Existence, uniqueness, and stability of Wasserstein gradient flows

First assume that $\mathrm{\Sigma }$ is countably generated. By [F3] and [F4], its countable generating family generates a countable algebra ${\mathcal{A}}_0$ such that each $E\in \mathrm{\Sigma }$ and each $\epsilon >0$ admit $A\in {\mathcal{A}}_0$ with $\sigma (E\mathrm{△}A)<\epsilon$. Relative Borel spaces $S\subseteq {\mathbb{R}}^k$ have countable generators obtained by intersecting $S$ with rational open boxes; every relative open set is a union of members of this countable base. Finite sigma-algebras are countably generated as well. These observations do not identify the label dimension $k$ with the output dimension $d$.

The functions ${\sum }_{j=1}^m{q}_j{\mathbf{1}}_{A_j}$, with $q_j\in \mathbb{Q}$ and $A_j\in {\mathcal{A}}_0$, form a countable subset of scalar $L^2(\sigma )$. It is dense: by [F2], approximate any $f$ by a finite-measure simple function ${\sum }_j{c}_j{\mathbf{1}}_{E_j}$; approximate each $E_j$ in measure using [F4], then approximate each coefficient by a rational. The error satisfies $\mathrm{\parallel }c{\mathbf{1}}_E-q{\mathbf{1}}_A{\mathrm{\parallel }}_2\le \mathrm{\mid }c-q\mathrm{\mid }\sqrt{\sigma (E)}+\mathrm{\mid }q\mathrm{\mid }\sqrt{\sigma (E\mathrm{△}A)}$, so the finitely many choices make the total error arbitrarily small.

The vector space $K$ is the finite product of $d$ copies of scalar $L^2(\sigma )$. By pointwise [F5], $2\mathrm{\mid }X\cdot Y\mathrm{\mid }\le \mathrm{\mid }X{\mathrm{\mid }}^2+\mathrm{\mid }Y{\mathrm{\mid }}^2$, so its displayed pairing is integrable and independent of representatives. Linearity of the integral and the Euclidean dot product give bilinearity and symmetry. Its value on $(X,X)$ is $\int \mathrm{\mid }X{\mathrm{\mid }}^2\ge 0$, and equality forces $\mathrm{\mid }X\mathrm{\mid }=0$ almost everywhere, precisely the zero class. Thus it is an inner product, with the stated norm. A Cauchy sequence for that norm is Cauchy in each of the finitely many scalar coordinates; [F1] gives coordinate limits, and the sum of their squared errors tends to zero. This proves completeness for every probability space, without a countable-generation hypothesis. When $\mathrm{\Sigma }$ is countably generated, the finite product of the countable dense scalar sets from step 2.1 is countable and dense in $K$, proving separability.

Pointwise Cauchy–Schwarz and scalar Cauchy–Schwarz show that the displayed pairing is integrable and bounded by $\mathrm{\parallel }u{\mathrm{\parallel }}_Z\mathrm{\parallel }v{\mathrm{\parallel }}_Z$. It is bilinear and symmetric; if $\int \mathrm{\parallel }u{\mathrm{\parallel }}_H^2=0$, then $\mathrm{\parallel }u(t){\mathrm{\parallel }}_H=0$ almost everywhere, so the equivalence class is zero. Thus it is a real inner product.

Let $(u_n)$ be Cauchy in $Z$. By [F4], choose a subsequence $(u_{n_k})$ with $\mathrm{\parallel }{u}_{n_{k+1}}-u_{n_k}{\mathrm{\parallel }}_Z\le 2^{-k}$. Use [F4] to choose strongly measurable representatives and modify them to zero on their exceptional null sets. Put $g_m(t)={\sum }_{k=1}^m\mathrm{\parallel }{u}_{n_{k+1}}(t)-u_{n_k}(t){\mathrm{\parallel }}_H$. Minkowski gives $\mathrm{\parallel }{g}_m{\mathrm{\parallel }}_2\le {\sum }_{k\ge 1}{2}^{-k}=1$; monotone convergence then gives $g={\sum }_k\mathrm{\parallel }{u}_{n_{k+1}}-u_{n_k}{\mathrm{\parallel }}_H\in L^2$. Hence $g<\mathrm{\infty }$ almost everywhere, and completeness of $H$ gives a pointwise limit $u(t)$ there; put $u=0$ on the null set.

The limit $u$ is strongly measurable: the chosen representatives are strongly measurable and converge pointwise off a measurable null set, so their simple approximations give a strongly measurable limit. The tail estimate $\mathrm{\parallel }u-u_{n_m}{\mathrm{\parallel }}_H\le {\sum }_{k\ge m}\mathrm{\parallel }{u}_{n_{k+1}}-u_{n_k}{\mathrm{\parallel }}_H$ and [F2] show $\mathrm{\parallel }u-u_{n_m}{\mathrm{\parallel }}_Z\le {\sum }_{k\ge m}{2}^{-k}\to 0$. Thus $u\in Z$ and the subsequence converges in norm. Since the original sequence is Cauchy, the full sequence converges to the same class; $Z$ is complete.

