A weak two-homogeneous projection limit is stationary on its support

Fix $T<\mathrm{\infty }$. W2 continuity bounds $M_2={\sup }_{0\le t\le T}\int \mathrm{\mid }u{\mathrm{\mid }}^{2}d{\mu }_t(u)<\mathrm{\infty }$. Positive two-homogeneity and continuity give a constant $C$ such that $\mathrm{\parallel }\mathrm{\Phi }(u)\mathrm{\parallel }\le C\mathrm{\mid }u{\mathrm{\mid }}^2$, $\mathrm{\parallel }d{\mathrm{\Phi }}_u\mathrm{\parallel }\le C\mathrm{\mid }u\mathrm{\mid }$, and $\mathrm{\mid }V(u)\mathrm{\mid }\le C\mathrm{\mid }u{\mathrm{\mid }}^2$; for the derivative bound, the neighborhood condition puts the unit sphere in some cutoff $Q_r$ where $d\mathrm{\Phi }$ is Lipschitz and bounded; differentiating homogeneity away from $0$ then gives $\mathrm{\parallel }d{\mathrm{\Phi }}_u\mathrm{\parallel }\le C\mathrm{\mid }u\mathrm{\mid }$. Thus $m_t=\int \mathrm{\Phi }d{\mu }_t$ is bounded, and $f_t=\mathrm{\nabla }R(m_t)$ is bounded on $[0,T]$. By [F2], choose a path law $\eta$ with marginals ${\mu }_t$ and $\dot{\gamma }(t)=v_t(\gamma (t))$ almost everywhere. Pathwise FTC for $\mathrm{\Phi }$, Cauchy--Schwarz, and scalar/Bochner Fubini [F6] give $m_t-m_s={\int }_s^t\int d{\mathrm{\Phi }}_u[v_q(u)]d{\mu }_q(u)dq,$ since ${\int }_0^T\int \mathrm{\parallel }d{\mathrm{\Phi }}_u\mathrm{\parallel }\mathrm{\mid }{v}_q(u)\mathrm{\mid }d{\mu }_{q}dq\le C(T{M}_2{)}^{1\mathrm{/}2}({\int }_0^{T}A(q)dq{)}^{1\mathrm{/}2}<\mathrm{\infty }$. Hence $m_t$ is absolutely continuous. For almost every $(t,u)$, write the selected subgradient as $p_t(u)=d{\mathrm{\Phi }}_u^{\ast }{f}_t+{\xi }_t(u)$ with ${\xi }_t(u)\in {\mathrm{\partial }}_{{\lambda }_V}V(u)$; subtracting the first-order expansion of the smooth term in the defining subdifferential inequality gives this decomposition. The flow definition gives $p_t=-v_t$. Along almost every path, $V\circ \gamma$ is absolutely continuous because $V$ is locally Lipschitz. At times when $\gamma$, $V\circ \gamma$ are differentiable and the supporting inequality holds, testing that inequality at $\gamma (t+h)$ for both signs of $h$ gives $(V\circ \gamma {)}^{\mathrm{\prime }}(t)=⟨{\xi }_t(\gamma (t)),\dot{\gamma }(t)⟩$. These derivatives are integrable: $\mathrm{\mid }{\xi }_t(u)\mathrm{\mid }\le \mathrm{\mid }{v}_t(u)\mathrm{\mid }+C\mathrm{\mid }{f}_t\mathrm{\mid }\mathrm{\mid }u\mathrm{\mid }$, so ${\int }_0^T\int \mathrm{\mid }{\xi }_t\mathrm{\mid }\mathrm{\mid }{v}_t\mathrm{\mid }d{\mu }_{t}dt<\mathrm{\infty }$. Taking expectations and using the marginals shows that $P_t=\int Vd{\mu }_t$ is absolutely continuous with $P_t^{\mathrm{\prime }}=\int {\xi }_t\cdot v_{t}d{\mu }_t$. The chain rule for $R(m_t)$ now yields $\frac{d}{dt}F({\mu }_t)=⟨f_t,m_t^{\mathrm{\prime }}⟩+P_t^{\mathrm{\prime }}=\int (d{\mathrm{\Phi }}_u^{\ast }{f}_t+{\xi }_t(u))\cdot v_t(u)d{\mu }_t(u)=-A(t)$ almost everywhere. Absolute integrability and the scalar FTC give the all-endpoint energy identity. Since $F\ge 0$, it follows that ${\int }_0^{\mathrm{\infty }}A(t)dt\le F({\mu }_0)<\mathrm{\infty }$.

