Is (square root) fidelity strictly concave?

We know that (square root) Fidelity which is defined as $$\text{F}(\rho,\sigma) = \| \sqrt{\rho} \sqrt{\sigma} \|_1 = \text{Tr}(\sqrt{\sqrt{\sigma} \rho \sqrt{\sigma}})$$ is satisfies the property of joint concavity. That is

$$\text{F}\left(\sum_{i=1}^{n} p_i\rho_i, \sum_{i=1}^{n} p_i\sigma_i \right) \geq \sum_{i=1}^{n} p_i \text{F}(\rho_i,\sigma_i) , \tag{1}$$

for an $$n$$ dimensional probability vector $$p$$ and ensembles of states $$\{\rho_i\}_{i=1}^n$$ and $$\{\sigma_i\}_{i=1}^n$$. It follows that $$\text{F} \left(\sum_{i=1}^{n} p_i\rho_i,\sigma \right) \geq \sum_{i=1}^{n} p_i \text{F}(\rho_i,\sigma) \tag{2},$$

for some state $$\sigma$$.

For now, assume all of the states and probabilities are distinct and all the states are full-rank. Can we show that (2) is strictly concave? That is for non-extremal probability vectors (extremal points being the $$n$$ dimensional standard basis vectors) the inequality of (2) becomes strict?

• Take all $\rho_{i}$ orthogonal to some rank $1$ $\sigma$ and it becomes an equality.
– JSdJ
May 6 at 14:22
• @JSdJ Thanks, that is a great example for the equality. Can we show this to be the case when all the states $\rho_i$ and $\sigma$ are full rank? May 8 at 11:05
• I'm not sure if I completely understand your question - do you mean that for $\rho_{i}$ and $\sigma$ orthogonal and full rank, it's necessarily an equality?
– JSdJ
May 11 at 17:02
• No, my intuition is that when all the $\rho_i$s and $\sigma$ and distinct and full rank, along with the probabilities $p_i$ being distinct, the inequality should be strict. May 13 at 5:16