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Suppose that we are given a non-negative even function $b\in C^\infty[-1,1]$ satisfying $b(0)=0$, $\sqrt{b(x+y)}\le \sqrt{b(x)}+\sqrt{b(y)}$ for any $x,y\in[-\frac12,\frac12]$. Can we always find a 3-dimensional smooth Riemannian manifold $(M,g)$ and a smooth curve $a:[-\frac12,\frac12]\to M$ parametrized by arc length, such that the square of the distance function $d_g(a(t),a(s))^2=b(t-s)$?

For example, if we set $(M,g)=(\mathbb{R}^3,\|\cdot\|)$, can we find a smooth curve $a:[-\frac12,\frac12]\to \mathbb{R}^3$ parametrized by arc length, such that the square of the distance function $\|a(t)-a(s)\|^2=b(t-s)$?

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    $\begingroup$ There is an obvious obstruction from the triangular inequality. $\endgroup$ – Benoît Kloeckner Jun 10 '17 at 15:16
  • $\begingroup$ @BenoîtKloeckner Thanks. I forgot to write a condition on $b$. $\endgroup$ – Right Jun 10 '17 at 15:28
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The answer is "no".

Take $b(x)=x^2-x^{1000}$.

Assume that there is a manifolds with the required curve $a$. Note that $a$ has vanishing geodesic curvature; in particular $a$ is a geodesic. On the other hand, no arc of $a$ is length-minimizing, a contradiction.

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