Suppose $M$ is a closed manifold with a Finsler metric $F$, and let $d$ be the induced distance on $M$. In general, due to the asymmetry of $F$, the distance $d$ is asymmetric as well.
Consider the sets,
$$B_{+}(p, r) = \{x \in M \;|\; d(p, x) < r\}, \quad \text{and} \quad B_{-}(p, r) = \{x \in M \;|\; d(x, p) < r\}.$$
For $r$ sufficiently small (e.g., smaller than both the forward and backward injectivity radii, which is positive due to the compactness of $M$), $B_{\pm}$ are diffeomorphic to euclidean balls.
Question: What can we say about the diameter of $B_{\pm}$? Can we at least claim that (for $r$ small) the diameter is a continuous monotone function of $r$, independent of the center $p$?
I did come across this MSE question, where the accepted answer claims that the diameter of a small Finsler ball is precisely twice the radius. I am aware of the Whitehead convexity theorem in the Finsler setting, which asserts that any sufficiently small ball is strongly geodesically convex. But I fail to see how to prove the claim in the linked post without assuming that the Finsler metric is symmetric.
Context: What I am really trying do is argue something like this. Consider the ball $B_-(p, r_0)$. Can I get some $0 < r < r_0$ so that for any $q \in B_-(p, r)$ we have $B_-(q, r) \subset B_-(p, r_0)$? In the presence of a symmetric metric, we could have taken any $r < \frac{r_0}{2}$. In general, I think I need to choose $r$ so that $\textrm{diam } B_-(q,r) < r_0$, which leads me to the above question.
Any comment or reference regarding this will be highly appreciated. Cheers!
EDIT: I might have found out one possible answer to this. Following Rademacher, one can define the reversibility of $(M, F)$ as $$\lambda = \lambda(M, F) = \sup \{ F(- \mathbf{v} ) \;|\; \mathbf{v} \in T_p M, \, F_p (\mathbf{v}) = 1, \, p \in M \}.$$ I think it follows that $d(p, q) \le \lambda d(q, p)$, for any $p, q \in M$, which leads to $\mathrm{diam} B_{\pm}(p, r) < (1 + \lambda)r$. Am I correct with this argument?