I have the following question but have no idea on its proof (one direction is trivial):



> 
> Let $A$ and $B$ be (bounded) positive self-adjoint operators on a complex Hilbert space $H$. Prove that
>  $$\limsup\_{n \to \infty} \|A^n x\|^{1/n} \le
> \limsup\_{n \to \infty} \|B^n x\|^{1/n}$$
>  holds for every $x \in H$ if and only if $A^n \le B^n$ for each positive integer $n$.
> 
> 
> 


Any suggestion?


Edit: I suspect the result maybe wrong, for example, if the two limits are equal, then it implies that $A=B$, too strong to be true; anyway, I don't know if the limit (in)equality is so strong. Maybe at most we can say $A^n \le B^n$ for large enough integer $n$.


And for the hard part, it suffices to show $A \le B$ by replacing $A$ with $A^n$ etc. and a similar limit inequality holds. A friend of mine using some trick arguments shows this holds when $H=\mathbb{C}^3$, a good evidence.



A question about comparison of positive self-adjoint operators
The condition $A^n \leq B^n$ for all $n$ defines the *spectral order* on the positive part of $B(H)$, usually written $A \preceq B$. It makes the positive part of any von Neumann algebra a complete lattice. It's equivalent to saying that $P\_{[0,t]}(B) \leq P\_{[0,t]}(A)$ for all $t > 0$, where
$P\_S(A)$ is the spectral projection of $A$ for $S$.


Suppose $\limsup \|A^nx\|^{1/n} \leq \limsup \|B^nx\|^{1/n}$ for all $x$. The set of $x$ for which the left limsup is $\leq t$ is precisely the range of $P\_{[0,t]}(A)$; this is easy to see if you take $A$ to be a multiplication operator. Thus the inequality implies $P\_{[0,t]}(B) \leq P\_{[0,t]}(A)$ for all $t$, i.e., $A \preceq B$.


(You can see that $P\_{[0,t]}(B) \subseteq P\_{[0,t]}(A)$ for all $t > 0$ implies $A \leq B$ by noting that $\langle f(A)x,x\rangle \leq \langle f(B)x,x\rangle$ for any simple function $f$ of the form $\sum a\_i\chi\_{[t\_i, t\_i + 1)}$. Taking a limit as $f$ approaches the function $t \mapsto t$ yields $\langle Ax,x\rangle \leq \langle Bx,x\rangle$. Also, $P\_{[0,t]}(B) \subseteq P\_{[0,t]}(A)$ for all $t$ implies the same for $B^n$ and $A^n$, so we actually get $A^n \leq B^n$ for all $n$.) 


