Smoothed margin monotonicity along homogeneous Clarke flow

We establish the identities of [F4] for the arbitrary margins allowed here. At any fixed parameter $x$, put $a_i=e^{-q_i(x)}$ and $f_i=e^{-q_i}$. Since $q_i=-\log f_i$, the two functions have exactly the same differentiability points, and $\mathrm{\nabla }{f}_i=-e^{-q_i}\mathrm{\nabla }{q}_i$ there. Taking gradient limits and closed convex hulls in [F1] gives ${\mathrm{\partial }}^{\circ }{f}_i(x)=-a_i{\mathrm{\partial }}^{\circ }{q}_i(x)$. Set $C={\sum }_i{\mathrm{\partial }}^{\circ }{f}_i(x)$, a nonempty compact convex set. For any direction $w$, the finite-sum limsup inequality and [F1] give ${\mathcal{L}}^{\circ }(x;w)\le {\sum }_i{f}_i^{\circ }(x;w)={\max }_{c\in C}⟨c,w⟩$. Thus ${\mathrm{\partial }}^{\circ }\mathcal{L}(x)\subseteq C$: otherwise, for $g\in {\mathrm{\partial }}^{\circ }\mathcal{L}(x)\setminus C$, let $c_0$ minimize $\mathrm{\parallel }g-c{\mathrm{\parallel }}_2$ on $C$, using [F5]. Convexity gives $⟨g-c_0,c-c_0⟩\le 0$ for every $c\in C$, whereas $⟨g,g-c_0⟩>⟨c_0,g-c_0⟩$, contradicting the directional bound. Consequently the flow inclusion supplies $\dot{\theta }={\sum }_i{e}^{-q_i}{h}_i$ with $h_i\in {\mathrm{\partial }}^{\circ }{q}_i(\theta )$ almost everywhere. The assumed arc chain rules give ${\mathcal{L}}^{\mathrm{\prime }}=-⟨{\sum }_i{e}^{-q_i}{h}_i,\dot{\theta }⟩=-\mathrm{\parallel }\dot{\theta }{\mathrm{\parallel }}_2^2$ almost everywhere. All these compositions are locally absolutely continuous, since the margins are locally Lipschitz. No graph realization is needed.

Put $\varphi (t)=\mathcal{L}(\theta (t))$, $r(t)=\mathrm{\parallel }\theta (t){\mathrm{\parallel }}_2$, and $z(t)=-\log \varphi (t)$. By [F1], $\theta$ is locally absolutely continuous; local Lipschitz continuity of the margins makes these compositions locally absolutely continuous on each compact time interval where $r>0$. Since $\varphi (t_0)<1$ and [step 1.1] gives ${\varphi }^{\mathrm{\prime }}=-\mathrm{\parallel }\dot{\theta }{\mathrm{\parallel }}_2^2\le 0$ almost everywhere, $\varphi (t)<1$ for all $t\ge t_0$. Thus every $q_i(\theta (t))>0$. Homogeneity gives $q_i(0)=0$, so $\theta (t)\mathrm{\ne }0$ and $r(t)>0$.

At almost every $t\ge t_0$, set $a_i=e^{-q_i(\theta (t))}$, $p_i=a_i\mathrm{/}\varphi$, and $\mu ={\sum }_i{p}_i{q}_i(\theta (t))$. By [step 1.1], choose $h_i\in {\mathrm{\partial }}^{\circ }{q}_i(\theta (t))$ with $v:=\dot{\theta }={\sum }_i{a}_i{h}_i$. Euler's identity [F3] yields $⟨\theta ,v⟩=L\varphi \mu$, hence $(\log r{)}^{\mathrm{\prime }}=L\varphi \mu \mathrm{/}{r}^2>0$. Also $q_i=z-\log p_i$, so $\mu =z-{\sum }_i{p}_i\log p_i\ge z>0$ because $p_i>0$, ${\sum }_i{p}_i=1$, and $-\log p_i\ge 0$.

The chain rule in [step 1.1] gives $z^{\mathrm{\prime }}=\mathrm{\parallel }v{\mathrm{\parallel }}_2^2\mathrm{/}\varphi$. Differentiating $z\mathrm{/}{r}^L$ and applying Cauchy–Schwarz, ${\tilde{\gamma }}^{\mathrm{\prime }}=r^{-L}(\mathrm{\parallel }v{\mathrm{\parallel }}_2^2\mathrm{/}\varphi -L^{2}z\varphi \mu \mathrm{/}{r}^2)\ge L^2\varphi \mu (\mu -z)\mathrm{/}{r}^{L+2}\ge 0$, because $\mathrm{\parallel }v{\mathrm{\parallel }}_2^2\ge ⟨\theta ,v{⟩}^2\mathrm{/}{r}^2=L^2{\varphi }^2{\mu }^2\mathrm{/}{r}^2$. Thus the smoothed margin is nondecreasing.

The function $u\mapsto q_{\min }(u):={\min }_i{q}_i(u)$ is continuous on the unit sphere, so [F5] gives a finite maximum $C$. Homogeneity and $\varphi \ge e^{-q_{\min }}$ imply $0<\tilde{\gamma }(t_0)\le z\mathrm{/}{r}^L\le \bar{\gamma }\le C$. Since $\tilde{\gamma }$ is nondecreasing by step 4.1, it has a finite positive limit. Writing $A={\sum }_i{e}^{-(q_i-q_{\min })}\in [1,N]$ gives $z=q_{\min }-\log A$ and $0\le \bar{\gamma }-z\mathrm{/}{r}^L=\log A\mathrm{/}{r}^L\le (\log N)\mathrm{/}{r}^L$.

The nonincreasing loss has a limit ${\varphi }_{\mathrm{\infty }}\ge 0$. If ${\varphi }_{\mathrm{\infty }}>0$, then $z\ge z(t_0)>0$ and $(r^2{)}^{\mathrm{\prime }}=2L\varphi \mu \ge 2L{\varphi }_{\mathrm{\infty }}z(t_0)>0$ almost everywhere by step 3.1, so $r\to \mathrm{\infty }$ while $z$ remains bounded. This forces $z\mathrm{/}{r}^L\to 0$, contradicting the positive lower bound in step 5.1. Hence $\varphi \to 0$. To show $r\to \mathrm{\infty }$, suppose otherwise. Then some $R<\mathrm{\infty }$ and times $t_k\to \mathrm{\infty }$ satisfy $r(t_k)\le R$. The closed ball of radius $R$ is compact by [F5], and each continuous $q_i$ is bounded above there. Since $N\ge 1$, this gives a common positive lower bound for $\varphi (t_k)={\sum }_i{e}^{-q_i(\theta (t_k))}$, contradicting $\varphi (t)\to 0$. Thus $r\to \mathrm{\infty }$. The error bound in step 5.1 now tends to zero, so the normalized and smoothed margins converge to the same positive limit.