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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 $H$)?

wemblem
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    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?? – DonAntonio Mar 18 '13 at 22:10
  • Maybe the question makes sense if it says "Let's say we have $||A|| \leq ||B||$. Is it then true that $||Ax|| \leq ||Bx||?$ – Euler....IS_ALIVE Mar 18 '13 at 22:23
  • Nope, all I have is that $A$ and $B$ are positive operators such that $A \leq B$. Is it possible to say anything at all about $||Ax||$ and $||Bx||$? – wemblem Mar 18 '13 at 22:24
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    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.

     – Ian Coley Mar 18 '13 at 22:27
  • The OP has repeated, by 2nd time, the weird condition $,A\le B,$ ...Unless he explains what does this mean I see no meaning to this question and perhaps I will vote to close it later. – DonAntonio Mar 18 '13 at 22:32
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    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? – wemblem Mar 18 '13 at 22:41
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    @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). – Julien Mar 18 '13 at 22:45
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    @wemblem Correct. Except that $A$ and $B$ need not be positive when we write $A\leq B$. Apparently it is a further assumption here. – Julien Mar 18 '13 at 22:48
  • @julien: I think saying that $ A \leq B $ requires both $ A $ and $ B $ to be at least self-adjoint operators. – Haskell Curry Mar 19 '13 at 02:04
  • @HaskellCurry True. This is a partial order on self-adjoint elements. I do not consider it otherwise. I made it clearer in my answer. – Julien Mar 19 '13 at 02:06
  • A simple counterexample comes from $ -2I \leq I $. Of course, if both $ A $ and $ B $ are required to be positive, then a counterexample is a little harder to find. – Haskell Curry Mar 19 '13 at 02:21
  • @HaskellCurry The title says: for positive operators... It is indeed trivial otherwise. – Julien Mar 19 '13 at 02:23

1 Answers1

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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 $B^2-A^2=C^2D^2=(CD)^2\geq 0$.

Julien
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