# Can Morse trajectories break if their first derivative is uniformly bounded?

Consider a compact Riemannian manifold endowed with a Morse function f. Fix two critical points x and y of f. Whenever you have a sequence $u_k$ of Morse trajectories connecting x and y it might happen that the $u_k$ "converge" to a broken trajectory. Suppose there is a bound $\|u_k'(s)\|\le C$ independently of k and s. Can the sequence still break?

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The answer is yes! I mixed up bubbling with breaking: breaking is if the derivative goes to 0! Stupid me. –  Orbicular Feb 17 '11 at 10:52
I somehow just saw this question. While you've already answered it yourself, you furthermore always have this inequality on a compact manifold $M$, since $|| \nabla f ||$ is continuous on $M$ and thus bounded, and $u'(s) = - \nabla f$. –  Sam Lisi Apr 10 '12 at 13:24