Conditional gradient inequality gives a length bound

For any $\beta >0$, the positive-base power has strictly positive derivative by [F2], so the mean value theorem makes $s\mapsto s^{\beta }$ strictly increasing on $(0,\mathrm{\infty })$. Given $\epsilon >0$, put $\delta ={\epsilon }^{1\mathrm{/}\beta }>0$. If $0<s<\delta$, the iterated-power law gives $0<s^{\beta }<{\delta }^{\beta }=\epsilon$; for $s=0$ the zero-base convention gives the same upper bound. Thus $s^{\beta }\to 0$ as $s\downarrow 0$, and the defined power is continuous on $[0,\mathrm{\infty })$. This applies to both $\beta =\rho$ and $\beta =1-\rho$.

If $a=b$, both the length and the stated numerator are zero. Hence assume $a<b$. Write $h(t)=f(x(t))-f(z)\ge 0$ and $g(t)=\mathrm{\parallel }\mathrm{\nabla }f(x(t))\mathrm{\parallel }$. The energy identity, or its interior chain-rule calculation, gives $h^{\mathrm{\prime }}=-g^2$ and $\mathrm{\parallel }{x}^{\mathrm{\prime }}\mathrm{\parallel }=g$; $h$ and $g$ are continuous on $[a,b]$. Where $h>0$, the power derivative, chain rule, and $g\ge c{h}^{\rho }$ give $-(h^{1-\rho }{)}^{\mathrm{\prime }}=(1-\rho )h^{-\rho }{g}^2\ge c(1-\rho )g$.

At any $q\in W$ with $f(q)=f(z)$, openness of $W$ and $f\ge f(z)$ make $q$ a local minimum. For each coordinate, the positive and negative difference quotients at $q$ have opposite weak signs; existence of the partial derivative forces it to be zero. Thus $\mathrm{\nabla }f(q)=0$ and the constant curve through $q$ solves the gradient equation. The gradient is locally Lipschitz: on a small closed ball in $U$, each continuous second partial derivative is bounded by compactness and extreme values; applying the chain rule and the one-variable mean value estimate along segments bounds each first-partial increment by a constant times the distance. Combining the finitely many coordinate bounds gives a Lipschitz bound for the gradient. ODE uniqueness therefore makes the given trajectory constant for the rest of $[a,b]$ after any time at which $h=0$. Its speed and all subsequent length increments are zero.

On a nondegenerate closed subinterval $[u,v]\subseteq [a,b]$ where $h>0$, both $g$ and $d(t)=(1-\rho )h(t{)}^{-\rho }g(t{)}^2$ are continuous, hence integrable. Newton–Leibniz applies to $-h^{1-\rho }$ with derivative extension $d$. Integral comparison and scalar linearity therefore give $c(1-\rho ){\int }_u^{v}g(t)dt\le {\int }_u^{v}d(t)dt=h(u{)}^{1-\rho }-h(v{)}^{1-\rho }$.

On the whole interval, continuity of $g$ gives integrability and a finite bound $0\le g\le K$. The integral function $L(t)={\int }_a^{t}g(s)ds$ is therefore defined and satisfies $\mathrm{\mid }L(v)-L(u)\mathrm{\mid }\le K\mathrm{\mid }v-u\mathrm{\mid }$. In particular, $L(t)\to L(\tau )$ as $t\uparrow \tau$, for every $a<\tau \le b$.

If $h(a)=0$, step 1.3 gives length zero and $h(b)=0$, proving the bound. Otherwise $h(a)>0$. If $h$ has no zero on $[a,b]$, use step 2.1 with $u=a$ and $v=b$. If it has a zero, its zero set is closed by continuity, bounded, and nonempty; compactness and the minimum of the identity function give its first element $\tau >a$. For $a<t<\tau$, step 2.1 gives $c(1-\rho )L(t)\le h(a{)}^{1-\rho }-h(t{)}^{1-\rho }$. As $t\uparrow \tau$, continuity of $h$, the zero-endpoint power limit in step 1.1, and step 2.2 give $c(1-\rho )L(\tau )\le h(a{)}^{1-\rho }$. Step 1.3 gives $L(b)=L(\tau )$ and $h(b)=0$. Dividing by $c(1-\rho )>0$ proves exactly the asserted estimate in every case.