Let $(M,g)$ be a Riemannian manifold with bounded geometry and $A\subset M$ be a set such that for any $x \in A$, $A$ is a subset of the normal neighborhood $A \subset B_{r}(x)$. Now let $y=exp_x v$, $v \in T_xM$ and $T_t:T_x \to T_{c(t)}$ be the parallel transport along $c(t)=exp_x tv $.
My question now is, if there is some constant $C$ that depends on the geometry of the manifold such that $$ | Ric_x(v,v)-Ric_y(T_{1}v,T_{1}v)| \leq C dist(x,y) $$ where $dist $ is the distance of $(M,g)$ and $dist(x,y)$ is very small.
My idea would be to define the function $f:[0,1]\to \mathbb{R}$ by $$ f(t):=Ric_{c(t)}(T_tv, T_tv) $$ and use a Taylor expansion to have $$ Ric_{y}(T_1v, T_1v)=f(1)= Ric_x(v,v)+ s < \nabla Ric_{c(s)}(T_sv, T_sv), T_s v> $$ for some $s \in [0,1]$. So that $$ |Ric_{y}(T_1v, T_1v) - Ric_x(v,v)| \leq \sup_{s \in[0,1]} < \nabla Ric_{c(s)}(T_sv, T_sv), T_s v> $$ But I don't know if there is a way to bound $$ |< \nabla Ric_{c(s)}(T_sv, T_sv), T_s v>| \leq C dist(x,y). $$ I would like to argue that since $||T_tv||^2=||v||^2$ is constant that $$ |< \nabla Ric_{c(s)}(T_sv, T_sv), T_s v>| \leq ||\nabla Ric_{c(s)}(T_sv, T_sv)|| \cdot||v|| \leq ||\nabla Ric_{c(s)}|| \cdot ||v||^2 \cdot ||v|| $$ but I don't know if that makes sense.