A lower-bounded semiconvex Hilbert energy generates a unique gradient flow

For $0<\tau$ with $1+\lambda \tau >0$ and $z\in H$, set ${\mathrm{\Psi }}_{\tau ,z}(y)=E(y)+\mathrm{\parallel }y-z{\mathrm{\parallel }}_H^2\mathrm{/}(2\tau )$. It is lower semicontinuous and $(\lambda +1\mathrm{/}\tau )$-strongly convex, with positive strong-convexity coefficient. It is bounded below and coercive because $E$ is bounded below. Choose a minimizing sequence $(y_n)$ using [F4]. The midpoint inequality ${\displaystyle {\mathrm{\Psi }}_{\tau ,z}\text{\hspace{0.17em}⁣}\left(\frac{y_n+y_m}{2}\right)\le \frac{{\mathrm{\Psi }}_{\tau ,z}(y_n)+{\mathrm{\Psi }}_{\tau ,z}(y_m)}{2}-\frac{1+\lambda \tau }{8\tau }\mathrm{\parallel }{y}_n-y_m{\mathrm{\parallel }}_H^2}$ and the lower bound by the infimum show that $(y_n)$ is Cauchy. Completeness of $H$ and lower semicontinuity give a minimizer, and strong convexity makes it unique. Call it $J_{\tau }z$. For $0<h<1$, set $y_h=(1-h)J_{\tau }z+hy$. Minimality at $y_h$, the $\lambda$-convex inequality, and expansion of the quadratic term give ${\displaystyle 0\le h(E(y)-E(J_{\tau }z))-\frac{\lambda }{2}h(1-h)\mathrm{\parallel }y-J_{\tau }z{\mathrm{\parallel }}_H^2-\frac{h}{\tau }⟨z-J_{\tau }z,y-J_{\tau }z{⟩}_H+\frac{h^2}{2\tau }\mathrm{\parallel }y-J_{\tau }z{\mathrm{\parallel }}_H^2\mathrm{.}}$ Divide by $h$ and let $h\downarrow 0$ to obtain ${\displaystyle E(y)\ge E(J_{\tau }z)+{⟨\frac{z-J_{\tau }z}{\tau },y-J_{\tau }z⟩}_H+\frac{\lambda }{2}\mathrm{\parallel }y-J_{\tau }z{\mathrm{\parallel }}_H^2,}$ so $(z-J_{\tau }z)\mathrm{/}\tau \in {\mathrm{\partial }}_{\lambda }E(J_{\tau }z)$.

Starting at $x_0$, iterate $x_n=J_{\tau }{x}_{n-1}$. Comparison with the admissible point $x_{n-1}$ gives ${\displaystyle E(x_n)+\frac{\mathrm{\parallel }{x}_n-x_{n-1}{\mathrm{\parallel }}_H^2}{2\tau }\le E(x_{n-1})\mathrm{.}}$ Thus $E(x_n)$ decreases and, for $m={\inf }_{H}E$, ${\sum }_n\tau \mathrm{\parallel }(x_{n-1}-x_n)\mathrm{/}\tau {\mathrm{\parallel }}_H^2\le 2(E(x_0)-m)$.

