Exercise in do Carmo: Length of smooth curve

Question

I am attempting an exercise in do Carmo's Differential Geometry of Curves and Surfaces. Here is the problem:

Let $\alpha:I\to\Bbb R^3$ be a differentiable curve and $[a,b]\subseteq I$. Consider any partition $P$ of $[a,b]$ by choosing $a=t_0\lt t_1\lt...\lt t_n=b$, and define $l(\alpha,P)=\sum_{i=1}^{n}\lVert\alpha(t_i)-\alpha(t_{i-1})\rVert$. Show that for any $\varepsilon\gt0$, there exists $\delta\gt0$ such that whenever $\max_i(t_i-t_{i-1})\lt\delta$, we have $$\bigg\lvert\int_a^b{\lVert\alpha'(t)\rVert dt}-l(\alpha,P)\bigg\rvert\lt\varepsilon$$

Here is my attempt:

The Riemann integral by definition is well-approximated by the sum $\sum_{i=1}^{n}(t_i-t_{i-1})\lVert\alpha'(c_i)\rVert$ whenever $\max_i(t_i-t_{i-1})$ is small enough, where $c_i\in[t_{i-1},t_i]$ is arbitrary. So it suffices to show that we can make $\big\lvert l(\alpha,P)-\sum_{i=1}^{n}(t_i-t_{i-1})\lVert\alpha'(c_i)\rVert\big\rvert$ as small as possible.

We have the following inequalities: $$\bigg\lvert l(\alpha,P)-\sum_{i=1}^{n}(t_i-t_{i-1})\lVert\alpha'(c_i)\rVert\bigg\rvert=\bigg\lvert\sum_{i=1}^{n}(t_i-t_{i-1})\bigg(\bigg\lVert\frac{\alpha(t_i)-\alpha(t_{i-1})}{t_i-t_{i-1}}\bigg\rVert-\lVert\alpha'(c_i)\rVert\bigg)\bigg\rvert$$

$$\le \sum_{i=1}^{n}(t_i-t_{i-1})\bigg\lvert\bigg\lVert\frac{\alpha(t_i)-\alpha(t_{i-1})}{t_i-t_{i-1}}\bigg\rVert-\lVert\alpha'(c_i)\rVert\bigg\rvert.$$

Then I am stuck here. I think I should make the values of each of $$\bigg\lvert\bigg\lVert\frac{\alpha(t_i)-\alpha(t_{i-1})}{t_i-t_{i-1}}\bigg\rVert-\lVert\alpha'(c_i)\rVert\bigg\rvert$$ small upon choosing $\delta$. I have thought of using reverse triangle inequality, but I have no idea how to control the size of $$\bigg\lVert\frac{\alpha(t_i)-\alpha(t_{i-1})}{t_i-t_{i-1}}-\alpha'(c_i)\bigg\rVert$$ even if I choose $c_i=t_i$ because then $\delta$ would have to depend on the choice of $t_i$, which is bad.

Mean value theorem for vector-valued function may be useful: link to Wikipedia.

Note that there is answer in the book and I have read it, but I am not satisfied with the proof. Specifically, he used mean value theorem for the following inequality:

$$\bigg\lvert \sum_{i=1}^{n}\lVert\alpha(t_i)-\alpha(t_{i-1})\rVert-\sum_{i=1}^{n}(t_i-t_{i-1})\lVert\alpha'(t_i)\rVert\bigg\rvert$$ $$\le\bigg\lvert\sum_{i=1}^{n}(t_i-t_{i-1})\sup_{s_i\in[t_{i-1},t_i]}\lVert\alpha'(s_i)\rVert-\sum_{i=1}^{n}(t_i-t_{i-1})\lVert\alpha'(t_i)\rVert\bigg\rvert.$$

This step seems wrong to me because of the term $\sum_{i=1}^{n}(t_i-t_{i-1})\lVert\alpha'(t_i)\rVert$ inside, unless I am mistaken.

Edit: I write a counterexample to do Carmo's claim. Consider $[a,b]\subseteq(0,\infty)$, $\alpha(t)=(t^2,0,0)$. Then $\sup_{s_i\in[t_{i-1},t_i]}\lVert\alpha'(s_i)\rVert=\lVert\alpha'(t_i)\rVert$ and we would have shown $$\bigg\lvert \sum_{i=1}^{n}\lVert\alpha(t_i)-\alpha(t_{i-1})\rVert-\sum_{i=1}^{n}(t_i-t_{i-1})\lVert\alpha'(t_i)\rVert\bigg\rvert$$ is equal to zero, which is not true, because by mean value theorem of real-valued functions, $\lVert\alpha(t_i)-\alpha(t_{i-1})\rVert=(t_i-t_{i-1})(2c_i)$ for some $c_i\in(t_{i-1},t_i)$, and $2c_i\lt 2t_i=\lVert\alpha'(t_i)\rVert$.

Answer 1

$\let\a\alpha$Do Carmo says in Section 1-2, and I paraphrase, that a function $ℝ→X$ is called "differentiable" if it is in $C^∞(ℝ,X)$.

which means that $\alpha$ was not just differentiable, but $C^1$.

