Energy dissipation along a downward gradient curve

Fix $t\ge 0$, put $p=q(t)$, choose a smooth chart $x$ around $p$, and write $y=x\circ q$ near $t$. By [F6], $y$ is $C^1$, with the right derivative at $0$. By [F7], $V_t={\sum }_i{y}_i^{\mathrm{\prime }}(t){\mathrm{\partial }}_i{\mathrm{\mid }}_p$ is a tangent vector. For any smooth local function $h$ near $p$, [F7], [F8] and the Euclidean chain rule give $d{h}_p(V_t)={\sum }_i{y}_i^{\mathrm{\prime }}(t){\mathrm{\partial }}_i(h\circ x^{-1})(y(t))=(h\circ q{)}^{\mathrm{\prime }}(t)$ when $t>0$. At zero, substitute $y(s)=y(0)+s{y}_{+}^{\mathrm{\prime }}(0)+o(s)$ as $s\downarrow 0$ into the first-order expansion of $h\circ x^{-1}$ to obtain the same identity with right derivatives. These identities determine the action of $V_t$ on every smooth germ by [F8], independently of the chart. Thus $V_t$ is the intrinsic tangent velocity denoted by $q^{\mathrm{\prime }}(t)$, including its right endpoint value.

Put $f(t)=U(q(t))$. Step 1.1 with $h=U$ gives $f^{\mathrm{\prime }}(t)=d{U}_{q(t)}(q^{\mathrm{\prime }}(t))$, with the right derivative at zero. The chartwise $C^1$ curve and the smooth potential make this derivative continuous, including from the right at zero. Substituting the downward gradient equation and the metric-dual identity yields $f^{\mathrm{\prime }}(t)=-g_{q(t)}({\mathrm{grad}}_{g}U,{\mathrm{grad}}_{g}U)=-\mathrm{\parallel }{\mathrm{grad}}_{g}U(q(t)){\mathrm{\parallel }}_g^2\le 0$.

For $0\le a<b$, the mean value theorem applied to $f$ on $[a,b]$ gives $f(b)-f(a)=f^{\mathrm{\prime }}(c)(b-a)\le 0$ for some $c\in (a,b)$. Thus $f$ is nonincreasing.

For $T>0$, extend $f$ to a continuously differentiable real function on an open interval containing $[0,T]$ by setting $f(t)=f(0)+t{f}_{+}^{\mathrm{\prime }}(0)$ for $t<0$ near zero. The fundamental theorem and step 2.1 give ${\int }_0^T\mathrm{\parallel }{\mathrm{grad}}_{g}U(q(t)){\mathrm{\parallel }}_g^{2}dt=f(0)-f(T)$.

Suppose $f$ is bounded below and set $L=\inf \{f(t):t\ge 0\}$. This is finite because the set is nonempty and bounded below. For every $\epsilon >0$ there is $s\ge 0$ with $f(s)<L+\epsilon$; for all $t\ge s$, monotonicity gives $L\le f(t)\le f(s)<L+\epsilon$. Hence $f(t)\to L$ as $t\to \mathrm{\infty }$. Taking the limit in step 3.2 proves the stated improper integral identity. Since $q^{\mathrm{\prime }}=-{\mathrm{grad}}_{g}U$, its squared speed is the integrand.