A compactly trapped gradient trajectory has a critical accumulation point

The set $K$ is nonempty because it contains $x(0)$. On every finite interval, the equation and continuity of $\mathrm{\nabla }f$ give the closed-interval $C^1$ trajectory convention of the energy supplier. Write $E=f(x(0))-{\min }_{K}f\ge 0$. The energy identity and the minimum of $f$ on compact $K$ give ${\int }_0^T\mathrm{\parallel }\mathrm{\nabla }f(x(t)){\mathrm{\parallel }}^{2}dt\le E$ for every $T>0$. Nonnegativity of the integrand also gives energy monotonicity on every subinterval.

For every integer $j\ge 1$ there is a time $t>j$ with $\mathrm{\parallel }\mathrm{\nabla }f(x(t))\mathrm{\parallel }<1\mathrm{/}j$. Otherwise continuity gives $\mathrm{\parallel }\mathrm{\nabla }f(x(t)){\mathrm{\parallel }}^2\ge 1\mathrm{/}{j}^2$ on every $[j,T]$. This integrand is continuous and therefore integrable, so [F5] gives ${\int }_j^T\mathrm{\parallel }\mathrm{\nabla }f(x(t)){\mathrm{\parallel }}^{2}dt\ge (T-j)\mathrm{/}{j}^2$. The energy identity on $[j,T]$ and step 1.1 bound this integral by $f(x(j))-{\min }_{K}f\le E$, contradicted by $T>j+j^{2}E$. Continuity makes the set of eligible times greater than $j$ open and nonempty. Density of the rationals supplies an eligible rational time; select $t_j$ as the first one in a fixed rational enumeration. This defines all $t_j$ without countable choice, with $t_j>j$ and thus $t_j\to \mathrm{\infty }$.

The sequence in bounded $K$ has a convergent subsequence by Bolzano–Weierstrass, and its limit $y$ belongs to closed $K$. Sequential continuity of the continuous gradient gives $\mathrm{\nabla }f(y)=0$, since the subsequence indices tend to infinity and the gradient norms are less than their reciprocal indices. Energy monotonicity and sequential continuity of $f$ give $f(y)\le f(x(0))$, proving the assertion.