TITLE: Can every curve on a Riemannian manifold be interpreted as a geodesic of a given metric?
QUESTION [32 upvotes]: Given a metric $g_{\mu\nu}$ it is possible to find the equations of the geodesic on the Riemannian manifold $M$ defined by the metric itself:
$$\frac{d^2x^a}{ds^2} + \Gamma^{a}_{bc}\frac{dx^b}{ds}\frac{dx^c}{ds} = 0$$ 
where:
$$\Gamma^a_{bc} = \frac{1}{2} g^{ad} \left( g_{cd,b} + g_{bd,c} - g_{bc,d} \right)$$ are the Christoffel symbols and $$g_{ab,c} = \frac{\partial {g_{ab}}}{\partial {x^c}}$$
Now, given a parametric equation of a curve, is it possible to find the metric of a Riemannian manifold which gives that curve as a geodesic? If the answer is 'Yes', is there a bijective correspondence between the curve and the metric? Or are there many metrics giving the same geodesic?
Thanks in advance.

REPLY [2 votes]: $$
\text{These are the geodesics: $\gamma_{a,b}$.}
$$
Let $M$ be normed-space with norm $||\cdot||$.
Fix $\gamma:[0,1]\times M \times M \to M$ continuous in the norm topology. Let $\gamma_{a,b}: [0,1] \to M$ be defined by $\gamma_{a,b}=\gamma(a,b,\cdot)$.
Define
$$
l_{a,b}(\Gamma) = \int_0^1 |\Gamma(t) - \gamma_{a,b}(t)|\, dt + \int_0^1 |\gamma_{a,b}(t)|\, dt,\\ 
\text{where } \Gamma \in C_{a,b} = [\text{Continuous } \Gamma:[0,1]\to M \text{ such that } \Gamma(0) = a, \Gamma(1) = b] \quad(a,b\in M).
$$
Define
$$
d(a,b) = \inf_{\Gamma \in C_{a,b}} l_{a,b}(\Gamma) \quad (a,b \in M).
$$
Define
$$
\text{len}(\Gamma) = \inf_{P \sqsubset [0,1]} \sum d(\Gamma(x_{n+1}),\Gamma(x_{n})) \quad (\Gamma\in C). \quad (\text{This is Just notation.})
$$
$$
\text{These are the geodesics: $\gamma_{a,b}$.}
$$
Remark. $l_{a,b}(\Gamma)$ is not the length of $\Gamma$. The definition of $\text{len}(\Gamma)$ includes a term account for the distance between $x_{n}$ and $x_{n+1}$. As stated, the right side is Just notation.
Extreme Competence Assumed.