Is $W_0(A_1^*,A_2^*)=\overline{ W_0(A_1,A_2)}$? 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!!
 A: 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. 
A: 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?
A: 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, $W_{0}(A_1^*,A_2^*)=\varnothing$.
