Consider the following nonlinear system of ODE $$ x'(t)=f(x(t))+\delta g(x(t)), x\in R^N $$ where $f$ and $g$ are two smooth vector fields and $\delta$ is a parameter. Suppose that as $\delta=0$, the system has a hyperbolic and asymptotically stable equilibrium $x_0$. That means eigenvalues of its linearization have negative real parts. Show that the equilibrium $x_\delta$ near $x_0$ is asymptotically stable by quadratic Lyapunov function if $\delta$ is small enough.
My attempt:
Based on the @copper.hat's answer, I edited my previous answer as follows. Define $$\phi(\delta, x):=f(x)+\delta g(x) $$ and define $y=x-x_\delta$.
Let $$y'=\frac{\partial \phi(x_\delta)}{\partial x} y=: Ay$$be the linearization of the the system $x'=\phi(\delta, x)$. And assume that $$ \phi(\delta, x)=Ay+h(y), $$ where $|h(y)|=o(|y|)$.
Since all eigenvalues of $A$ have negative real part [Question: why does this one hold?], then there exists a positive definite matrix $B$ so that $A^TB+BA+I=0$. Consider $$ V(x)=y^TBy $$
Then $$ \dot{V}(x)=(x'-x'_\delta)^TB(x-x_\delta)+(x-x_\delta)^TB(x'-x'_\delta) $$ $$ =((x-x_\delta)^TA^T+h^T(x-x_\delta))B(x-x_\delta)+(x-x_\delta)^TB(A(x-x_\delta)+h) $$ $$ =(x-x_\delta)^T(-I)(x-x_\delta)+2(x-x_\delta)^TBh(x-x_\delta) $$ $$ <-\|x-x_\delta\|^2+2\gamma\|B\|\|x-x_\delta\|^2=-(1+2\gamma \|B\|)\|x-x_\delta\|^2 $$ where we assume that for any $\gamma>0$, there is $r>0$ so that $\|x-x_\delta\|<r$ we have $$ |h(x-x_\delta)|\le \gamma\|x-x_\delta\| $$
Thus, we choose $\gamma< \frac{1}{2\gamma \|B\|}$, then as $\|x-x_\delta\|<r$ $$ \dot{V}(x)<0 $$
If you just want to establish asymptotic stability, it would be simpler to note that the implicit function theorem with $\phi(\delta,x) = f(x)+ \delta g(x)$ shows that for small $\delta$ the perturbed system has an equilibrium point $x_\delta$, and $\lim_{\delta \to 0} x_\delta = x_0$. By continuity, for small(er) $\delta$, the eigenvalues of ${\partial \phi(\delta, x_\delta) \over \partial x}$ also have negative real parts and hence is asymptotically stable.