The localized lifted subdifferential is a pointwise constrained subdifferential

The set $C_r$ is convex because $Q_r$ is convex. To show it is closed, take $X_n\in C_r$ with $X_n\to X$ in $K$. Countable Choice [F3] gives a subsequence with $\mathrm{\parallel }{X}_{n_k}-X{\mathrm{\parallel }}_K^2\le 2^{-k}$ and strongly measurable representatives; discard their countably many exceptional null sets. Tonelli [F6] implies ${\sum }_k\mathrm{\mid }{X}_{n_k}(s)-X(s){\mathrm{\mid }}^2<\mathrm{\infty }$ for almost every $s$, so $X_{n_k}(s)\to X(s)$ there; closedness of $Q_r$ gives $X\in C_r$. For lower semicontinuity of $E_r$, consider a sequence with finite energy liminf and pass to a subsequence of finite energies, so every term lies in $C_r$ and its limit lies there by the closure just proved. For any $u\in Q_r$, a vector $\zeta \in {\mathrm{\partial }}_{{\lambda }_V}V(u)$ supplied by [F2] gives $V(y)\ge V(u)+\zeta \cdot (y-u)+\frac{{\lambda }_V}{2}\mathrm{\mid }y-u{\mathrm{\mid }}^2$ for $y\in \mathrm{\Omega }$, proving lower semicontinuity of $V$ at $u$. Fatou [F4] makes the potential integral lower semicontinuous along this sequence, and the smooth term is continuous by [F1]. A constant $Q_r$-valued function has finite energy because $Q_r$ is nonempty and $\mathrm{\Phi },V$ are bounded on it, while $R,V\ge 0$ make $E_r$ bounded below. Thus $E_r$ is proper, lower semicontinuous, and lower bounded.

Let $f=V+{\iota }_{Q_r}$. Since every point of $Q_r$ has a subgradient by [F2], the supporting inequality implies that $f$ is lower semicontinuous at every point of $Q_r$. The function $g(u)=f(u)-\frac{{\lambda }_V}{2}\mathrm{\mid }u{\mathrm{\mid }}^2$ is proper, lower semicontinuous and convex, with $\mathrm{\partial }g(u)={\mathrm{\partial }}_{{\lambda }_V}f(u)-{\lambda }_{V}u$. For $\epsilon >0$ small enough that $1+\epsilon {\lambda }_V>0$, minimize $f(y)+\mathrm{\mid }x-y{\mathrm{\mid }}^2\mathrm{/}(2\epsilon )$. This is strongly convex and coercive, so it has a unique minimizer $J_{\epsilon }x$. Its optimality inequality says $z_{\epsilon }(x)=(x-J_{\epsilon }x)\mathrm{/}\epsilon \in {\mathrm{\partial }}_{{\lambda }_V}f(J_{\epsilon }x)$; this follows by testing minimality at $(1-h)J_{\epsilon }x+hy$, using semiconvexity of $f$, dividing by $h$, and letting $h\downarrow 0$. Comparing the two optimality inequalities at $x$ and $x^{\mathrm{\prime }}$ and adding gives $\mathrm{\parallel }{J}_{\epsilon }x-J_{\epsilon }{x}^{\mathrm{\prime }}\mathrm{\parallel }\le (1+\epsilon {\lambda }_V{)}^{-1}\mathrm{\parallel }x-x^{\mathrm{\prime }}\mathrm{\parallel }$, so $z_{\epsilon }$ is continuous. For fixed $x\in Q_r$, comparison with $y=x$ shows $J_{\epsilon }x\to x$. Monotonicity of $\mathrm{\partial }g$ between $z_{\epsilon }(x)-{\lambda }_V{J}_{\epsilon }x\in \mathrm{\partial }g(J_{\epsilon }x)$ and any $z-{\lambda }_{V}x\in \mathrm{\partial }g(x)$ gives $(1+\epsilon {\lambda }_V)\mathrm{\parallel }{z}_{\epsilon }(x){\mathrm{\parallel }}^2\le ⟨z,z_{\epsilon }(x)⟩$. Thus $z_{\epsilon }(x)$ is bounded by every candidate norm in the limit. Passing to a limit in the defining subgradient inequalities shows the graph of $\mathrm{\partial }g$ is closed; every convergent subsequence therefore limits to an element of ${\mathrm{\partial }}_{{\lambda }_V}f(x)$. The norm inequality makes that limit the unique minimum-norm element. Finite-dimensional compactness gives convergence of the full sequence ${\epsilon }_n\downarrow 0$, so the minimum section is a pointwise limit of continuous maps and hence Borel.

The selected section has norm at most $M_r$ by [F2], so composition with a strongly measurable $X$ is measurable and square integrable; call it ${\zeta }_X\in K$. For any $Y\in C_r$, its pointwise supporting inequality integrates to ${\displaystyle \int f(Y)d\sigma \ge \int f(X)d\sigma +⟨{\zeta }_X,Y-X{⟩}_K+\frac{{\lambda }_V}{2}\mathrm{\parallel }Y-X{\mathrm{\parallel }}_K^2\mathrm{.}}$ Together with [F1], this proves that $\mathrm{\nabla }{G}_r(X)+{\zeta }_X\in {\mathrm{\partial }}_{{\lambda }_E}{E}_r(X)$ and establishes nonemptiness.

