[proofplan] We reduce the metric problem to countably many real-valued absolutely continuous functions of the form $t\mapsto d(\gamma(t),x)$. A countable [dense subset](/page/Dense%20Subset) of the compact image of $\gamma$ recovers the metric distance by taking suprema of these scalar distance differences. The pointwise supremum of the scalar derivatives is an $L^1$ admissible control, and the [Lebesgue differentiation theorem](/theorems/74) then identifies it with the two-sided metric derivative almost everywhere. Finally, any other admissible control bounds all scalar derivatives, so it bounds this supremum almost everywhere. [/proofplan]

[step:Choose an integrable control and a countable dense family of distance functions] Since $\gamma$ is absolutely continuous, there exists a nonnegative function \begin{align*} m:(0,1)\to[0,\infty),\qquad m\in L^1((0,1),\mathcal L^1), \end{align*} such that \begin{align*} d(\gamma(s),\gamma(t))\le \int_s^t m(r)\,d\mathcal L^1(r) \end{align*} for every $0\le s\le t\le 1$. The curve $\gamma$ is continuous, because the preceding estimate sends $d(\gamma(s),\gamma(t))$ to $0$ as $s\to t$ by absolute continuity of the [Lebesgue integral](/page/Lebesgue%20Integral). Hence $K:=\gamma([0,1])$ is compact in $X$. The compact [metric space](/page/Metric%20Space) $K$ is separable: for each integer $j\ge 1$, choose a finite $1/j$-net in $K$, and take the union of these finite nets. Let $D=\{x_k:k\in\mathbb N\}\subset K$ be a countable dense subset of $K$. For each $k\in\mathbb N$, define the real-valued function \begin{align*} f_k:[0,1]\to\mathbb R,\qquad f_k(t):=d(\gamma(t),x_k). \end{align*} For $0\le s\le t\le 1$, the [reverse triangle inequality](/theorems/2300) gives \begin{align*} |f_k(t)-f_k(s)|\le d(\gamma(s),\gamma(t))\le \int_s^t m(r)\,d\mathcal L^1(r). \end{align*} Thus each $f_k$ is absolutely continuous on $[0,1]$. By the real-valued theorem on absolutely continuous functions, $f_k'(t)$ exists for $\mathcal L^1$-a.e. $t\in(0,1)$ and \begin{align*} |f_k'(t)|\le m(t) \end{align*} for $\mathcal L^1$-a.e. $t\in(0,1)$. [/step]

[step:Recover the metric distance from the dense family] For every $a,b\in K$, \begin{align*} d(a,b)=\sup_{k\in\mathbb N}|d(a,x_k)-d(b,x_k)|. \end{align*} Indeed, the reverse triangle inequality gives \begin{align*} |d(a,x_k)-d(b,x_k)|\le d(a,b) \end{align*} for every $k\in\mathbb N$. Conversely, because $D$ is dense in $K$, there exists a sequence $(x_{k_j})_{j=1}^{\infty}$ in $D$ such that $x_{k_j}\to a$ in $X$. The distance function $z\mapsto d(z,b)$ is continuous, so \begin{align*} |d(a,x_{k_j})-d(b,x_{k_j})|\to d(a,b). \end{align*} Taking the supremum over $k\in\mathbb N$ gives the reverse inequality. Applying this identity with $a=\gamma(s)$ and $b=\gamma(t)$ gives \begin{align*} d(\gamma(s),\gamma(t))=\sup_{k\in\mathbb N}|f_k(t)-f_k(s)| \end{align*} for every $s,t\in[0,1]$. [/step]

