Let $E$ be a complex Hilbert space. Let $T\_1,T\_2\in \mathcal{L}(E)$. Let
\begin{align}
W\_{\max}(T\_1,T\_2)
=\big\{ (\lambda\_1,\lambda\_2)\in \mathbb{C}^2; & \;\exists\,(x\_n)\_n;\;\|x\_n\|=1,\;(\langle T\_1 x\_n\; ,\;x\_n\rangle,\,\langle T\_2 x\_n\; ,\;x\_n\rangle)\to (\lambda\_1,\lambda\_2),\\
& \text{ and }\displaystyle\lim\_{n\rightarrow+\infty}(\|T\_1x\_n\|^2+\|T\_2x\_n\|^2)\rightarrow \|T\_1\|^2+\|T\_2\|^2\;\big\}.
\end{align}


Why $W\_{max}(T\_1,T\_2)$ is not convex?



Solving this problem of convexity
Consider the matrices
$$
T\_1 = \left[\begin{array}{ccc} 1&0&0 \\ 0& 0&0 \\0&1&0 \end{array}\right] \ \ \textrm{and} \ \ T\_2 = \left[\begin{array}{ccc} 0&0&0 \\ 1& 0&0 \\0&1&0 \end{array}\right].
$$
Note that because we are working in finite-dimensional Hilbert space we do not need the limits in the definition of $W\_\max(T\_1,T\_2)$.


Any unit vector $x \in \mathbb C^3$ that norms both $T\_1$ and $T\_2$ will be of the form $x = \left( \begin{smallmatrix}a\\ b\\ 0\end{smallmatrix}\right)$ where $|a|^2 + |b|^2 = 1$. Hence,
$$
(\langle T\_1 x \ | \ x\rangle, \langle T\_2 x \ | \ x\rangle) = (|a|^2, a\overline b)
$$
and we see that $(1,0)$ and $(0,0)$ are in $W\_\max(T\_1,T\_2)$. If $|a| = 1/2$ then $|b| = \sqrt 3/2$ and so $(|a|^2, a\overline b) = (1/4, \lambda \sqrt 3/4)$ for some $|\lambda| = 1$. Thus, it is clear that $(1/4, 0) \notin W\_\max(T\_1,T\_2)$ even though $(1/4,0) = 1/4(1,0) + 3/4(0,0)$. Therefore $W\_\max(T\_1,T\_2)$ is not convex.


