Point convergence for compact analytic periodic gradient flows

At every point choose an angular chart and put $m=\mathrm{\mid }V\mathrm{\mid }\ge 1$. The coordinate potential is a real analytic germ in $m$ variables. By [F7], its absolutely convergent series defines a holomorphic complexification on a smaller complex polydisc. The smoothness supplier applies on that open polydisc. Restricting its successive real coordinate difference quotients to the real slice shows that all real coordinate derivatives of the potential exist and are continuous there. Thus $U$ is smooth. In the flat angular metric its gradient coordinates are the first partial derivatives of $U$, by [F3], so the gradient is smooth as well. Consequently $U$ and its gradient are continuous globally, and the smooth-potential hypothesis of [F3] holds.

For $\alpha >0$, [F5] gives $(x^{\alpha }{)}^{\mathrm{\prime }}=\alpha x^{\alpha -1}>0$ on $(0,\mathrm{\infty })$. Applying the mean value theorem in [F9] on each closed positive interval proves strict monotonicity. Given $\epsilon >0$, set $\delta ={\epsilon }^{1\mathrm{/}\alpha }>0$. The iterated-power law gives ${\delta }^{\alpha }=\epsilon$, so $0<x<\delta$ implies $0<x^{\alpha }<\epsilon$. Together with $0^{\alpha }=0$, this proves continuity from the right at zero and $x^{\alpha }\to 0$ as $x\downarrow 0$. It applies in particular to $\alpha =1-\rho >0$.

By compactness and [F4], $U$ is bounded below. Put $E(t)=U(q(t))$ and $L={\lim }_{t\to \mathrm{\infty }}E(t)$. By [F3], $E(t)\ge L$ and ${\int }_0^{\mathrm{\infty }}\mathrm{\parallel }{q}^{\mathrm{\prime }}(t){\mathrm{\parallel }}^{2}dt<\mathrm{\infty }$. For each positive integer $n$, the integral over $[n,n+1]$ tends to zero as $n\to \mathrm{\infty }$; choose $s_n\in [n,n+1]$ with $\mathrm{\parallel }{q}^{\mathrm{\prime }}(s_n){\mathrm{\parallel }}^2$ no larger than that integral. Such a point exists by continuity, since otherwise the integral would exceed itself.

By sequential compactness, a subsequence of $q(s_n)$ converges to some $p\in T_V$. Since $q^{\mathrm{\prime }}=-\mathrm{grad}U$ and the gradient is continuous, step 2.1 gives $\mathrm{grad}U(p)=0$. Continuity of $U$ and $s_n\to \mathrm{\infty }$ give $U(p)=L$. Restricting to this subsequence, write its times as $t_n$.

Choose an angular chart centered at $p$ and a closed metric ball of radius $r>0$ around $p$ inside that chart and inside a neighborhood where [F2] applies to the coordinate representation of $U$. Because the chart metric is Euclidean, the Euclidean gradient norm in [F2] equals the flat gradient norm. Thus there are $c>0$ and $0<\rho <1$ such that $\mathrm{\parallel }\mathrm{grad}U(x)\mathrm{\parallel }\ge c\mathrm{\mid }U(x)-L{\mathrm{\mid }}^{\rho }$ throughout this ball.

Write $e(t)=E(t)-L\ge 0$. While $q$ stays in the ball and $e>0$, the chain rule, [F3], [F5], and step 4.1 give $-\frac{d}{dt}e(t{)}^{1-\rho }\ge c(1-\rho )\mathrm{\parallel }{q}^{\mathrm{\prime }}(t)\mathrm{\parallel }$. By [F8], integration therefore bounds the length between any two such times $a<b$ by $[e(a{)}^{1-\rho }-e(b{)}^{1-\rho }]\mathrm{/}[c(1-\rho )]$. If $e$ first reaches zero, monotonicity and $e\ge 0$ make it zero thereafter; [F3] then makes the gradient and speed zero thereafter. By step 1.2 the energy-power term tends to zero as the upper endpoint approaches that first zero $T$. Continuity of the speed at $T$ gives a bound $K$ on a small one-sided neighborhood, so ${\int }_b^T\mathrm{\parallel }{q}^{\mathrm{\prime }}(t)\mathrm{\parallel }dt\le K(T-b)\to 0$ as $b\uparrow T$. Taking these limits extends the length bound across that time. If $e(a)=0$ initially, the curve is already stationary and the bound is zero.

Since $q(t_n)\to p$ and $e(t_n)\to 0$, choose $n$ so that $d(q(t_n),p)+e(t_n{)}^{1-\rho }\mathrm{/}[c(1-\rho )]<r$. If the curve ever left the ball after $t_n$, continuity would give a first time on its boundary. The distance from $q(t_n)$ to that boundary point is at most the intervening length, which step 5.1 bounds by $e(t_n{)}^{1-\rho }\mathrm{/}[c(1-\rho )]$; the triangle inequality then contradicts its distance $r$ from $p$. Hence the whole tail stays in the ball. Taking the supremum of its finite-interval lengths gives $\mathrm{\ell }(t_n,\mathrm{\infty })\le e(t_n{)}^{1-\rho }\mathrm{/}[c(1-\rho )]<\mathrm{\infty }$.

The estimate in step 5.1 applies to every later pair of times, so $\mathrm{\ell }(a,\mathrm{\infty })$ for $a\ge t_n$ is at most $e(a{)}^{1-\rho }\mathrm{/}[c(1-\rho )]$, which tends to zero by step 1.2. For any $t\ge t_n$ and any later subsequence time $t_j\ge t$, the distance $d(q(t),q(t_j))$ is at most this remaining length. Letting $j\to \mathrm{\infty }$ and using $q(t_j)\to p$ proves $d(q(t),p)\to 0$ as $t\to \mathrm{\infty }$. Thus the trajectory converges to the critical point $p$ and has finite tail length.