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Let $E$ be a complex Hilbert space. Let $A_1,A_2\in \mathcal{L}(E)$. Let \begin{eqnarray*} W_0(A_1,A_2) &=&\{(\lambda_1,\lambda_2)\in \mathbb{C}^2;\;\exists\,(x_n)_n;\;\|x_n\|=1,\;(\langle A_1 x_n\; ,\;x_n\rangle,\,\langle A_2 x_n\; ,\;x_n\rangle)\to (\lambda_1,\lambda_2),\\ &&\phantom{++++++++++}\;\hbox{and}\;\displaystyle\lim_{n\rightarrow+\infty}(\|A_1x_n\|^2+\|A_2x_n\|^2)=\|A_1\|^2+\|A_2\|^2\;\}. \end{eqnarray*}

How to show that $$W_0(A_1^*,A_2^*)=\overline{ W_0(A_1,A_2)}:=\{(\overline{\lambda_1},\overline{\lambda_2});\;(\lambda_1,\lambda_2)\in W_0(A_1,A_2)\,\}?$$

I try as follows:

$(\lambda_1,\lambda_2)\in W_0(A_1^*,A_2^*)$ if and only if there exists $(y_n)_n$ such that $\|y_n\|=1$, $(\langle A_1 y_n\; ,\;y_n\rangle,\,\langle A_2 y_n\; ,\;y_n\rangle)\to (\overline{\lambda_1},\overline{\lambda_2})$ and $\displaystyle\lim_{n\rightarrow+\infty}(\|A_1^*y_n\|^2+\|A_2^*y_n\|^2)\rightarrow \|A_1\|^2+\|A_2\|^2$

I stuck here, because I think that $\|A_1^*y_n\|$ is not in general equal to $\|A_1y_n\|$.

If the result is false, I want to construct a counter-example. I think it is true only for normal operators.

Thank you!!

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  • $\begingroup$ Perhaps the result is true for normal operators but not in general. I haven't thought about a counter-example. If A is normal then $||A_1 ^{*}x||^{2}$=$||A_1 x||^{2}$. $\endgroup$ – Kavi Rama Murthy Nov 27 '17 at 11:29
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For $A_1,A_2$ normal, the result is trivial because $\|A_j^*x_n\|=\|A_jx_n\|$.

Without normality, the result is not true. For instance, with $E=\mathbb C^2$, let $A_1=E_{21}$, $A_2=E_{11}$. Because of the finite dimension, the limits in the definition of $W_0$ can be taken as equalities.

If $\|x\|=1$, $\|E_{21}x\|^2+\|E_{11}x\|^2=2$, then $x=(\mu,0)$, with $|\mu|=1$. So $$ \langle E_{21}x,x\rangle=0,\ \ \ \langle E_{11}x,x\rangle=1. $$ So $$W_0(A_1,A_2)=\{(0,1)\}. $$ For the adjoints, the two equalities $\|x\|=1$, $\|E_{12}x\|^2+\|E_{11}x\|^2=2$ are inconsistent, since $\|E_{12}x\|^2+\|E_{11}x\|^2=\|x\|^2=1$. So $$ W_0(A_1^*,A_2^*)=\varnothing. $$

Note that this example can be embedded in any Hilbert space of dimension 2 or more.

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I think it is not true: let $E=\mathbb{C}^1$, $A_{1}=i\cdot\operatorname{id}$ and $A_{2}=0$. Then $A_{1}^*=-i\cdot\operatorname{id}$,

$W_0(A_{1},A_{2})=\{(-i,0)\}$ but $W_0(A_{1}^*,A_{2}^*)=\{(i,0)\}$.

Perhaps You forgot complex conjugate?

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Let us consider $A_1=\left(\begin{array}{cc}1&0\\0&0\end{array}\right)$ and $A_2=\left(\begin{array}{cc}0&0\\1&0\end{array}\right)$. We get $W_{0}(A_1,A_2)=\{(1,0)\}$. However, $W_{0}(A_1^*,A_2^*)=\varnothing$.

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