sufficient condition for Lyapunov stability theory

Question

I am reading a paper:

I am confused about the following (see equations (2) and (3) in the paper): http://ieeexplore.ieee.org/document/7039413/
or
http://groups.csail.mit.edu/robotics-center/public_papers/Majumdar14a.pdf

Consider the following condition:

$$V(x) \leq \rho \Longrightarrow \dot{V}(x) <0$$

The paper says that the sufficient condition for the above is:
$$(x^Tx)(V(x)-\rho)+L(x)\dot{V} \geq 0$$

My question is:

Obviously $x^Tx \geq 0$ and $(V(x)-\rho)\leq 0$. So the first part is less than or equal to $0$.

$\dot{V}<0$. So even though $L(x)<0$, there is no guarantee that the whole function is greater than or equal to $0$. I have no idea why it is the sufficient condition for the first inequality.

Please advise, thanks!

Answer 1

You got the logic backwards. The author is saying that Equation (3) is sufficient for Equation (2). In other words, Equation (3) implies (2), not the other way around. Specifically,

$$\Big(\big((x^Tx)(V(x)-\rho)+L(x)\dot{V} \geq 0 \implies -L(x)\dot{V}\le (x^Tx)(V(x)-\rho)\big)\bigwedge -L(x)\ge 0\Big) \implies \big(V(x)-\rho\le 0\implies \dot V\le 0\big).$$