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Is the convexity radius $\mathop{\mathrm{conv.rad}}(p)$ at a point $p\in M$ of a complete Riemannian manifold $M$ always smaller than the injectivity radius $\mathop{\mathrm{inj}}(p)$ at that point?

For compact manifolds with non-positive curvature, there is a bound for global values: $\mathop{\mathrm{conv.rad}}(M)=\frac12\mathop{\mathrm{inj}(M)}$ (Petersen, page 259), but not for local values at a point $p$.

For other compact manifolds, I could find only bounds from below, such as $$\mathop{\mathrm{conv.rad}}(M)\ge\min\left\{\frac12\mathop{\mathrm{inj}(M)},\frac\pi{2\sqrt K}\right\},$$ where $K=\sup\sec(M)$, and this is also for global values, while my question is about local values.

In "A Panoramic View of Riemannian Geometry", Berger (page 328 in this file, page 278 in the printed book) says "It is always less or than equal to half of the injectivity radius," but this should be an error.

For the non-complete case (or for a manifold with boundary) $\mathop{\mathrm{conv.rad}}(p)>\mathop{\mathrm{inj}}(p)$ is certainly possible: for example, consider an Euclidean unit disk and a point near its boundary.

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  • $\begingroup$ Please, spell out your definition of conv.rad: With some definitions, the inequality $conv.rad\le in.rad$ is automatic. (With this definition, conv.rad is the largest $r$ such that the exponential map on $B(0,r)$ is injective and its image is geodesically convex in the sense that any two points of the image are connected by a unique minimizing geodesic and the latter is contained in the image of $B(0,r)$.) $\endgroup$ – Moishe Kohan Feb 6 at 23:20

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