If $f:\mathbb{R}^n\rightarrow\mathbb{R}_+$ is a nonnegative real analytic function and $g:\mathbb{R}^n\rightarrow\mathbb{R}$ is a strongly convex smooth function with a surjective gradient $\nabla g:\mathbb{R}^n\rightarrow\mathbb{R}^n$. Is it possible that the solution $x:\mathbb{R}_+\rightarrow\mathbb{R}^n$ to $$\dot{x}=-\nabla f(x), \quad x(0)=x_0$$ is bounded for all $x_0\in\mathbb{R}^n$, i.e., $\|x(t)\|\leq c(x_0)$ for all $t\in\mathbb{R}_+$ for some constant $c(x_0)>0$ (assume $c(x_0)$ is smooth in $x_0$), but the solution to the Hessian Riemannian gradient flow $$\dot{x}=-\left(\nabla^2g(x)\right)^{-1}\nabla f(x), \quad x(0)=x_0,$$ where $\nabla^2g$ denotes the Hessian of $g$, is unbounded for some $x_0$.

  • $\begingroup$ Re, exactly, and I apologise for being inadvertently obscure. Since my other comments are now irrelevant, I have deleted them. $\endgroup$
    – LSpice
    Commented Jun 25, 2023 at 0:51
  • $\begingroup$ so do we have any information on $c(x_{0})$? Does it diverge as $\|x_{0}\|_{2}\to +\infty$? $\endgroup$ Commented Jun 25, 2023 at 4:50
  • $\begingroup$ @ThomasKojar Yes, we can imagine if we initialize farther and farther from the origin, then even if the corresponding solution is still bounded (or even converges), the bound may be larger and larger. But I guess we can assume $c(x_0)$ is a smooth function of $x_0$. $\endgroup$ Commented Jun 25, 2023 at 4:53
  • $\begingroup$ so are you worried that the solution of the Hessian-flow will diverge? And are you also trying to get a $\tilde{c}(x_{0})$ bound for $x(t)$ solving the Hessian-flow? $\endgroup$ Commented Jun 25, 2023 at 4:54
  • $\begingroup$ @ThomasKojar Yes, like I know the gradient flow under Euclidean metric will always be bounded (or converges if you like), I am worried that if I use such a nice $g$ to induce a new Riemannian metric, and under this new metric, the solution goes to infinity as $t\to\infty$. For the $\tilde{c}(x_0)$ bound I don't care. I just need qualitative analysis result. By the way, I know the general case may be too hard and it would be very nice to have some results for a special $g$ like I listed in the post. $\endgroup$ Commented Jun 25, 2023 at 4:57


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.