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 $N_1=N_2?$


1 Answer 1


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.

  • $\begingroup$ 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

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