Let $f(s)= (f_{1}(s),\ldots,f_{d}(s)), s \in \mathbb{R}^d, \textbf{o} \le s \le \textbf{1}$, be vector of probability generating functions, were each entry has finite third moments. Consider matrix of partial derivatives $$M(s)=\begin{Vmatrix}M_{kl}(s)\end{Vmatrix},M_{kl}(s)=\frac{\partial f_{k}(s)}{\partial s_l},{k,l=\overline{1,d}}.$$ For some fixed point $s_0$, there exists natural $n$ such that matrices $M^n(s)$ have entries limited from below by positive $c(s_0)$, for all $s_0 \le s \le\textbf{1}$. Then Perron-Frobenius theorem guarantees that each matrix $M(s)$ has positive eigenvalue $\rho(s)$, that is biggest in magnitude and is simple. Let $v(s)$ and $u(s)$ be left and right eigenvectors, corresponding to that eigenvalue, which we choose so that $v'(s)u(s)=u(s)\textbf{1}=1,$ and therefore they are unique.
Now let $s_1 \to s_2$, $s_0 \le s_1, s_2 \le\textbf{1}$. In given conditions eigenvectors will be continious and differentiable functions of matrix coefficients and therefore of $s$. It is possible to show that $$\lvert u(s_1)-u(s_2) \rvert \le K \lvert s_1-s_2 \rvert,$$ for all $s_0 \le s_1, s_2 \le\textbf{1}$. Therefore $\overline{\lim\limits_{s_1 \to s_2}} \frac{\lvert u(s_1)-u(s_2) \rvert}{\lvert s_1-s_2 \rvert} \le K$.
My question can we expect $\lim\limits_{s_1 \to s_2} \frac{\lvert u(s_1)-u(s_2) \rvert}{\lvert s_1-s_2 \rvert} = 0$? Can it possibly happen and if so what would be necessary conditions?