I am trying better understand the description of a Lyapunov function. Say we're considering the definition of asymptotic stability of the equilibrium point $x=0$ for the system $\dot{x}=f(x)$ (abbreviated version of Theorem 4.1 from Khalil):
If $V(x)$ is a continuously differentiable function with \begin{equation} V(0) = 0 \end{equation} \begin{equation} V(x) > 0 \text{ for }x\ne0 \end{equation} \begin{equation} \dot{V}(x) \le 0 \end{equation} then $x=0$ is stable. If additionally \begin{equation} \dot{V}(x) < 0 \text{ for }x\ne0 \end{equation} then $x=0$ is asymptotically stable.
I am curious to know what the implication of replacing $V(0)=0$ with $V(0)>0$ in the above statement would be. This certainly seems problematic from an energy storage perspective. As an example, what could be said about $x=0$ using \begin{equation} V(x)=x^TPx+c^2 \end{equation} where $P>0$ and $c$ is a positive constant?
Edit: Trying out the suggest shift
Starting from $V(x)=x^TPx+c^2$, rewrite as \begin{equation} V(\bar{x}) = \bar{x}^2 \end{equation} where \begin{equation} \bar{x} = (x^TPx+c^2)^{\frac{1}{2}} \end{equation} so that V(0)=0. Taking the derivative \begin{align} \dot{V}(\bar{x}) &= 2\bar{x}\dot{\bar{x}} \\ &=2\bar{x}\left[\frac{1}{2}\left(x^TPx+c^2\right)^{-\frac{1}{2}}\left(2x^TP\dot{x}\right)\right] \\ &= 2x^TP\dot{x}. \end{align} Then say that $\dot{x}$ is such that \begin{equation} \dot{V}(\bar{x}) \le -x^TQx \end{equation} with $Q>0$. Is it correct to say that $\dot{V}(\bar{x})\le0$ from here? I can't claim that $\dot{V}(\bar{x})<0$ since it's clear that $\dot{V}(\bar{x})=0$ when $x=0$. It is not actually even possible for $\bar{x}=0$. Regardless, does this imply $\bar{x}$ is stable and thus $x$ is stable due to their relationship?