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I would like to know if the following holds:

Let $(M,g_1)$ be a smooth Riemannian manifold, $f:M \longrightarrow \mathbb{R}$ a smooth function with gradient $\nabla^1f$ and $x:\mathbb{R} \longrightarrow M$ a curve satisfying the negative gradient flow equation, i.e. $\dot x(s) = -\nabla^1f(x(s)).$ Furthermore, let's assume that $x(s) \xrightarrow{s \to \infty} p$, where $p$ is a critical point of $f$. Let now $g_2$ be another Riemannian metric on $M$ which might be completely unrelated to $g_1$. $f$ has a different gradient under $g_2$ denoted by $\nabla^2f$. My conjecture is now the following: For $s$ large enough, there is a constant $C >0$ such that $$||\nabla^2f(x(s))||_2^2 \leq C ||\nabla^1f(x(s))||_1^2,$$ where $|| \cdot ||_i$ denotes the norm with respect to $g_i$. I would be glad if someone could prove this (sketch should be enough) or give a counter example.

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Your conjecture is true. Recall that for any differentiable function $f:M\to\mathbb{R}$ and $p\in M$ we have $$\|\nabla f(p)\|=\|df(p)\|,$$ where $df$ denotes the derivative of $f$ and the norm $\|\cdot\|$ is induced by a Riemannian metric $g$. Hence, the inequality you are after can be stated as $$\|df(x(s))\|_2^2\leq C\|df(x(s))\|_1^2.$$ Now, as all norms on a finite dimensional space are equivalent to one another, for $p\in M$ there is some $C_p>0$, such that $$\|\cdot\|_{1,p}\leq C_p\|\cdot\|_{2,p}.$$ Furtheremore, for a compact subset $S\subset M$ the constant $C$ can be chosen uniformly. Your trajectory converges to a point, and so its image indeed lies in some compact subset of the limit point, at least for $s$ large enough.

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