LaSalle statement - Find a positive function I am looking for a function with certain properties. They follow:


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*Consider a mapping $\mathbf{f}: \mathbb{R}^2 \to \mathbf{R}$, positive definite, and $\mathbf{x} = \mathbf{0}$ is the greatest compact which $\mathbf{f}(\mathbf{x}) = 0$. The question is if exists a mapping $\mathbf{g}: \mathbb{R}^2 \to \mathbb{R}$ such that $\mathbf{f}(\mathbf{x}) = x_1 \mathbf{g}(\mathbf{x})$. 


It is all I know. I thank in advance.
Best regards.
-- Edit --
Consider the dynamical system below. Fo so, find u that stabilizes the system on $(0, 0)$.
\begin{eqnarray*}
\dot x_1 &=& x_1^3 + \sin(x_2) \cos(x_1) + u \\
\dot x_2 &=& -x_1 x_2 \cos{x_1}
\end{eqnarray*}
My approach was to find a Lyapunov control function that stabilizes the system on the origin according to Ljapunov function $V(x) = x_1^2 + x_2^2$. Than.
\begin{equation}
u = \frac{1}{\frac{\partial V}{\partial x}^{\intercal} g} (- \frac{\partial V}{\partial x}^{\intercal} f + \phi(x))
\end{equation}
Then, $\dot V(x) = -\phi(x)$. The function may satisfy the LaSalle conditions i.e. the greatest compact set of $\dot V(x)$ is 0. Furthermore, $u$ is always finite.
 A: Let $V(x) = x^2_1 + x^2_2$, the time derivative of $V(x)$ is 
$$
\begin{align}
\frac{\partial V(x)}{\partial x} &= \nabla V \cdot f(x)  \\
 &= \left[\frac{\partial V(x)}{\partial x_1},\frac{\partial V(x)}{\partial x_2}\right] \left[f_1(x),f_2(x)\right]^{T} \\
 &= \left[2x_1,2x_2\right][x^3_1+\sin x_2\cos x_1 + u \ , \ -x_1x_2\cos x_1]^{T} \\
&= 2x^4_1 +2x_1\sin x_2\cos x_1 + 2x_1u - 2x_1x^2_2\cos x_1 \\
&= 2x^4_1 +2x_1\cos x_1(\sin x_2-x^2_2) + 2x_1u  \tag{1}
\end{align}
$$
Now let us choose $u=-2x^3_1-\cos x_1(\sin x_2 - x^2_2)$ and substitute it in Eq(1), we get 
$$
\begin{align}
\dot{V}(x) &= 2x^4_1 +2x_1\cos x_1(\sin x_2-x^2_2) + 2x_1u  \\
&= 2x^4_1 +2x_1\cos x_1(\sin x_2-x^2_2) + 2x_1 \Big[-2x^3_1-\cos x_1(\sin x_2 - x^2_2)\Big] \\
&= 2x^4_1 +2x_1\cos x_1(\sin x_2-x^2_2) - 4x^4- 2x_1\cos x_1(\sin x_2-x^2_2) \\
&= -2x^4_1 \tag{2}
\end{align}
$$
With our choice of $u$, indeed $\dot{V}(x) \leq 0$ plus $V(x)$ is positive definite, therefore, the origin is stable in the sense of Lyapunov. Unfortunately, Lyapunov theorem doesn't allow us to draw any conclusion about the asymptotic stability because $\dot{V}(x)\leq 0$ (i.e negative semi definite). Using LaSalle theorem, we can show asymptotic stability (i.e. having $V(x)$ positive definite and $\dot{V}(x)$ negative semi definite) if we can show that $\dot{V}(x)$ does not vanish identically along any trajectory in $\mathbb{R}$, other than the null solution $x=0$. We can show that by setting $\dot{V}(x)$ to zero, hence
$$
\dot{V}(x) = -2x^4_1 \implies 0 = -2x^4_1  \implies x_1 = 0 \implies \dot{x}_1 = 0 \tag{3}
$$
From the system's equations, we check 
$$
\begin{align}
\dot{x}_1 &= x^3_1 + \sin x_2 \cos x_1 + u  \\
\dot{x}_1 &= x^3_1 + \sin x_2 \cos x_1 + \Big[-2x^3_1-\cos x_1(\sin x_2 - x^2_2)\Big]  \\
\end{align}
$$
From Eq(3), we see that if $x_1=0$ then $\dot{x}_1=0$, hence:
$$
\begin{align}
0 &= 0 + \sin x_2 (1) + \Big[0-(1)(\sin x_2 - x^2_2)\Big]  \\
0 &= 0 + \sin x_2  - \sin x_2 + x^2_2   \\
0 &= x_2
\end{align}
$$
$\dot{V}(x)$ does not vanish identically along any solution other than $x = 0$,
and the origin is asymptotically stable by LaSalle Theorem (i.e. the stability here is locally). Moreover, since $V(x)$ is radially unbounded (i.e. $V(x) \rightarrow \infty$ as $||x|| \rightarrow \infty$), the origin is globally asymptotically stable. 
