Show the equivalence of arc length definitions

Question

Definition 1:

Let $r: [a,b] \to \Bbb R^d$ be a continuous differentiable function. Then the arc length is given by $$L(r) = \int_a^b || r'(t) || \, dt$$

Definition 2:

Let $r: [a,b] \to \Bbb R^d$ be a continuous function. Then the arc length is given by $$ V(r) = \sup_P \sum_{k=1}^n || r(x_k)-r(x_{k-1}) ||$$ where the supremum is taken over all partitions $P = \{a=x_0 \lt x_1 \lt \ldots \lt x_n = b \}$ of $[a,b]$.

How can I show that for a continuous differentiable $r(t)$ the two definitions are equivalent, i.e. $L(r)=V(r)$?

What I've done so far:

I found this question, which shows that I can convert the supremum to a limit

$$V(r) = \sup_P \sum_{k=1}^n || r(x_k)-r(x_{k-1}) || = \lim_{n \to \infty} \sum_{k=1}^n || r(x_k)-r(x_{k-1}) ||$$

by choosing an appropriate sequence of partitions $P_n$ of which I take the $x_k$'s. This gives

$$ \lim_{n \to \infty} \sum_{k=1}^n || r(x_k)-r(x_{k-1}) || = \lim_{n \to \infty} \sum_{k=1}^n || \frac{r(x_k)-r(x_{k-1})}{x_k-x_{k-1}} || (x_k-x_{k-1})$$

Now I somehow need to show that

$$\lim_{n \to \infty} \sum_{k=1}^n || \frac{r(x_k)-r(x_{k-1})}{x_k-x_{k-1}} || (x_k-x_{k-1}) = \int_a^b ||r'(t)|| \, dt$$

How can I justify this step of converting the sum to an intergral and taking the limit of the inside simultaneously?

Answer 1

Claim. For a parametrized curve $\gamma \in \mathcal C^1([a,b], \Bbb R^d)$, we have $$\bbox[5px,border:2px solid #C0A000]{ \lim_{n\to\infty} \sum_{i=1}^n \| \gamma(x_{i,n})-\gamma(x_{i-1,n})\| = \int_a^b \|\gamma'(x)\| \,\mathrm dx, }$$

where $x_{i,n} := a \cdot (1-\frac in) + b\cdot\frac in$.

Proof. Note that the left hand side equals \begin{equation} \lim_{n\to\infty} \underbrace{ \sum_{i=1}^n \left\| \frac{\gamma(x_{i,n})-\gamma(x_{i-1,n})}{\frac{b-a}n}\right\|\cdot {\frac{b-a}n} }_{=: \kappa_n} \end{equation} and observe that $x_{i,n}-x_{i-1,n}=\frac{b-a}n$.

The derivative of $\gamma$ is assumed to be continuous on a compact interval and it is thus also uniformly continuous (cf. Heine-Cantor Theorem), thus there exists a $\delta>0$ for every $\varepsilon > 0$ such that whenever $\lvert t-s\rvert< \delta$, we have $$ \lVert\gamma'(t)-\gamma'(s)\rVert<\epsilon. $$ Using the reverse triangle inequality and the fundamental Theorem of calculus, we have $$\left\lvert\bigg\lVert\frac{\gamma(x_{i,n})-\gamma(x_{i-1,n})}{\frac{b-a}n}\bigg\rVert-\bigg\lVert\gamma'(x_{i,n})\bigg\rVert\right\rvert\le\left\lVert \frac{\gamma(x_{i,n})-\gamma(x_{i-1,n})}{\frac{b-a}n} - \gamma'(x_{i,n})\right\rVert = \frac n{b-a}\left\lVert\int_{x_{i-1,n}}^{x_{i,n}} (\gamma'(t)- \gamma'(x_{i,n}))\,\mathrm dt \right\rVert.$$

By Jensen's inequality, $$ \frac n{b-a}\left\lVert\int_{x_{i-1,n}}^{x_{i,n}} (\gamma'(t)- \gamma'(x_{i,n}))\right\rVert \le\frac n{b-a}\int_{x_{i-1,n}}^{x_{i,n}} \lVert \gamma'(t)- \gamma'(x_{i,n})\rVert\,\mathrm dt. $$ Using uniform continuity, the latter term is now $<\epsilon$ as long as $\frac{b-a}n<\delta$.

In particular, for any $\varepsilon > 0$, we have \begin{equation}\tag{*} \label{*} \left\|\rule{0cm}{1cm} \underbrace{ \sum_{i=1}^n \left\| \frac{\gamma(x_{i,n})-\gamma(x_{i-1,n})}{\frac{b-a}n} \right\| \cdot\frac{b-a}n }_{=\kappa_n} -\underbrace{ \sum_{i=1}^n \Big\|\gamma'(x_{i,n})\Big\|\cdot\frac{b-a}n }_{=:\rho_n} \right\| < \varepsilon \cdot (b-a) \end{equation} whenever $n>\frac{b-a}\delta$ (note that $\delta$ depends on $\varepsilon$).

Since the $\rho_n$ are Riemann approximation sums, we have $\lim_{n\to\infty} \rho_n = \int_a^b \|\gamma'(x)\| \,\mathrm dx$. By \eqref{*}, we can conclude that $\lim_{n\to\infty} \kappa_n = \lim_{n\to\infty} \rho_n$, which proves our claim. $\square$