If every divergent curve has unbounded lenght then the manifold is complete.

A divergent curve on a noncompact Riemannian manifold $M$ is a curve $\alpha:[0,\infty)\to M$ where given any compact $K\subset M$ we have there exists $t_{0}$ such that $\alpha(t) \notin K$ for $t>t_{0}$. Subsequently the length is defined in the usual sense for curves on a manifold:

$L(\alpha)=\int_{0}^{\infty} ||\alpha′(t)||dt$.

The problem is

Prove that a noncompact manifold is complete if and only if every divergent curve has unbounded (i.e. infinite) length.

I proved that if M is Complete ⟹ Every divergent curve has unbounded length. But i don't know how i prove the reciprocal.

Thanks

Every divergent curve has unbounded length $\implies$ Completeness:

Suppose the manifold was not complete. Then, there exists a geodesic $\gamma:[0,T) \to M$ which cannot be extended further than $T$. It is clear that this curve has bounded length (specifically, $k.T$ for some constant $k$). However, $\gamma$ is divergent. This is a standard result of maximal solutions of ODEs, but we can make a particular simple proof here: Suppose it was not divergent. Then it is contained in some compact set. Consider the sequence $\gamma(t_n)$, with $t_n \to T$. It has a convergent subsequence $\gamma(t_{n_i})\to p$. Consider this point $p$, and a strongly normal neighbourhood chart domain(*) around it. Now, take a point $\gamma(\xi)$ ($\xi$ sufficiently close to $T$) such that it is in this chart. Since $\gamma(t_{n_i}) \to p$, the line segment (in the local chart on $T_{\gamma(\xi)}M$) emanating from $\gamma(\xi)$ must pass through $p$. By passing a little more through $p$ with the segment (which is possible because we are in an open set), we would be able to extend the geodesic, which is a contradiction.

(*) Strongly normal in the sense that it is a normal neighbourhood of each one of its points.

Completeness $\implies$ Every divergent curve has unbounded length:

Note that if the manifold is complete, Hopf-Rinow holds. Therefore, the closed ball of radius $N$ ($N$ a natural number) is compact. Your hypothesis says that the curve must leave this ball. Note that this holds for any $N$. What does this imply?

By definition of the metric on a Riemannian manifold, this implies that, for a divergent curve $\alpha$, the $t_0$ given by the definition of divergence (when considering the compact being the closed ball of radius $N$) is such that the length of $\alpha|_{[0,t_0]}$ is greater than $N$. Therefore, $$\int_0^{\infty} \Vert \alpha '\Vert \geq \int_0^{t_0} \Vert \alpha '\Vert \geq N .$$

I'd like to give a slightly more direct and generalizable proof that every divergent curve having unbounded length implies completeness.

Let $$\gamma:[0,T) \to M$$ ($$T \leq \infty$$) be a continuously (equivalent to smoothly) inextendible geodesic; we wish to show that $$\gamma$$ has infinite length. If $$\gamma$$ is divergent, we are done by hypothesis. Otherwise, the end of $$\gamma$$ is contained in a compact set. Thus there exists a sequence $$(t_n) \subset [0,T)$$ satisfying $$t_n \to T$$ with $$(\gamma(t_n))$$ convergent to some point $$p \in M$$. Since $$\gamma$$ is inextendible, there must exist another such sequence $$(s_n)$$ with $$(\gamma(s_n))$$ convergent to another point $$q \in M, q \neq p$$. Passing to subsequences, we may assume $$t_n \leq s_n \leq t_{n+1}$$. Choosing $$\epsilon = \frac{d(p,q)}{3}$$, for sufficiently large $$n$$ we have $$\gamma(t_n) \in B_\epsilon(p)$$, $$\gamma(s_n) \in B_\epsilon(q)$$, so that the triangle inequality gives $$L(\gamma|_{[t_n,s_n]}) \geq d(\gamma(t_n),\gamma(s_n)) \geq d(p,q)-d(p,\gamma(t_n)) - d(q,\gamma(s_n)) \geq 3 \epsilon - \epsilon - \epsilon = \epsilon$$ So that $$L(\gamma) \geq \sum_n L(\gamma|_{[t_n,s_n]})$$, which diverges, as desired.

Note this proof in fact shows more generally that every continuously inextendible curve (not necessarily a geodesic) of finite length in a length space is divergent. By constructing a path of finite length through (a subsequence of) a Cauchy sequence, one can show your condition implies completeness in the case of a length space. Hopf-Rinow partially generalizes to length spaces as well, and it says that a complete and locally compact length space has the property that closed and bounded sets are compact, giving that such a space has no divergent curves of finite length.