3

This is problem 3 from Spivak's calculus 4th edition, appendix of chapter 13. before stating problem 3, problem 2 is relevant.

Problem 2 asks to prove that if $f$ and $g$ are continuous and non negative on $[a,b]$ and $P = \{t_0, \dots , t_n\}$ is a partition of $[a,b]$ and we choose sets of points $x_i, u_i \in [t_{i-1}, t_i]$ we can make the sum $\sum_{i=1}^n \sqrt{f(x_i) + g(u_i)}\Delta t_i $ within $\epsilon$ of $\int_{a}^b \sqrt{f + g}$ provided that all $\Delta t_i$ are small enough(there exists $\delta$ such that mesh $P < \delta$ ).

Problem 3 Consider a curve $c$ given parametrically by two functions $u$ and $v$ on $[a,b]$ For a partition $P = \{t_0, \dots, t_n\}$ of $[a,b]$ we define

$$\mathcal{l}(c,P) = \sum_{i = 1}^n \sqrt{[u(t_i) -u(t_{i-1})]^2+ [v(t_i) -v(t_{i-1})]^2} $$.

We define the length of $c$ to be the least upper bound of all $\mathcal{l}(c,P)$ if it exists. Prove that if $u'$ and $v'$ are continuous then the length of $c$ is $$\int_a^b \sqrt{u'^2+v'^2}$$

What I've done is this:

By the mean value theorem, there exist sets of points $x_i, y_i \in [t_{i-1}, t_i]$ with $$\sum_{i = 1}^n \sqrt{u'^2(x_i) + v'^2(y_i)} = \sum_{i=1}^n \sqrt{\bigg{[}\frac{u(t_i) - u(t_{i-1})}{\Delta t_i}\bigg{]}^2 + \bigg{[}\frac{v(t_i)-v(t_{i-1})}{\Delta t_i}\bigg{]}^2} $$

Thus $$ \mathcal{l}(c,P) = \sum_{i= 1}^n \sqrt{u'^2(x_i) + v'^2(y_i)}\Delta t_i$$

Let $m_i, m'_i \in [t_{i-1}, t_i]$ such that $u'(m_i)$ is the minimum value of $u'$ in $[t_{i-1}, t_i]$ and similarly for $v'(m'_i)$ and $M_i, M'_i \in [t_{i-1}, t_i]$ such that $u'(M_i)$ is the maximum value of $u'$ in $[t_{i-1}, t_i]$ and similarly for $v'(M'_i)$

Thus

$$ \sum_{i = 1}^n \sqrt{u'^2(m_i) + v'^2(m'_i)}\Delta t_i \leq \mathcal{l}(c,P) \leq \sum_{i = 1}^n \sqrt{u'^2(M_i) + v'^2(M'_i)}\Delta t_i $$

By problem 2, there exists $\delta > 0$ such that for all partitions P with mesh $ P < \delta$ we have $$-\epsilon < \sum_{i = 1}^n \sqrt{u'^2(m_i) + v'^2(m'_i)}\Delta t_i - \int_a^b \sqrt{u'^2 + v'^2} \leq \mathcal{l}(c,P) - \int_a^b \sqrt{u'^2 + v'^2} \leq \sum_{i = 1}^n \sqrt{u'^2(M_i) + v'^2(M'_i)}\Delta t_i - \int_a^b \sqrt{u'^2 + v'^2} <\epsilon$$

So for all $\epsilon > 0$ there exists $\delta > 0$ such that for all partitions P with mesh $P < \delta$ we get $$|\mathcal{l}(c,P) - \int_a^b \sqrt{u'^2 + v'^2} | < \epsilon \tag{1}$$

This is where I am stuck, if $A = \sup\{\mathcal{l}(c,P), P \text{ a partition of } [a,b]\}$ I've been trying to show that for all $\epsilon > 0 $ $|A - \int_a^b \sqrt{u'^2 + v'^2} |<\epsilon$ but I don't know how to prove it from(1). Any ideas?

Donlans Donlans
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  • Suppose a contradiction, that $l(c,P)$ has (if I could say) "limit" $I=$ the integral, but supremum of $l(c,P)$ not equal to $I$, thus $A$ and $I$ have a "gap" between them, take $\varepsilon=\frac{|A-I|}{2}$ and yeild a contradiction with the proven "limit" -- does that work?) – Alexey Burdin May 25 '20 at 01:58
  • No, not that way. Build a sequence of "bad" points (I mean partitions), that "tends to" c (I mean in terms of $mesh P$), but lim of $l(c,P)$ of that sequence is not $I$ (because it's equal to $\sup l(c,P)$) so there will be a contradiction between Heine and Cauchy limit definitions, but there's not. "Bad" point sequence construction: $A=\sup l(c,P)\Leftrightarrow \forall \varepsilon \exists P: A\ge l(c,P)>A-\varepsilon$, so take $\varepsilon=\frac{1}{2^n}$ and for every $\varepsilon$ take $P(\varepsilon)$ from supremum definition. the $l(c,P)$ on that sequence will tend to $A$, not $I$ – Alexey Burdin May 25 '20 at 02:20
  • A more general result holds : see this answer for details. – Paramanand Singh May 25 '20 at 03:32
  • Essentially this means that if the integrand consists of two parts $f, g$ like $\phi(f(x), g(x)) $ where $\phi$ is continuous then we can choose different tags for $f, g$ while forming a Riemann sum and yet get the same integral when taking limit. This is quite essential for deriving arc-length formula as the tag points for $u'^2$ and $v'^2$ obtained by mean value theorem can be different. This is what makes Spivak different. It does not miss even the finer details. – Paramanand Singh May 25 '20 at 03:37
  • @AlexeyBurdin Thanks for the help, it took me a while to grasp it though, in the book, spivak hasn't introduced heine limit definition, so I think he had in mind another solution. so I will try to find another one – Donlans Donlans May 25 '20 at 03:42
  • @ ParamanandSingh I don't understand the general result, I think that's above my current level. I haven't learn measures yet :( – Donlans Donlans May 25 '20 at 03:44
  • @DonlansDonlans: well the result does not involve measures (read the answer again). Anyway have you proved problem 2? If yes then problem 3 is trivial and you don't need to put so much effort into that. You need to stop at the step where you get $x_i, y_i$ and then use problem 2. Done!! – Paramanand Singh May 25 '20 at 03:51
  • @ ParamanandSingh Yes, I have proven problem 2, problem 1 was similar but with $fg$ instead of $ \sqrt{f + g}$. I have also proven it. now that I think about it, that part of choosing $m_i$ etc is unnecessary, problem 2 already provides that. wow what was I thinking – Donlans Donlans May 25 '20 at 04:13
  • There is another fine point which you need to take care of. The integral is the limit of such sums under consideration. You have to prove that it is also the supremum of such sums. A limit is not necessarily a supremum but in this case it is. – Paramanand Singh May 25 '20 at 04:20
  • @ParamanandSingh I was thinking about using the fact that the sums are increasing, I would have to prove that they are bounded and therefore they would converge to the supremum by a theorem(monotonic bounded sequence I think it's called) but in the book, convergent sequences and their theorems aren't discussed until chapter 22 So using those results feels like cheating – Donlans Donlans May 25 '20 at 05:03

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