If $V(x)$ is positive definite, then $\Omega_c$ is bounded

Asked Jan 07 '18 at 23:38

Active Jan 09 '18 at 14:23

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$x(t)$ satisfies the ODE: $\frac{d}{dt}x(t)=f(x)$, $f$ is a local Lipschitz map $\mathbb{R}^n \rightarrow \mathbb{R}^n$. $V(x):\mathbb{R}^n \rightarrow \mathbb{R}$ is a positive definite and continuously differentiable function. Denote $\Omega_C=\{ x \in \mathbb{R}^n : V(x) \le c \}$, where $c>0$. Is the following statement true and how to prove it?

If $\frac{d}{dt}V(x) \le 0$ in $\Omega_c$, then $\Omega_c$ is bounded for sufficiently small $c>0$.

Note: A function $V: \mathbb{R}^n \rightarrow \mathbb{R}$ is called positive defnite if $V(0)=0$ and $V(x) > 0, ~x\ne 0$.

edited Jan 09 '18 at 14:23

asked Jan 07 '18 at 23:38

winston

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1

Do you know anything else about $V$? Continuity...? I shoudl note that, as stated, the question is definitely not true. – Fimpellizzeri Jan 08 '18 at 00:02
Even differentiability will not help: Think of a continuous bimodal distribution with $0$ being the unique minimizer. – max_zorn Jan 08 '18 at 04:51
@Fimpellizieri Please see the new edit. – winston Jan 08 '18 at 16:48
@max_zorn Please see the new edit. – winston Jan 08 '18 at 16:48
What is $f$? Notice that $\Omega_c$ does not depend on $t$. – Fimpellizzeri Jan 08 '18 at 17:07
@Fimpellizieri new edit – winston Jan 09 '18 at 14:23
If $f$ is locally Lipschitz, but not globally so, then the solution $x$ need not exist everywhere. – Fimpellizzeri Jan 09 '18 at 16:21
The question is ill posed. Do you mean to say perhaps: "Suppose there is some $k>0$ such that $\frac{d}{dt}V(x(t))\leq 0$ in $\Omega_k$. Is it true that there is some $c>0$ such that $\Omega_c$ is bounded?" – Fimpellizzeri Jan 09 '18 at 16:28
Moreover, $V(x(t))$ is defined in (a subset) of $\mathbb R$, so to say that $\frac{d}{dt}V(x(t))\leq 0$ in $\Omega_k$ is nonsensical. – Fimpellizzeri Jan 09 '18 at 16:35

If $V(x)$ is positive definite, then $\Omega_c$ is bounded

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