I will start with an example to motivate my question, and then ask it more generally.
Example. Let us consider the Hilbert spaces $H^1 \subset L^2$, and a Frechet-differentiable functional $E:H^1\to\mathbb{R}$, e.g. $E(f)=\int (f')^2$. The gradient flow equation for this functional has the form $$\frac{d}{dt}f=-\mathrm{grad}_V E(f),$$ where $\mathrm{grad}_V E$ is such that $dE(f)(u)=\langle\mathrm{grad}_V E(f), u\rangle_V$. Choosing $V=L^2$ and $V=H^1$ gives different gradient-flow equations. Let us choose an initial value $f|_{t=0}\in H^1$ and look at the solutions for both choices. For the Dirichlet energy, they will naturally converge to the same steady-state solution, because there is only one.
Question. How far can this behavior be generalized? Assume I have two Hilbert spaces $U\subset V$, a functional $E:U\to\mathbb{R}$, and I prescribe $f|_{t=0}\in U$. Assume furthermore that the gradient flows for $\mathrm{grad}_U E$ and $\mathrm{grad}_V E$, starting from the same initial value, both converge in $U$. Does it follow that they converge to the same solution? To frame it in more intuitive terms: If the functional has more than one local minimum, can the choice of inner product "redirect" the flow enough, so it converges to different local minima from the same initial point?
I would assume that at least in finite-dimensional spaces, the choice of inner product does not matter. Does the same hold in infinite-dimensional spaces? If not, can we get there with additional assumptions?
Yes, as a minimal example consider $H=\Bbb R^2$. Let $\langle, \rangle_1$ be the usual euclidean scalar product and consider an additional product $$\langle (x_1, y_1) , (x_2,y_2)\rangle_2 :=2x_1x_2+x_1y_2+x_2y_1+2y_1y_2 $$ Now look at the functional $$E:\Bbb R^2\to\Bbb R, \quad (x,y) \mapsto \sin(x)^2$$ then the respective gradients are (exercise): $$\nabla_1E = 2\sin(x)\cos(x)\cdot (1,0),\qquad \nabla_2 E = 2\sin(x)\cos(x)\cdot (2/3, -1/3)$$ Now start at $f=(1,0)$. The first gradient will move you to $(0,0)$, the second will however have you end up in $(0,1/2)$. For both cases the value of the functional itself doesnt change but you can smoothly modify the function close to $(0,1/2)$ if you want so that the second gradient falls into a shallower (or deeper) hole.