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Let $ B (H) $ indicate the set of all bounded linear operators on a complex separable Hilbert space $ H $. Let $ A \in B(H) $, where $ A $ is a positive semi-definite operator in $ H $ (i.e. $ \langle Ax, x \rangle >0 $ for all $x \in H$) Then $ \|A^n\|^m \leq \|A^m\|^n $, for all $ m>n>0$.

How can prove this inequality? Or any hint it.

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  • $\begingroup$ @ToniMhax For self-adjoint operators it's actually equal, the LHS is $(\sup_{\lambda\in \sigma(A)}\lambda^n)^m$ and the RHS the same with $m$ and $n$ swapped. $\endgroup$
    – MaoWao
    Commented Jun 6, 2023 at 17:34
  • $\begingroup$ Since $n >m$, you may have $A^m=0$ whereas $A^n \ne 0$, so the proposed inequality cannot be true in generality. $\endgroup$
    – Christophe Leuridan
    Commented Jun 6, 2023 at 17:42
  • $\begingroup$ @ChristopheLeuridan Look the operator is positive semi definite not equal zero. Others thanks. $\endgroup$
    – Anas Abbas H.
    Commented Jun 6, 2023 at 21:50
  • $\begingroup$ It is still false. Take $$A = \left(\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}\right).$$ Then it is definite positive, the operator norm of $A$ is strictly larger than $1$, but the operator norm of $$A^n = \left(\begin{array}{cc} 1 & n \\ 0 & 1 \end{array}\right).$$ is equivalent to $n$ as $n$ goes to infinity. $\endgroup$
    – Christophe Leuridan
    Commented Jun 7, 2023 at 20:32

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