Fix $t$ and $u\mathrm{\ne }0$. Write $P_t(u)=d{R}_{m_{{\mu }_t}}[\mathrm{\Phi }(u)]+V(u)$. If $\xi \in \mathrm{\partial }{P}_t(u)$, the local subdifferential definition applied along $(1+s)u$ gives ${\mathrm{lim\hspace{0.17em}inf}}_{s\to 0}\frac{P_t((1+s)u)-P_t(u)-s⟨\xi ,u⟩}{\mathrm{\mid }s\mathrm{\mid }\mathrm{\mid }u\mathrm{\mid }}\ge 0.$ Since $P_t((1+s)u)=(1+s{)}^2{P}_t(u)$, the one-sided limits for $s\downarrow 0$ and $s\uparrow 0$ are respectively $2P_t(u)-⟨\xi ,u⟩$ and its negative. Both must be nonnegative, so $⟨\xi ,u⟩=2P_t(u)$. By [F1], $v_t(u)$ is the negative of one such vector, giving $⟨v_t(u),u⟩=-2P_t(u)$.

For $s>0$, substitute $z=sy$ in the local subdifferential definition of $\mathrm{\partial }{P}_t(su)$. Two-homogeneity gives $P_t(z)-P_t(su)-s\xi \cdot (z-su)=s^2(P_t(y)-P_t(u)-\xi \cdot (y-u))$, while $\mathrm{\mid }z-su\mathrm{\mid }=s\mathrm{\mid }y-u\mathrm{\mid }$. Thus $\xi \in \mathrm{\partial }{P}_t(u)$ iff $s\xi \in \mathrm{\partial }{P}_t(su)$; the reverse implication follows by scaling by $1\mathrm{/}s$. Hence the minimum-norm selection scales by $s$, and $v_t(s\theta )=s{v}_t(\theta )$ wherever the canonical pointwise selection is defined.

Apply [F3] to the weakly convergent projections. Then $m_{{\mu }_t}\to m_{\nu }$ in Hilbert norm. The differentials $d{R}_{m_{{\mu }_t}}$ converge in operator norm to $d{R}_{m_{\nu }}$ by local Lipschitz continuity; since $\mathrm{\Phi }$ and $V$ are bounded on the sphere, $G_t(\theta ):=P_t(\theta )$ converges uniformly to $G_{\nu }(\theta )$. In particular, weak convergence and boundedness of the masses give $a(t):={\int }_{{\mathbb{S}}^{d-1}}{G}_t(\theta {)}^{2}d{h}^2({\mu }_t)(\theta )⟶{\int }_{{\mathbb{S}}^{d-1}}{G}_{\nu }(\theta {)}^{2}d\nu (\theta )=:a_{\mathrm{\infty }}\mathrm{.}$

By [F5] and the uniform convergence in step 3.1, $a(t)$ is measurable. By [F3] and [F4], positive one-homogeneity of the velocity gives $\int \mathrm{\mid }{v}_t(u){\mathrm{\mid }}^{2}d{\mu }_t(u)=\int \mathrm{\mid }{v}_t(\theta ){\mathrm{\mid }}^{2}d{h}^2({\mu }_t)(\theta )$. Step 1.2 on the unit sphere gives $2G_t(\theta )=-⟨v_t(\theta ),\theta ⟩$, so $4a(t)\le \int \mathrm{\mid }{v}_t(\theta ){\mathrm{\mid }}^{2}d{h}^2({\mu }_t)(\theta )$. Integrating and using step 1.1 yields ${\int }_0^{\mathrm{\infty }}a(t)dt\le F({\mu }_0)\mathrm{/}4<\mathrm{\infty }$. A nonnegative function with a finite integral cannot converge to a positive limit; hence $a_{\mathrm{\infty }}=0$. Since $G_{\nu }^2$ is nonnegative, its zero integral implies $G_{\nu }=0$ $\nu$-almost everywhere.