Closed almost geodesics in a Riemannian manifold

Question

Let $M$ be a smooth Riemanniann manifold. For $\varepsilon \geq 0$ we call an $\varepsilon$-geodesic (I am not sure that this is a standard name) a smooth map $$\gamma\colon [a,b]\to M$$ such that for any $t\in [a,b]$ there is a neighborhood $[u,v]$ of $t$ such that for any $x,y\in [u,v]$ one has $$(1-\varepsilon)dist_M(\gamma(x),\gamma(y))\leq length[\gamma(x),\gamma(y)]\leq (1+\varepsilon)dist_M(\gamma(x),\gamma(y)),$$ where in the middle one has the length of the path $\gamma$ between the points $x$ and $y$, and $dist_M$ denotes the distance in $M$.

Note that $0$-geodesics are the usual geodesics.

Given a point $p\in M$, does there exist a neighborhood $U$ of $p$ and positive $\varepsilon >0$ such that there are no closed $\varepsilon$-geodesics contained in $U$?

Remark. For $\varepsilon=0$ this result is well known (the injectivity radius is positive).

Answer 1

Any curve $\gamma:[a,b]\to M$ parametrized by arc length is an $\varepsilon$-geodesic for any $\varepsilon>0$.

The inequality $(1-\varepsilon)dist_M(\gamma(x),\gamma(y))\leq length[\gamma(x),\gamma(y)]$ is obvious (for $\varepsilon=0$).

To prove the other inequality, for each $t\in [a,b]$ we will pick the geodesic $\alpha_t$ based at $\gamma(t)$ and with direction $\gamma'(t)$.

As $\gamma$ and $\alpha_t$ coincide up to first order in $t$, we have that $d(\gamma(t+\delta),\alpha_t(t+\delta))$ is $o(\delta)$. By compactness we can find for every $\varepsilon$ some $\delta$ such that, $\forall t\in[a,b]$ and $\forall s$ with $|s|\leq\delta$, $d(\alpha_t(t+s),\gamma(t+s))<\varepsilon |s|$.

This means that, if for some $t_0$ we pick the neighborhood $[u,v]$ of $t_0$ with $v-u<\delta$, then $\forall x,y\in [u,v]$ we have

$$dist_M(\gamma(x),\gamma(y))\geq dist_M(\gamma(x),\alpha_x(y))-dist_M(\alpha_x(y),\gamma(y))\geq |y-x|-\varepsilon|y-x|=(1-\varepsilon)length[\gamma(x),\gamma(y)].$$

So $length[\gamma(x),\gamma(y)]\leq (1+2\varepsilon)dist_M(\gamma(x),\gamma(y))$, as we wanted.