[step:Define the candidate speed as the supremum of scalar derivatives] Let $N\subset(0,1)$ be a Borel set of $\mathcal L^1$-measure zero outside which every derivative $f_k'(t)$ exists and the inequality $|f_k'(t)|\le m(t)$ holds for every $k\in\mathbb N$. Define \begin{align*} h:(0,1)\to[0,\infty),\qquad h(t):=\sup_{k\in\mathbb N}|f_k'(t)| \end{align*} for $t\in(0,1)\setminus N$, and set $h(t):=0$ for $t\in N$. The function $h$ is measurable as the pointwise supremum of countably many [measurable functions](/page/Measurable%20Functions), after defining each $f_k'$ arbitrarily on its null exceptional set. Since $h(t)\le m(t)$ for $\mathcal L^1$-a.e. $t\in(0,1)$, we have $h\in L^1((0,1),\mathcal L^1)$. For every $k\in\mathbb N$ and every $0\le s\le t\le 1$, the fundamental theorem for absolutely continuous real-valued functions gives \begin{align*} |f_k(t)-f_k(s)|\le \int_s^t |f_k'(r)|\,d\mathcal L^1(r)\le \int_s^t h(r)\,d\mathcal L^1(r). \end{align*} Taking the supremum over $k\in\mathbb N$ and using the metric recovery identity yields \begin{align*} d(\gamma(s),\gamma(t))\le \int_s^t h(r)\,d\mathcal L^1(r). \end{align*} Thus $h$ is an admissible control for $\gamma$. [/step]

[step:Identify the metric derivative with the candidate speed at almost every time]Let $G\subset(0,1)$ be the set of points $t$ such that $t$ is a Lebesgue point of $h$, every $f_k$ is differentiable at $t$, and $h(t)=\sup_{k\in\mathbb N}|f_k'(t)|<\infty$. By the Lebesgue differentiation theorem and the countability of the family $(f_k)_{k\in\mathbb N}$, the complement of $G$ has $\mathcal L^1$-measure zero. Fix $t\in G$. For $q\ne0$ with $t+q\in[0,1]$, the admissibility of $h$ gives \begin{align*} \frac{d(\gamma(t+q),\gamma(t))}{|q|}\le \frac{1}{|q|}\int_{\min\{t,t+q\}}^{\max\{t,t+q\}} h(r)\,d\mathcal L^1(r). \end{align*} Since $t$ is a Lebesgue point of $h$, the right-hand side converges to $h(t)$ as $q\to0$. Therefore \begin{align*} \limsup_{q\to0}\frac{d(\gamma(t+q),\gamma(t))}{|q|}\le h(t). \end{align*} For the reverse inequality, fix $\varepsilon>0$. By the definition of $h(t)$ as a supremum, choose $k\in\mathbb N$ such that \begin{align*} |f_k'(t)|>h(t)-\varepsilon. \end{align*} For all admissible $q\ne0$, the reverse triangle inequality gives \begin{align*} \frac{d(\gamma(t+q),\gamma(t))}{|q|}\ge \left|\frac{f_k(t+q)-f_k(t)}{q}\right|. \end{align*} Because $f_k$ is differentiable at $t$, the right-hand side converges to $|f_k'(t)|$ as $q\to0$. Hence \begin{align*} \liminf_{q\to0}\frac{d(\gamma(t+q),\gamma(t))}{|q|}\ge h(t)-\varepsilon. \end{align*} Letting $\varepsilon\downarrow0$ gives the reverse inequality. Thus the metric derivative exists at $t$ and equals $h(t)$.[/step]

