02
  • $\begingroup$ @LeonardTristanHuang Interesting, thanks! Do you have a reference? $\endgroup$ – Julien Mar 20 '13 at 3:03
  • $\begingroup$ Yes, I do. A proof may be found in the article On Characterizations of Commutativity of C$ ^{\ast} $-Algebras, published in Proceedings of the AMS, Volume $ 131 $, Number $ 12 $. $\endgroup$ – Leonard Huang Mar 20 '13 at 3:14
  • $\begingroup$ @LeonardTristanHuang Thanks a lot. $\endgroup$ – Julien Mar 20 '13 at 3:18
  • $\begingroup$ @LeonardHuang the result you mention is known as Ogasawara's theorem, and it was known already in the fifties (T. Ogasawara, A theorem on operator algebras, J. Sci. Hiroshima Univ. Ser. A, vol. 18 (1955), pp. 307–309). See also this question, where I give a proof based on representation theory of $C^*$-algebras. $\endgroup$ – Josse van Dobben de Bruyn Jun 25 '18 at 18:36

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