Proof of continuously differentiable curve arc length

Question

I was reading Terence Tao's lecture on complex analysis. This is a question regarding his "continuity" solution to the proposition:

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My question is: Why is $\Omega_{\varepsilon}$ closed? I have quoted the part of the proof. The (7) referred to is as follows.

Answer 1

By hypothesis, the function $\gamma': [a, b] \to \mathbb{C}$ is continuous, therefore $|\gamma'|: [a, b] \to \mathbb{R}_{\geqslant 0}, t \mapsto |\gamma'(t)|$ is continuous. Since $[a, b]$ is compact, $|\gamma'|$ is bounded above, say by $M \geqslant 0$.

At the start of the proof of Proposition 10, Tao obtains the upper bound $$ \tag{3}\label{eq:ub} |\gamma| \leqslant \int_a^b |\gamma'(t)|\,dt. $$

If $a \leqslant T \leqslant T' \leqslant b$, then $\gamma_{[a, T']} = \gamma_{[a, T]} + \gamma_{[T, T']}$, therefore by the result of Exercise 7, and by \eqref{eq:ub}, $$ |\gamma_{[a, T']}| - |\gamma_{[a, T]}| = |\gamma_{[T, T']}| \leqslant \int_T^{T'} |\gamma'(t)|\,dt \leqslant M(T' - T). $$

The left hand side of the claimed inequality (4) is therefore a (uniformly) continuous function of $T$. Rather more obviously, the right hand side of (4) is also a (uniformly) continuous function of $T$.

If we denote by $\Omega'_\epsilon$ the set of $T \in [a, b]$ such that (4) holds, then this is the inverse image of a closed set under a continuous function, and so is closed.

The set $\Omega_\epsilon$ is defined as the union of all intervals of the form $[a, T]$ contained in $\Omega'_\epsilon$, and so $\Omega_\epsilon$ is itself an interval: either it is the closed interval $[a, c]$, or it is the half-open interval $[a, c)$, where $c \in [a, b]$ is its least upper bound. In either case, $[a, c) \subseteq \Omega'_\epsilon$, therefore $c \in \Omega'_\epsilon$, because $\Omega'_\epsilon$ is closed. It follows that $[a, c] \subseteq \Omega'_\epsilon$, i.e. $c \in \Omega_\epsilon$, by the definition of $\Omega_\epsilon$. So $\Omega_\epsilon = [a, c]$.

It's hard to believe that this is what Tao had in mind, however, because if it were, it would mean that he throws away this strong information about $\Omega_\epsilon$, only to use the weaker fact that it is topologically closed. The rest of his argument could have been used to derive a contradiction from the hypothesis that $c < b$, and so to prove that $c = b$; but instead of doing this, Tao appeals to the continuity method, i.e. the principle that a non-empty open and closed subset of an interval of $\mathbb{R}$ is necessarily the whole interval. To set this up, he establishes in Exercise 9 that the interval $[a, b]$ is topologically connected. But the proof he suggests is along similar lines to the proof given here that $\Omega_\epsilon$ is closed:

Exercise 9 Let $a\leq b$ be real numbers. Show that the interval $[a, b]$ is topologically connected, that is to say the only two subsets of $[a, b]$ that are both open and closed relative to $[a, b]$ are the empty set and all of $[a, b]$. (Hint: if $E$ is a non-empty set that is both open and closed in $[a, b]$ and contains $a$, consider the supremum of all $T_* \in [a, b]$ such that $[a, T_*] \subset E$.)

I don't know why this redundancy has occurred, but most likely it's because of some silly mistake I've made. I offer the above proof, anyway, because at least it seems to establish that $\Omega_\epsilon$ is closed - even if it's at far too heavy a cost!

Answer 2

I don't know why Tao is going through all that to prove $\int_a^b|\gamma '| \le V_a^b(\gamma).$ I would do it this way: First note that because $\gamma'$ is continuous on $[a,b],$ it is uniformly continuous there. So let $\epsilon>0.$ Choose $\delta > 0$ such that $|t-s| < \delta$ implies $|\gamma'(t)-\gamma'(s)| < \epsilon.$

Claim: If $a\le s<t \le b,$ with $t-s < \delta,$ then

$$\tag 1 \left | \,|\gamma(t) - \gamma(s)| - |\gamma'(t)|(t-s) \, \right | <\epsilon (t-s).$$

Proof: Using $||x|-|y|| \le |x-y|, $ we have the left side of $(1)$ bounded above by

$$ | \gamma(t) - \gamma(s) - \gamma'(t)(t-s)| = \left | \int_s^t (\gamma'(u) - \gamma'(t))\,du \right| \le \int_s^t |\gamma'(u) - \gamma'(t)|\,du < \epsilon(t-s).$$

So now let $a=s_0 < s_1 < \cdots <s_n=b$ be a partition of $[a,b]$ with mesh size $<\delta.$ Then

$$ V_a^b(\gamma) \ge \sum_{k=1}^{n}|\gamma(s_k) - \gamma(s_{k-1})| = \sum_{k=1}^{n}|\gamma'(s_k)|(s_k - s_{k-1}) + \sum_{k=1}^{n}\left (|\gamma(s_k) - \gamma(s_{k-1})| - |\gamma'(s_k)|(s_k - s_{k-1})\right ).$$

In the last sum, the claim shows each summand, in absolute value, is less than $\epsilon(s_k-s_{k-1}).$ Thus this sum in absolute value is no more than $\epsilon(b-a).$ It follows that $$ V_a^b(\gamma) \ge \sum_{k=1}^{n}|\gamma'(s_k)|(s_k - s_{k-1}) - \epsilon(b-a).$$ Now let the mesh size go to $0.$ We get $V_a^b(\gamma) \ge \int_a^b|\gamma'| - \epsilon(b-a).$ Since $\epsilon$ is arbitrary, we have $V_a^b(\gamma) \ge \int_a^b|\gamma'|$ as desired.