# Why existence of Lyapunov function implies Lyapunov stability at the equilibrium point

Why existence of Lyapunov function (locally positive definite and the time derivative of the Lyapunov-candidate-function is locally negative semidefinite) implies Lyapunov stability (i.e for any $\epsilon >0$, there exist $\delta >0$ such that any trajectory initiated at the delta neighbourhood of the equilibrium point remains in the $\epsilon$ neighbourhood) at the equilibrium point ?

Assume that we are interested in stability of the trivial solution $x=0$. We have that there is function $V(x)$ such that $V(x)>0,\,x\in U\backslash\{0\}$, and $\dot V(x)\leq 0,\,x\in U$. Take a ball $B_\epsilon\colon |x|\leq \epsilon\subset U$. Since $V$ is continuous and positive, it reaches a minimum on $\partial B_\epsilon$, call it $k$. Now take another ball $B_\delta\colon|x|\leq \delta$, such that $\delta<\epsilon$. Due to the continuity of $V$ and condition $V(0)=0$ this ball can be chosen such that $V(x)<k$ when $x\in B_\delta$. Now pick any initial condition $x_0\in B_\delta$. Since $\dot V\leq 0$ then $V$ along the orbits is non-increasing, and in particular $V(x(t;x_0))<k$, i.e., $x(t;x_0)$ does not leave $B_\epsilon$, which literally means that $0$ is Lyapunov stable.