Gradient conjecture

Let $U \subseteq \mathbb{R}^n$ be an open set, let $f : U \to \mathbb{R}$ be real analytic, and let $x_0 \in U$ be an isolated critical point of $f$, i.e. $\nabla f(x_0) = 0$ and $\nabla f(x) \neq 0$ for all $x$ in some punctured neighborhood of $x_0$.

Let $x : [0, \tau) \to U$, with $0 < \tau \le \infty$, be a trajectory of the gradient vector field of $f$, that is, a solution of

$$\dot{x}(t) = \nabla f\bigl(x(t)\bigr), \qquad t \in [0,\tau),$$

such that $x(t) \neq x_0$ for all $t$ and $x_0$ is a limit point of $x$, so that

$$\lim_{t \to \tau^-} x(t) = x_0 .$$

For $t \in [0,\tau)$ let

$$\ell(t) = \left[\, x(t) - x_0 \,\right] \in \mathbb{P}\mathbb{R}^{n-1}$$

denote the point of the real projective space $\mathbb{P}\mathbb{R}^{n-1}$ determined by the secant line joining $x_0$ to $x(t)$.

Then the limit

$$\lim_{t \to \tau^-} \ell(t)$$

exists in $\mathbb{P}\mathbb{R}^{n-1}$; equivalently, the unit secant directions

$$\frac{x(t) - x_0}{\lVert x(t) - x_0 \rVert}$$

converge in the unit sphere $S^{n-1}$ as $t \to \tau^-$.

The same assertion holds, with the same conclusion, for trajectories of the negative gradient field $\dot{x}(t) = -\nabla f(x(t))$ converging to $x_0$.