Lyapunov function, Continuously Differentiable Lecturer states that to show that a system is stable (not asymptotically stable) we need to find a Lyapunov function that is continuous only, second source says that it must be continuously differentiable.
Which is correct?
Since differentiable implies continuity, im unsure of the meaning of continuously differentiable, if someone could clarify that also.
 A: Lyapunov theorems can be proved using only the assumption of continuity, where one says that a Lyapunov function is a continuous function $V$ which decreases (weakly) along every orbit (see for example Zehnder's Lectures on Dynamical Systems, Section IV.2). The thing is that the usual way of checking that a function is decreasing is by using the derivative, so for simplicity most sources add the assumption that $V$ is differentiable and that $\dot V \le 0$.
Regarding the phrase continuously differentiable, it does not mean “continuous and differentiable” (which would indeed be redundant), but that the function can be differentiated and that the result of the differentiation is continuous. The ending “-ly” turns the adjective continuous into an adverb, indicating that the function is “differentiable in a continuous way”. To be explicit: we say that a multivariate function $V(x_1,\dots,x_n)$ is continuously differentiable if all the partial derivatives $\partial V/\partial x_i(x_1,\dots,x_n)$ exist and are continuous.
(This is a simple sufficient condition for $V$ to be differentiable in the multivariate sense.)
A: According to the proof of Lyapnuov's theorem given in [1], the assumption of continuity of partial derivatives is necessary only to prove asymptotic stability. For simple stability it is not necessary.
[1] Khalil, Hassan K. Nonlinear control. Vol. 406. New York: Pearson, 2015.
