Consider a system of the form
\begin{equation}
\Sigma\colon\ \dot{x}=f(x),
\end{equation}
where $x\in \mathbb{R}^n$ and $f\colon \mathbb{R}^n\to \mathbb{R}^n$ is a sufficiently smooth function vanishing at the origin. Assume that $\Sigma$ has a globally exponentially stable equilibrium at $x=0$. Now apply a global coordinate transformation
\begin{equation}
z=T(x),
\end{equation}
where $T\colon \mathbb{R}^n\to \mathbb{R}^n$, $T(0)=0$ is a global $\mathcal{C}^1$ diffeomorphism. Accordingly, we obtain the transformed system
\begin{equation}
\Sigma^\prime \colon\ \dot{z}=g(z):=\frac{\partial T(x)}{\partial x}f(x)\bigg|_{x=T^{-1}(z)}.
\end{equation}
Question: Is $\Sigma^\prime$ globally exponentially stable at $z=0$?
[Math] Exponential stability under global diffeomorphism
control theorydynamical systemsnonlinear systemstability-theory
Related Solutions
I think you misunderstood the paper. The property "global asymptotic stability" is defined to be the combination of local stability and global attractivity. There is nothing to prove here; it's a definition. Here is the relevant part:
There is no asymptotic stability. Taken the system
$$ \cases{ \dot x_1 = x_2\\ \dot x_2 = -x_1^3 } $$
and multiplying by $x_1^3, x_2$ as
$$ \cases{ x_1^3\dot x_1 = x_1^3x_2\\ x_2\dot x_2 = -x_1^3x_2 } $$
and adding we have
$$ \frac 12x_1^4+x_2^2 = C $$
so those orbits characterize a center around the origin.
Best Answer
Yes, it is. Moreover, let $x(t)$ and $y(t) = \Phi(x(t))$ be solutions of the systems $$ \dot{x}=f(x,t), \quad x\in\Omega\subseteq\mathbb{R}^n, $$ and $$ \dot{y}=h(y,t), \quad y\in\Phi(\Omega)\subseteq\mathbb{R}^n, $$ respectively (the second system is the transformed first system, $\quad y=\Phi(x)$ is a diffeomorphism); suppose that the solution $y(t)$ is bounded and exponentially stable; finally, exists an $\epsilon>0$ such that a closed $\epsilon$-neighborhood of any point of $y(t)$ is a subset of $\Phi(\Omega)$. Then the solution $x(t)$ is exponentially stable.
The proof of this fact can be found in https://link.springer.com/article/10.1134/S0012266107110043
Update: The main idea of the (simplified, for an equilibrium point and for $\Omega=\Phi(\Omega)=\mathbb R^n$) proof is as follows. Let the zero solution of the system $$\tag{1} \dot y= g(y),\quad g(0)=0 $$ be exponentially stable. It means, by definition, that there exists $\delta>0$ such that $\forall y_0 \in D_{\delta}=\{y:\; \|y\|< \delta\}$ the solution of the initial value problem $$ \dot y= g(y),\quad y(0)=y_0 $$ satisfies $$\tag{2} \|y(t)\|\le M e^{-\gamma (t-t_0)}\|y_0\|, $$ where $M$ and $\gamma$ are some positive constants.
Let $\Phi(x)$ be a diffeomorphism, $\Phi(0)=0$. Since $\Phi$ is a diffeomorphism and, consequently, $\Phi^{-1}$ is also a diffeomorphism, they fulfill the Lipschitz property $$\tag{3} \|\Phi(x_1)-\Phi(x_2)\|\le L \|x_1-x_2\| $$ for any $x_1,x_2\in \Phi^{-1}\left(D_{\delta}\right)$ and, respectively, $$\tag{4} \|\Phi^{-1}(y_1)-\Phi^{-1}(y_2)\|\le L' \|y_1-y_2\| $$ for any $y_1,y_2\in D_{\delta}$, where $L$,$L'$ are some constants.
Now one can combine (2), (3) and (4) and obtain $\forall x_0 \in \Phi^{-1}\left(D_{\delta}\right)$, $y(t)=\Phi(x(t))$, $y_0=\Phi(x_0)$ $$\tag{5} \|x(t)\|=\|x(t)-0\|=\| \Phi^{-1}\left(y(t)\right)-\Phi^{-1}(0) \|\le L' \|y(t)-0\| $$ $$ \le L' M e^{-\gamma (t-t_0)}\|y_0\|= L' M e^{-\gamma (t-t_0)}\|\Phi(x_0)-\Phi(0)\|\le L' M L e^{-\gamma (t-t_0)}\|x_0\| $$ Thus, the zero solution of the system in variables $x$ is exponentially stable.
As for the global exponential stability, well, (3) and (4) are not, in general, satisfied in $\mathbb R^n$ for an arbitrary diffeomorphism; but if the solutions $x(t)$ and $y(t)$ are contained in some compact sets (which is a common case for control theory problems), then one can determine the constants $L$ and $L'$ and establish (5).