Finite partition energy yields Hilbert-valued AC2 paths

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.