Conversely, let $\eta \in {\mathrm{\partial }}_{{\lambda }_E}{E}_r(X)$. Put $C=L_r{A}_r^2+N_r{B}_r$ and ${\lambda }_E={\lambda }_V-C$. Then $E_r-\frac{{\lambda }_E}{2}\mathrm{\parallel }\cdot {\mathrm{\parallel }}_K^2$ is the sum of the convex integral $X\mapsto \int g(X)d\sigma$ and the differentiable convex function $G_r(X)+\frac{C}{2}\mathrm{\parallel }X{\mathrm{\parallel }}_K^2$. The latter has gradient $\mathrm{\nabla }{G}_r(X)+CX$. The convex sum rule here follows directly from one-sided directional derivatives: if $q$ is convex and $h$ is differentiable convex, a supporting vector $w$ of $q+h$ satisfies $q^{\mathrm{\prime }}(X;H)\ge ⟨w-\mathrm{\nabla }h(X),H⟩$ for every feasible direction $H$; convex secant slopes dominate $q^{\mathrm{\prime }}(X;H)$, so $w-\mathrm{\nabla }h(X)$ supports $q$. Apply this with $w=\eta -{\lambda }_{E}X$ to conclude that $\eta -\mathrm{\nabla }{G}_r(X)-{\lambda }_{V}X$ supports the convex integral. Testing with $Y=X$ off a measurable set $A$ and $Y=y$ on $A$ gives a pointwise supporting inequality for every $y$ in a countable dense subset of $Q_r$, outside a common null set by [F3]. The relative interior is nonempty by [F5]; on it, continuity [F5] extends the inequality from the dense subset. For a boundary point $y$, choose $y_0$ in the relative interior and let $B(y_0,\rho )$ be a relative ball contained in $Q_r$. For $y_n=(1-1\mathrm{/}n)y+(1\mathrm{/}n)y_0$, convexity gives a relative ball of radius $\rho \mathrm{/}n$ around $y_n$ contained in $Q_r$, so $y_n$ is in the relative interior. Convexity gives $g(y_n)\le (1-1\mathrm{/}n)g(y)+(1\mathrm{/}n)g(y_0)$, and lower semicontinuity gives $g(y)\le {\mathrm{lim\hspace{0.17em}inf}}_{n}g(y_n)$; hence $g(y_n)\to g(y)$ and the supporting inequality passes to $y$. Undoing the quadratic shift proves $\eta -\mathrm{\nabla }{G}_r(X)$ is a $K$-section of ${\mathrm{\partial }}_{{\lambda }_V}f(X(s))$, proving the reverse inclusion.

For the shifted selection, apply step 1.2 to $f_a(u)=⟨a,\mathrm{\Phi }(u)⟩+V(u)+{\iota }_{Q_r}(u)$, whose semiconvexity coefficient is ${\alpha }_a={\lambda }_V-B_r\mathrm{\parallel }a\mathrm{\parallel }$. Its resolvent $J_{\epsilon }^{a}x$ is jointly continuous in $(a,x)$ where $1+\epsilon {\alpha }_a>0$: comparison of the two optimality inequalities gives $(1+\epsilon {\alpha }_a)\mathrm{\mid }{J}_{\epsilon }^{a}x-J_{\epsilon }^{a\mathrm{\prime }}x\mathrm{\prime }\mathrm{\mid }\le \mathrm{\mid }x-x\mathrm{\prime }\mathrm{\mid }+\epsilon A_r\mathrm{\parallel }a-a\mathrm{\prime }\mathrm{\parallel }$ when the coefficient is taken as a common lower bound for the two parameters. Hence $(x-J_{\epsilon }^{a}x)\mathrm{/}\epsilon$ is jointly continuous there. On each bounded parameter ball, step 1.2 proves its pointwise limit at $u\in Q_r$ is the minimum of $\mathrm{\partial }{f}_a(u)=b_a(u)+{\mathrm{\partial }}_{{\lambda }_V}f(u)$; the sum formula follows by subtracting the smooth Taylor remainder in the local supporting inequality. The parameter balls exhaust $\mathcal{H}$, so the limit is jointly Borel. A bounded potential subgradient gives an admissible shifted vector of norm $A_r\mathrm{\parallel }a\mathrm{\parallel }+M_r$, proving the bound. The pointwise projection inequality $\mathrm{\mid }S(a,u)\mathrm{\mid }\le \mathrm{\mid }z\mathrm{\mid }$ for every $z$ in the shifted set integrates over labels. The exact subdifferential formula from step 3.1 therefore proves that this measurable section is the unique minimum-norm Hilbert subgradient.