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In the Euclidean plane, for a closed smooth curve of length $\ell$ whose curvature is bounded above by $\epsilon$ we have the inequality $$ \ell \ge 2\pi \epsilon^{-1} $$ which follows from the fact that the total curvature is $2|k|\pi$ with $k \not= 0$ the winding number.

Is there a known generalisation of this to closed curves in a CAT(0) (simply connected with negative sectional curvature) Riemannian manifold? That is given such a manifold $X$, is there a function $f : [0, 1] \to [0, +\infty[$ (say) so that for any closed smooth curve in $X$ of length $\ell$ and curvature bounded above by $\epsilon$ we have $\ell \ge f(\epsilon)$? (of course we want $\lim_{\epsilon\to 0} f(\epsilon) = 0$ and for pinched negative curvature I would expect it to go faster than $1/\epsilon$).

We can also ask a purely metric version of this question: given a CAT(0)-space $X$, is there a function $f : [0, 1] \to [0, +\infty[$ (say) so that for any closed curve in $X$ of length $\ell$ which is a local $(1+\epsilon)$-quasi-geodesic we have $\ell \ge f(\epsilon)$?

(By "local $(1+\epsilon)$-quasi-geodesic" I mean a curve $\gamma$ such that for any two points $x, y \in X$ such that $d(x, y) \le C$ and $x, y$ lie on $\gamma$, if $a$ is the length of the (shortest) arc between $x$ and $y$ on $\gamma$ then $a - d(x, y) \le \epsilon$. Here $C$ is a constant depending on $X$.)

In the Euclidean plane, for a closed smooth curve of length $\ell$ whose curvature is bounded above by $\epsilon$ we have the inequality $$ \ell \ge 2\pi \epsilon^{-1} $$ which follows from the fact that the total curvature is $2|k|\pi$ with $k \not= 0$ the winding number.

Is there a known generalisation of this to closed curves in a CAT(0) (simply connected with negative sectional curvature) Riemannian manifold? That is given such a manifold $X$, is there a function $f : [0, 1] \to [0, +\infty[$ (say) so that for any closed smooth curve in $X$ of length $\ell$ and curvature bounded above by $\epsilon$ we have $\ell \ge f(\epsilon)$? (of course we want $\lim_{\epsilon\to 0} f(\epsilon) = 0$ and for pinched negative curvature I would expect it to go faster than $1/\epsilon$).

We can also ask a purely metric version of this question: given a CAT(0)-space $X$, is there a function $f : [0, 1] \to [0, +\infty[$ (say) so that for any closed curve in $X$ of length $\ell$ which is a local $(1+\epsilon)$-quasi-geodesic we have $\ell \ge f(\epsilon)$?

(By "local $(1+\epsilon)$-quasi-geodesic" I mean a curve $\gamma$ such that for any two points $x, y \in X$ such that $d(x, y) \le C$ and $x, y$ lie on $\gamma$, if $a$ is the length of the (shortest) arc between $x$ and $y$ on $\gamma$ then $a - d(x, y) \le \epsilon$. Here $C$ is a constant depending on $X$.)

Edit following comments: in Gromov-hyperbolic spaces it seems that the condition of being a "local-quasi-geodesic" implies (for sufficiently small $\epsilon$) that the curve is a global quasi-geodesic, in particular it cannot close. I think the proof of Theorem 1.13, p. 405 of Bridson--Haefliger can be immediately adapted to do this (and for this particular problem points (1) and (2) of the theorem are sufficient). So we can take $f = +\infty$ in a neighbourhood of 0 (depending on the hyperbolicity constant). The comment by shurtados shows that for the hyperbolic plane we can take the neighbourhood to be $[0, 1[$.

As noted in Ycor's comment the "purely metric" version is not optimal for CAT(0) spaces (as opposed to hyperbolic) and for these a version that would include non-Riemannian $X$ and singular curves should probably involve some sort of "curvature measure" on the curve whose integral would be computable. This question seems interesting even for euclidean space.