Let $x_{\tau }$ be the linear interpolation between the iterates and ${\bar{x}}_{\tau }(t)=x_n$ on $(t_{n-1},t_n]$; then ${\dot{x}}_{\tau }=-{\xi }_{\tau }$ and ${\xi }_{\tau }(t)\in {\mathrm{\partial }}_{\lambda }E({\bar{x}}_{\tau }(t))$. The action bound in step 2.1 gives $\mathrm{\parallel }{\xi }_{\tau }{\mathrm{\parallel }}_{L^2(0,T;H)}^2\le 2(E(x_0)-m)$ and $\mathrm{\parallel }{x}_{\tau }-{\bar{x}}_{\tau }{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(0,T;H)}\le \sqrt{\tau }\mathrm{\parallel }{\xi }_{\tau }{\mathrm{\parallel }}_{L^2(0,T;H)}$. For two uniform meshes $\tau ,\sigma$, put ${\delta }_{\tau }=x_{\tau }-{\bar{x}}_{\tau }$ and $q(t)=\mathrm{\parallel }{x}_{\tau }(t)-x_{\sigma }(t){\mathrm{\parallel }}_H^2$. The subgradient monotonicity obtained by adding the defining inequalities for ${\mathrm{\partial }}_{\lambda }E$ gives ${\displaystyle \frac{1}{2}{q}^{\mathrm{\prime }}(t)\le -\lambda \mathrm{\parallel }{\bar{x}}_{\tau }(t)-{\bar{x}}_{\sigma }(t){\mathrm{\parallel }}_H^2+(\mathrm{\parallel }{\xi }_{\tau }(t){\mathrm{\parallel }}_H+\mathrm{\parallel }{\xi }_{\sigma }(t){\mathrm{\parallel }}_H)(\mathrm{\parallel }{\delta }_{\tau }(t){\mathrm{\parallel }}_H+\mathrm{\parallel }{\delta }_{\sigma }(t){\mathrm{\parallel }}_H)\mathrm{.}}$ Using $\mathrm{\parallel }{\delta }_{\tau }(t){\mathrm{\parallel }}_H\le \tau \mathrm{\parallel }{\xi }_{\tau }(t){\mathrm{\parallel }}_H$ on each step and Cauchy–Schwarz in time, the interpolation-error terms obey ${\displaystyle {\int }_0^T(\mathrm{\parallel }{\xi }_{\tau }\mathrm{\parallel }+\mathrm{\parallel }{\xi }_{\sigma }\mathrm{\parallel })(\mathrm{\parallel }{\delta }_{\tau }\mathrm{\parallel }+\mathrm{\parallel }{\delta }_{\sigma }\mathrm{\parallel })dt\le 2(\tau +\sigma )(\mathrm{\parallel }{\xi }_{\tau }{\mathrm{\parallel }}_{L^2}^2+\mathrm{\parallel }{\xi }_{\sigma }{\mathrm{\parallel }}_{L^2}^2),{\int }_0^T\mathrm{\parallel }{\delta }_{\tau }-{\delta }_{\sigma }{\mathrm{\parallel }}^{2}dt\le 2{\tau }^2\mathrm{\parallel }{\xi }_{\tau }{\mathrm{\parallel }}_{L^2}^2+2{\sigma }^2\mathrm{\parallel }{\xi }_{\sigma }{\mathrm{\parallel }}_{L^2}^2\mathrm{.}}$ When $\lambda \ge 0$ its term in the differential inequality is nonpositive; when $\lambda <0$, $\mathrm{\parallel }{\bar{x}}_{\tau }-{\bar{x}}_{\sigma }{\mathrm{\parallel }}^2\le 2q+2\mathrm{\parallel }{\delta }_{\tau }-{\delta }_{\sigma }{\mathrm{\parallel }}^2$ bounds it by $2\mathrm{\mid }\lambda \mathrm{\mid }q+2\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\parallel }{\delta }_{\tau }-{\delta }_{\sigma }{\mathrm{\parallel }}^2$. Since the action norms are uniformly bounded by step 2.1, for $0<\tau ,\sigma \le 1$ the differential inequality has the form $q^{\mathrm{\prime }}(t)\le 4\mathrm{\mid }\lambda \mathrm{\mid }q(t)+r(t)$, $q(0)=0$, with ${\int }_0^{T}r(t)dt\le C(\tau +\sigma )$ and $C$ depending only on $T,\lambda ,E(x_0)-m$. Multiplying by $e^{-4\mathrm{\mid }\lambda \mathrm{\mid }t}$ and integrating yields ${\sup }_{t\le T}q(t)\le C_T(\tau +\sigma )$, so the interpolants converge uniformly on every bounded time interval.

Let $x$ be the uniform limit. Every finite partition $P$ satisfies ${\sum }_{[a,b]\in P}\mathrm{\parallel }x(b)-x(a){\mathrm{\parallel }}_H^2\mathrm{/}(b-a)\le {\mathrm{lim\hspace{0.17em}inf}}_{\tau \downarrow 0}{\int }_0^T\mathrm{\parallel }{\dot{x}}_{\tau }{\mathrm{\parallel }}_H^2\le 2(E(x_0)-m)$, by Cauchy–Schwarz on each mesh interval and uniform convergence. [F2] gives $x\in A{C}^2$ and a derivative. The step functions ${\xi }_{\tau }$ are bounded in the separable Hilbert space [F1]; apply [F7] successively to their inner products with a countable dense family and take a diagonal subsequence. The limits define a bounded linear functional on the dense span, hence on the whole space, and [F3] represents its extension by $\xi \in L^2(0,T;H)$. Thus ${\xi }_{\tau }\rightharpoonup \xi$. Since $x_{\tau }(t)-x_0=-{\int }_0^t{\xi }_{\tau }(r)dr$ and the Bochner integral on $(0,t)$ is a bounded linear map, weak convergence and uniform convergence imply $\dot{x}=-\xi$. Also ${\bar{x}}_{\tau }\to x$ uniformly.