We can therefore use the integral MVT (which is just FTC in disguise) $$ \a(t_i) - \a(t_{i-1}) = \int_{t_{i-1}}^{t_{i}}\a'(τ)\ dτ = (t_i-t_{i-1})\int_0^1 \a'(t_{i-1}+(t_i-t_{i-1})u) \ du$$

Hence,

$$\left| \frac{\a(t_i) - \a(t_{i-1})}{ t_i-t_{i-1}} - \a'(c_i)\right| \\ ≤ ∫_0^1 \left| \a'(t_{i-1}+(t_i-t_{i-1})u) - \a'(c_i) \right| \ du \\ \leq \sup_{s∈[t_{i-1},t_i]}|\a'(s) - \a'(c_i)| $$ So by prescribing $\max_i |t_i-t_{i-1}| < \delta$, uniform continuity of $\a'$ (again using $\a∈ C^1$) implies that this quantity is arbitrarily small, independent of the choice of $c_i\in[t_{i-1},t_i]$.

Answer 2

Rudin gives a proof in Principles of Mathematical Analysis, page 136-137, that a continuously differentiable curve $\gamma:[a,b]\to\Bbb R^n$ is rectifiable and the length such a curve is precisely $\int_a^b\lVert\gamma'(t)\rVert dt$. However, the definition of rectifiable curves as in Rudin's book is different from the definition in do Carmo's book. Since I don't know a source showing their equivalence, I state here the two definitions and show part of their equivalence. Let $\gamma:[a,b]\to\Bbb R^n$ be continuous.

$(1)$. (definition in do Carmo's book) We say that $\gamma$ is rectifiable if there exists a real number $r$ having the property that for any $\varepsilon\gt0$, there exists $\delta\gt0$ such that for any partition $P=\{t_0\lt...\lt t_n\}$ of $[a,b]$ with $\max(t_i-t_{i-1})\lt\delta$, we have $$\bigg\lvert r-\sum_{i=1}^n\lVert\gamma(t_i)-\gamma(t_{i-1})\rVert\bigg\rvert\lt\varepsilon.$$ The number $r$ is the unique number having that property if it exists and is called the length of $\gamma$.

$(2)$. (definition in Rudin's book) We say that $\gamma$ is rectifiable if $$l:=\sup\bigg\{l(\gamma,P)\rVert:P\text{ is a partition of} [a,b]\bigg\}\lt\infty.$$ The number $l$ is called the length of $\gamma$.

Since Rudin has proved that a smooth curve is rectifiable in the sense of $(2)$ with $l=\int_a^b\lvert\gamma'(t)\rvert dt$, it suffices to show that $(2)$-rectifiable implies $(1)$-rectifiable and $r=l$ has the desired property.

Assume $\gamma$ is $(2)$-rectifiable. Let $\varepsilon\gt0$ be given. Choose a partition $P=\{t_0\lt...\lt t_n\}$ such that $l-l(\gamma,P)\lt\frac{\varepsilon}{2}$. Choose $\delta\gt0$ by uniform continuity of $\gamma$, so that for any $s,t\in[a,b]$, $\lvert s-t\rvert\lt\delta$ implies $\lVert\gamma(s)-\gamma(t)\rVert\lt\frac{\varepsilon}{4n}$, where $n$ is the number of intervals partitioned by $P$. Note that $n$ depends only on $\varepsilon$. So $\delta$ indeed depends only on $\varepsilon$. We now show that for any partition $P'=\{s_0\lt...\lt s_m\}$ of $[a,b]$ with $\max(s_i-s_{i-1})\lt\delta$, we have $l-l(\gamma,P')\lt\varepsilon$ (no need for absolute value because $l$ is always no less than $l(\gamma,P')$), proving $\gamma$ is $(1)$-rectifiable with $(1)$-length equal to $(2)$-length.

Define a common refinement $Q=P\cup P'$ of $P$ and $P'$. By triangle inequality, we see that $l(\gamma,Q)$ is no less than both of $l(\gamma,P)$ and $l(\gamma,P')$. We now give an upper bound to $l(\gamma,Q)-l(\gamma,P')$. For a refinement by one extra point $p$ to $P'$ (where $s_i\lt p\lt s_{i+1}$), we have $$l(\gamma,P'\cup\{p\})-l(\gamma,P')=\lVert\gamma(p)-\gamma(t_i)\rVert+\lVert\gamma(t_{i+1})-\gamma(p)\rVert-\lVert\gamma(t_{i+1})-\gamma(t_i)\rVert$$ $$\le\lVert\gamma(p)-\gamma(t_i)\rVert+\lVert\gamma(t_{i+1})-\gamma(p)\rVert$$ $$\le\frac{\varepsilon}{4n}+\frac{\varepsilon}{4n}=\frac{\varepsilon}{2n}$$ where the second inequality is due to uniform continuity. By induction on the number of extra points put into $P'$ to make $Q$, since $Q$ has at most $n$ more points than $P'$, we have $l(\gamma,Q)-l(\gamma,P')\le\frac{\varepsilon}{2n}\times n=\frac{\varepsilon}{2}$. Then observe the following inequalities: $$l-l(\gamma,P')\le l-l(\gamma,P')+l(\gamma,Q)-l(\gamma,P)=\big(l-l(\gamma,P)\big)+\big(l(\gamma,Q)-l(\gamma,P')\big)\le\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon,$$ which is the desired inequality.