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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.

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    $\begingroup$ What is the source of this question? $\endgroup$
    – Fedor Petrov
    Commented May 2, 2020 at 13:31
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    $\begingroup$ You can find it here: artofproblemsolving.com/community/c7h1976172p13713536 $\endgroup$
    – Lao-tzu
    Commented May 2, 2020 at 13:38
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    $\begingroup$ The $\limsup$'s are actually limits, and they are equal to the maximum of the smallest closed support of the spectral measure of $x$ if $\|x\|=1$. $\endgroup$
    – Christian Remling
    Commented May 2, 2020 at 14:43
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    $\begingroup$ @Lao-tzu: This follows from the spectral theorem, by writing $\|A^n x\|^2 = \int t^{2n}\, d\rho(t)$, with $d\rho(t) = d\| E(t)x\|^2$ being the spectral measure of $x$. (In other words, we are computing the $n$ norm of the function $t$ in $L^2(\rho)$, and $\|t\|_p \to \|t\|_{\infty}$ as $p\to\infty$.) $\endgroup$
    – Christian Remling
    Commented May 2, 2020 at 15:01
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    $\begingroup$ @Lao-tzu: No, I don't think it does, I thought about it and couldn't make any progress. The "spectral measure" is the $\rho$ from above, and what I'm saying is that $\|t\|_{L^{\infty}(\rho)}$ is the same as $\max S$, with $S$ defined as the smallest closed set with $\rho(\mathbb R\setminus S)=0$ (this exists and is unique). $\endgroup$
    – Christian Remling
    Commented May 2, 2020 at 15:09

1 Answer 1

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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 $n$.)

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    $\begingroup$ Nik, thanks for your answer! I've never heard about "spectral order". I trust your argument makes good sense but I cannot make sense of all of your argument. Any reference I can find the related material? Maybe you can add that in your answer. $\endgroup$
    – Lao-tzu
    Commented May 3, 2020 at 13:05
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    $\begingroup$ Just to record for others' convenience: Lemma 4 and Theorem 3 of this article ams.org/journals/proc/1971-028-02/S0002-9939-1971-0276788-1/… answer the question completely, but uses other results which I haven't check. $\endgroup$
    – Lao-tzu
    Commented May 3, 2020 at 14:09

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