For each fixed $y$ with $E(y)<\mathrm{\infty }$, integrate the discrete inequality from step 1.1 over any measurable $A\subset (0,T)$. Weak convergence of ${\xi }_{\tau }$ and uniform convergence of ${\bar{x}}_{\tau }$ pass the linear term to its limit; lower semicontinuity of $E$ and Fatou's lemma [F6], applied to $E({\bar{x}}_{\tau })-m\ge 0$, pass the energy term. Thus the subgradient inequality holds for almost every $t$, first for a countable dense subset of the graph (choose one graph point from each nonempty member of a countable base of $H\times \mathbb{R}$, using [F4]) $\{(y,E(y)):E(y)<\mathrm{\infty }\}$ and then for every such $y$ by continuity in the graph variables. Hence $\xi (t)\in {\mathrm{\partial }}_{\lambda }E(x(t))$ almost everywhere.

Work on $[0,T]$, choosing the construction on $[0,T+1]$ to allow small time translations. Put $u=\dot{x}=-\xi$ and $e(t)=E(x(t))$. The discrete bound $E({\bar{x}}_{\tau }(t))\le E(x_0)$, uniform convergence and lower semicontinuity give $m\le e(t)\le E(x_0)$ for every $t$, where $m={\inf }_{H}E$. As a lower-semicontinuous function of time, $e$ is measurable and bounded. Extend $u$ and $\xi$ by zero outside $(0,T+1)$. Time translations are continuous in $L^2(\mathbb{R};H)$: this is immediate for interval step functions because the measure of the symmetric difference of an interval with its translate tends to zero, and holds for every $L^2$ function by the density in [F1] and the isometry of translations. For $0<a<b<T$ and sufficiently small $h>0$, put $q_h(r)=(x(r+h)-x(r-h))\mathrm{/}(2h)=(2h{)}^{-1}{\int }_{-h}^{h}u(r+s)ds$ for $a<r<b$. Cauchy–Schwarz in the averaging variable and translation continuity give $q_h\to u$ in $L^2(a,b;H)$ and ${\sup }_h\mathrm{\parallel }{q}_h{\mathrm{\parallel }}_{L^2(a,b;H)}<\mathrm{\infty }$. Indeed, ${\int }_a^b\mathrm{\parallel }{q}_h-u{\mathrm{\parallel }}_H^{2}dr\le (2h{)}^{-1}{\int }_{-h}^h{\int }_a^b\mathrm{\parallel }u(r+s)-u(r){\mathrm{\parallel }}_H^{2}drds\to 0$.

For each sufficiently small fixed $h$, the subgradient inequalities at $x(r-h)$ and $x(r+h)$ hold for almost every $r\in (a,b)$ and imply $⟨\xi (r-h),q_h(r){⟩}_H+\lambda h\mathrm{\parallel }{q}_h(r){\mathrm{\parallel }}_H^2\le (e(r+h)-e(r-h))\mathrm{/}(2h)\le ⟨\xi (r+h),q_h(r){⟩}_H-\lambda h\mathrm{\parallel }{q}_h(r){\mathrm{\parallel }}_H^2$. Both pairings converge in $L^1(a,b)$ to $g(r)=⟨\xi (r),u(r){⟩}_H$, since translations of $\xi$ and $q_h$ converge in $L^2$ and Cauchy–Schwarz bounds their products. The quadratic remainder has $L^1$ norm at most $\mathrm{\mid }\lambda \mathrm{\mid }h\mathrm{\parallel }{q}_h{\mathrm{\parallel }}_{L^2(a,b;H)}^2\to 0$. The sandwich therefore implies convergence of the entire centered energy quotient to $g$ in $L^1(a,b)$: if $L_h\le d_h\le U_h$, then $\mathrm{\mid }{d}_h-g\mathrm{\mid }\le \mathrm{\mid }{L}_h-g\mathrm{\mid }+\mathrm{\mid }{U}_h-g\mathrm{\mid }$ almost everywhere.

For every $0<a<b<T$ the exact integral identity is ${\int }_a^b(e(r+h)-e(r-h))\mathrm{/}(2h)dr=(2h{)}^{-1}({\int }_{b-h}^{b+h}e(r)dr-{\int }_{a-h}^{a+h}e(r)dr)$. By [F8] there is a full-measure set $B\subset (0,T)$ on which centered averages recover $e$. For any $a,b\in B$ with $a<b$, Step 6.1 and this exact identity give $e(b)-e(a)={\int }_a^{b}g(r)dr$. Fix $a_{\ast }\in B$ and define $\tilde{e}(t)=e(a_{\ast })+{\int }_{a_{\ast }}^{t}g(r)dr$ on $[0,T]$. Here $g\in L^1(0,T)$ by Cauchy–Schwarz. Thus [F9] makes $\tilde{e}$ absolutely continuous and [F10] gives ${\tilde{e}}^{\mathrm{\prime }}=g$ almost everywhere, while the preceding equality implies $e=\tilde{e}$ on $B$. This proves an absolutely continuous representative by an integral identity, without inferring it from an almost-everywhere derivative.