[guided]The purpose of this step is to convert the integral control obtained in the previous step into an actual pointwise speed. We work at a point $t$ where all the one-dimensional ingredients behave well. More precisely, let $G\subset(0,1)$ be the set of times $t$ for which $t$ is a Lebesgue point of $h$, every real-valued distance function $f_k$ is differentiable at $t$, and $h(t)=\sup_{k\in\mathbb N}|f_k'(t)|<\infty$. The Lebesgue differentiation theorem applies to $h\in L^1((0,1),\mathcal L^1)$, and each $f_k$ is absolutely continuous, hence differentiable almost everywhere. Since there are only countably many $k$, the intersection of these full-measure sets still has full measure. Fix $t\in G$. First we prove the upper bound on the metric difference quotient. If $q\ne0$ and $t+q\in[0,1]$, the admissibility of $h$ gives the estimate over the interval with endpoints $t$ and $t+q$: \begin{align*} d(\gamma(t+q),\gamma(t))\le \int_{\min\{t,t+q\}}^{\max\{t,t+q\}} h(r)\,d\mathcal L^1(r). \end{align*} Dividing by $|q|$ gives \begin{align*} \frac{d(\gamma(t+q),\gamma(t))}{|q|}\le \frac{1}{|q|}\int_{\min\{t,t+q\}}^{\max\{t,t+q\}} h(r)\,d\mathcal L^1(r). \end{align*} The expression on the right is the average of $h$ over a shrinking interval containing $t$. Since $t$ is a Lebesgue point of $h$, these averages converge to $h(t)$ as $q\to0$. Therefore \begin{align*} \limsup_{q\to0}\frac{d(\gamma(t+q),\gamma(t))}{|q|}\le h(t). \end{align*} Now we prove the lower bound. The number $h(t)$ is the supremum of the scalar speeds $|f_k'(t)|$. Given $\varepsilon>0$, choose $k\in\mathbb N$ with \begin{align*} |f_k'(t)|>h(t)-\varepsilon. \end{align*} The scalar function $f_k$ records distance from the fixed point $x_k$. A metric movement of $\gamma$ must be at least as large as the change in this scalar distance, because the reverse triangle inequality gives \begin{align*} |f_k(t+q)-f_k(t)|=|d(\gamma(t+q),x_k)-d(\gamma(t),x_k)|\le d(\gamma(t+q),\gamma(t)). \end{align*} After division by $|q|$, this becomes \begin{align*} \frac{d(\gamma(t+q),\gamma(t))}{|q|}\ge \left|\frac{f_k(t+q)-f_k(t)}{q}\right|. \end{align*} Since $f_k$ is differentiable at $t$, the right-hand side converges to $|f_k'(t)|$ as $q\to0$. Hence \begin{align*} \liminf_{q\to0}\frac{d(\gamma(t+q),\gamma(t))}{|q|}\ge h(t)-\varepsilon. \end{align*} Because $\varepsilon>0$ was arbitrary, the lower bound is at least $h(t)$. The upper and lower bounds agree, so the two-sided metric derivative exists at $t$ and equals $h(t)$.[/guided]

[step:Prove the distance estimate and minimality] Define $|\gamma'|:(0,1)\to[0,\infty)$ by setting $|\gamma'|(t)=h(t)$ at points where the metric derivative exists and by setting $|\gamma'|(t)=h(t)$ on the remaining null set as well. Then $|\gamma'|\in L^1((0,1),\mathcal L^1)$, and the admissibility estimate already proved for $h$ gives \begin{align*} d(\gamma(s),\gamma(t))\le \int_s^t |\gamma'|(r)\,d\mathcal L^1(r) \end{align*} for every $0\le s\le t\le 1$. It remains to prove minimality. Let $g\in L^1((0,1),\mathcal L^1)$ be nonnegative and suppose that \begin{align*} d(\gamma(s),\gamma(t))\le \int_s^t g(r)\,d\mathcal L^1(r) \end{align*} for every $0\le s\le t\le 1$. For each $k\in\mathbb N$, the same reverse-triangle argument gives \begin{align*} |f_k(t)-f_k(s)|\le \int_s^t g(r)\,d\mathcal L^1(r) \end{align*} for every $0\le s\le t\le 1$. At every common point where $g$ is a Lebesgue point and $f_k$ is differentiable, dividing this inequality by the interval length and letting the interval shrink to the point gives \begin{align*} |f_k'(t)|\le g(t). \end{align*} After intersecting over the [countable set](/page/Countable%20Set) of indices $k\in\mathbb N$, this inequality holds for every $k$ for $\mathcal L^1$-a.e. $t\in(0,1)$. Taking the supremum over $k$ yields \begin{align*} |\gamma'|(t)=h(t)=\sup_{k\in\mathbb N}|f_k'(t)|\le g(t) \end{align*} for $\mathcal L^1$-a.e. $t\in(0,1)$. Therefore $|\gamma'|$ is the minimal admissible control, completing the proof. [/step]