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?
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.