Lower semicontinuity of $e$ and continuity of $\tilde{e}$ give $e(t)\le \tilde{e}(t)$ for every $t\in [0,T]$ by approaching $t$ through $B$. Fix $t\in (0,T]$. For sufficiently large $n$, put $I_n=(t-2^{-n},t-2^{-n-1})\subset (0,T)$ and choose $s_n\in I_n\cap B$ where the subgradient inclusion holds and $\mathrm{\parallel }\xi (s_n){\mathrm{\parallel }}_H^2\le 2\mathrm{\mid }{I}_n{\mathrm{\mid }}^{-1}{\int }_{I_n}\mathrm{\parallel }\xi (r){\mathrm{\parallel }}_H^{2}dr$. Such points exist since the excluded set is null and the upper threshold exceeds the mean, with the zero-integral case treated by choosing a point where $\xi =0$. Since $(t-s_n)\mathrm{/}\mathrm{\mid }{I}_n\mathrm{\mid }\le 2$, one has $(t-s_n)\mathrm{\parallel }\xi (s_n){\mathrm{\parallel }}_H^2\le 4{\int }_{I_n}\mathrm{\parallel }\xi {\mathrm{\parallel }}_H^2\to 0$. Consequently $\mathrm{\mid }⟨\xi (s_n),x(t)-x(s_n){⟩}_H\mathrm{\mid }\le ((t-s_n)\mathrm{\parallel }\xi (s_n){\mathrm{\parallel }}_H^2{)}^{1\mathrm{/}2}({\int }_{s_n}^t\mathrm{\parallel }u{\mathrm{\parallel }}_H^2{)}^{1\mathrm{/}2}\to 0$. The subgradient inequality at $s_n$ gives $e(t)\ge e(s_n)+⟨\xi (s_n),x(t)-x(s_n){⟩}_H+(\lambda \mathrm{/}2)\mathrm{\parallel }x(t)-x(s_n){\mathrm{\parallel }}_H^2$; using $e(s_n)=\tilde{e}(s_n)$ and continuity of $x,\tilde{e}$ yields $e(t)\ge \tilde{e}(t)$. At $t=0$, the bound $e(r)\le E(x_0)$ for all $r$ and equality on $B$ give $\tilde{e}(0)\le E(x_0)=e(0)$, proving equality there too. Thus $e=\tilde{e}$ on $[0,T]$ and, for every $0\le s\le t\le T$, $E(x(t))-E(x(s))={\int }_s^t⟨\xi (r),u(r){⟩}_{H}dr=-{\int }_s^t\mathrm{\parallel }\dot{x}(r){\mathrm{\parallel }}_H^{2}dr$. Since $T$ is arbitrary this holds for all finite endpoints.

At almost every $t$, let $\zeta \in {\mathrm{\partial }}_{\lambda }E(x(t))$. Applying its defining inequality to $x(t+h)$ for positive and negative $h$, then using differentiability of $x$ and $e$, gives $⟨\zeta ,\dot{x}(t){⟩}_H=-\mathrm{\parallel }\dot{x}(t){\mathrm{\parallel }}_H^2$. Since $\xi =-\dot{x}\in {\mathrm{\partial }}_{\lambda }E(x(t))$, this says $⟨\zeta -\xi ,\xi {⟩}_H=0$ and hence $\mathrm{\parallel }\zeta {\mathrm{\parallel }}_H^2=\mathrm{\parallel }\xi {\mathrm{\parallel }}_H^2+\mathrm{\parallel }\zeta -\xi {\mathrm{\parallel }}_H^2$. Thus $\xi$ is the unique minimum-norm subgradient almost everywhere.

For two solutions, $\lambda$-semiconvexity implies $⟨\xi -\tilde{\xi },x-\tilde{x}{⟩}_H\ge \lambda \mathrm{\parallel }x-\tilde{x}{\mathrm{\parallel }}_H^2$. Therefore $\frac{1}{2}\frac{d}{dt}\mathrm{\parallel }x-\tilde{x}{\mathrm{\parallel }}_H^2=-⟨\xi -\tilde{\xi },x-\tilde{x}{⟩}_H\le -\lambda \mathrm{\parallel }x-\tilde{x}{\mathrm{\parallel }}_H^2$ almost everywhere. Grönwall proves the stated estimate and uniqueness.