0
$\begingroup$

Let $E$ be a complex Hilbert space and $\mathcal{L}(E)$ the algebra of all bounded linear operators on $E$.

On $\mathcal{L}(E)^n$, we have two equivalent norms: \begin{eqnarray*} N_1({\bf A}) &=&\sup\left\{\bigg(\displaystyle\sum_{k=1}^n\|A_kx\|^2\bigg)^{\frac{1}{2}},\;x\in E,\;\|x\|=1\;\right\}, \end{eqnarray*} and $$N_2({\bf A})=\bigg(\displaystyle\sum_{k=1}^n\|A_k\|^2\bigg)^{1/2},$$ for every ${\bf A} = (A_1,...,A_n) \in \mathcal{L}(E)^n$

In general $N_1\neq N_2$. If $A_iA_j=A_jA_i$ for all $i,j$, is $$N_1=N_2?$$ If the claim is false, under which conditions we have $N_1=N_2?$

$\endgroup$

1 Answer 1

1
$\begingroup$

As quite an obvious counterexample, take $A_k$ to be the orthoprojector $E\ni x:=(x_1,\dots,x_n)\mapsto x_ke_k\in E$ on $E:=\mathbb{C}^n$. Then $A_iA_j=A_jA_i=\delta_{ij}A_i$ but $N_1(A)=1$ and $N_2(A)=\sqrt{n}$.

On the positive side, I think $A_k:=f_k(A)$ with $A$ self-adjoint, and increasing functions $f_k:\mathbb{R}\mapsto \mathbb{R}$ should work, even if they don't necessarily commute between them.

$\endgroup$
2
  • $\begingroup$ Please in the case of two normal operators which commute, is the two norms are equal? Thanks. $\endgroup$
    – Student
    Commented Nov 28, 2019 at 10:10
  • $\begingroup$ You may take $n=2$ in the above example. $\endgroup$ Commented Nov 28, 2019 at 11:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.