Recall (see the note on JNR parameters) that the joint numerical range of $k$ Hermitian matrices $A_1, \ldots, A_k \in \mathbb{C}^{d\times d}$ is
with $\mathcal{M}_d = {\rho : \rho \succeq 0,\; \Tr \rho = 1}$ the set of $d$-dimensional density matrices.
One natural generalization is to restrict the rank of the admissible states. Let $\mathcal{M}^{(r)}_d = {\rho \in \mathcal{M}_d : \operatorname{rk}\rho \le r}$; equivalently, these are mixtures of at most $r$ pure states. The associated rank-restricted JNR is
Many other variants exist; see numericalshadow.org.
$L$ is convex quite obviously: it is the image of the convex set $\mathcal{M}_d$ under the linear map $\rho \mapsto (\Tr \rho A_1, \ldots, \Tr \rho A_k)$. For $L^{(r)}$ this is no longer clear. A basic result of Hausdorff 1 shows that $L^{(1)}(A_1, A_2)$ is always convex, but simple generalization fails, see e.g. $L^{(1)}(\sigma_x, \sigma_y, \sigma_z)$, which is a hollow sphere.
A theorem of Au-Yeung and Poon from 1979 2 gives a partial answer.
Theorem (Au-Yeung–Poon, 1979). Let $A_1, \ldots, A_k \in \mathbb{C}^{d\times d}$ be Hermitian. If $1 \le r \le d-1$ and
- $k < (r+1)^2$ when $r < d-1$, or
- $k < (r+1)^2 - 1$ when $r = d-1$,
then $L^{(r)}(A_1, \ldots, A_k)$ is convex.
Since
which itself follows from $\mathcal{M}_d = \operatorname{conv}{|\psi\rangle\langle\psi|}$, we can say that if the conditions of the above theorem are met, for each tuple of expectation values in $L$ there exists a state of rank at most $r$ realizing it.
It has some practical use, as many expressions in quantum mechanics are defined via expectation values. Take for instance
These expectation values are sufficient to specify the covariance matrix
Now let's employ the above observation. Here $k=5$, so for $d\ge 3$ the Au-Yeung–Poon condition with $r=2$ is satisfied ($5 < 9 = (r+1)^2$). Therefore, every expectation value tuple in $L$ is realizable by a state of rank at most 2, and consequently, in finite-dimensional cases with $d\ge 3$, every covariance matrix is realized by some rank-2 state. Naturally, for qubits $d=2$ and every state is a rank-2 state, so this is a fully general observation.