Fix the given countable dense set $D\subset H$. From strong measurability choose finite-valued simple functions $s_n$ converging pointwise almost everywhere to a representative $u$. Replace $s_n$ by $t_n=s_n{\mathbf{1}}_{\{\mathrm{\parallel }{s}_n{\mathrm{\parallel }}_H\le 2\mathrm{\parallel }u{\mathrm{\parallel }}_H\}}$; then $t_n\to u$ pointwise and $\mathrm{\parallel }u-t_n{\mathrm{\parallel }}_H^2\le 9\mathrm{\parallel }u{\mathrm{\parallel }}_H^2$, so dominated convergence makes finite-valued simple functions dense in $Z$. Let $\mathcal{A}$ be the algebra generated by rational intervals restricted to $(0,T)$. It is countable and generates the Borel sigma-algebra. To verify approximation by this particular algebra, let $\mathcal{M}$ consist of Borel sets approximable in symmetric-difference measure by members of $\mathcal{A}$. It contains $\mathcal{A}$, is closed under complements because symmetric differences are preserved by complements, and is closed under finite unions by subadditivity. For a countable union, first approximate it by a finite subunion: the measure of the omitted part tends to zero since $(0,T)$ has finite measure. Then approximate that finite union by $\mathcal{A}$. Thus $\mathcal{M}$ is a sigma-algebra and contains all Borel sets. For a simple function on a finite measurable partition, approximate its cells by $\mathcal{A}$, disjointify the approximating cells, and assign the complement to the last cell. The new and old partitions disagree only inside the union of the cellwise symmetric differences. Replace each value by a nearby element of $D$. The integral squared error is bounded by a constant times the measure of that disagreement plus $T$ times the maximal squared value error. It tends to zero. The resulting functions take values in $D$ and have finitely many rational breakpoints within $(0,T)$; there are only countably many of them. They are dense, proving separability.

Let $S$ be the vector space of all finite $H$-valued interval step functions. It contains the dense family in [F1], hence is dense in $Z$. If $f(t)={\sum }_{i=0}^{N-1}{c}_i{\mathbf{1}}_{(t_i,t_{i+1})}(t)$ on a partition $P$, define $\mathrm{\Lambda }(f)={\sum }_i⟨c_i,x(t_{i+1})-x(t_i){⟩}_H$. Refining $P$ leaves this sum unchanged, so $\mathrm{\Lambda }$ is well-defined and linear on $S$. Finite-sum Cauchy–Schwarz gives $\mathrm{\mid }\mathrm{\Lambda }(f)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_Z{Q}_P(x{)}^{1\mathrm{/}2}\le \sqrt{C}\mathrm{\parallel }f{\mathrm{\parallel }}_Z$. By [F1], $\mathrm{\Lambda }$ extends uniquely to a bounded linear functional on $Z$, and [F2] represents it by $u\in Z$.

Fix $0\le s<t\le T$ and $c\in H$. The function $c{\mathbf{1}}_{(s,t)}$ belongs to $S$, so its defining value is $\mathrm{\Lambda }(c{\mathbf{1}}_{(s,t)})=⟨c,x(t)-x(s){⟩}_H$. By the Riesz representation in step 1.1 and commutation of bounded linear functionals with Bochner integrals, this equals ${\int }_s^t⟨c,u(r){⟩}_{H}dr$. The identity holds for every $c\in H$; take $c$ equal to the difference of $x(t)-x(s)$ and ${\int }_s^{t}u(r)dr$ to see that its squared norm is zero. Thus $x(t)-x(s)={\int }_s^{t}u(r)dr$ in $H$.

By [F3], choose simple $H$-valued functions $s_n$ with ${\int }_0^T\mathrm{\parallel }u-s_n{\mathrm{\parallel }}_{H}dt<2^{-3n}$ and put $g_n(t)=\mathrm{\parallel }u(t)-s_n(t){\mathrm{\parallel }}_H$. Markov's inequality gives $\mathrm{Leb}\{g_n>2^{-n}\}\le 2^{-2n}$; [F7] shows that outside a null set $g_n(t)\le 2^{-n}$ eventually. Remove also the countable union of differentiation exceptional sets for all scalar functions $g_n$ and the finitely many indicators defining each $s_n$, using [F5] and [F6]. At each remaining interior $t$, the average of $g_n$ over either one-sided interval with endpoint $t$ has limsup at most $2g_n(t)$, since that interval lies in the centered interval of twice its length. The average of $\mathrm{\parallel }{s}_n(q)-s_n(t){\mathrm{\parallel }}_H$ tends to zero: $s_n$ has finitely many values and each level-set indicator has density $1$ or $0$ at $t$. Therefore ${\displaystyle \underset{h\to 0}{\mathrm{lim\hspace{0.17em}sup}}{\parallel \frac{1}{h}{\int }_t^{t+h}u(q)dq-u(t)\parallel }_H\le 3g_n(t)}$ for every $n$, and letting $n\to \mathrm{\infty }$ proves the limit is zero. Step 2.1 identifies this average with $(x(t+h)-x(t))\mathrm{/}h$, proving norm differentiability of $x$ with derivative $u$ almost everywhere.

The norm of $\mathrm{\Lambda }$ is exactly $\sqrt{C}$. The upper bound is step 1.1. For any partition $P$ with $Q_P(x)>0$, take the step function $f$ with $c_i=(x(t_{i+1})-x(t_i))\mathrm{/}(t_{i+1}-t_i)$ on each interval. It belongs to $S$ and satisfies $\mathrm{\Lambda }(f)=\mathrm{\parallel }f{\mathrm{\parallel }}_Z^2=Q_P(x)$, so $\mathrm{\parallel }\mathrm{\Lambda }{\mathrm{\parallel }}_Z\ge \mathrm{\mid }\mathrm{\Lambda }(f)\mathrm{\mid }\mathrm{/}\mathrm{\parallel }f{\mathrm{\parallel }}_Z=\sqrt{Q_P(x)}$. If $Q_P(x)=0$, this lower bound is automatic. Taking the supremum over partitions gives $\mathrm{\parallel }\mathrm{\Lambda }{\mathrm{\parallel }}_Z^2=C$. By [F2], $\mathrm{\parallel }u{\mathrm{\parallel }}_Z=\mathrm{\parallel }\mathrm{\Lambda }{\mathrm{\parallel }}_Z$, proving the displayed energy identity.

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.