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Let $f$ be a function from $[a,b] \to \mathbb{R} $ and $a = t_0 < t_1 .. < t_n = x$ be a partition (called $P$) of $[a,x] \subseteq [a,b]$.

Define $l(f,P) = \sum\limits_{i=1}^{n} \sqrt{\left(t_i - t_{i-1}\right)^2 + \left(f(t_i) - f(t_{i-1})\right)^2}$

Let $L(x) = \sup l(f,P)$ where the supremum ranges over all partitions of $[a,x]$.

$L$ here represents arc length of the graph of f

Let $d(x)$ be the distance between $(a,f(a))$ and $(x,f(x))$

Suppose $L(b)$ is defined show $\lim_{x \to a} \frac{L(x)}{d(x)} = 1$

I managed to show if supposing f is differentiable and the derivative is continuous at a but I don't know how to show it in the general case, or if it even is true in general? (I suspect Spivak wants the reader to assume differentiability)

davik
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2 Answers2

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We are considering$$\lim_{x \to a} {{\int_a^x \sqrt{1 + (f')^2}}\over{\sqrt{(x - a)^2 + (f(x) - f(a))^2}}}.$$By the mean value theorem,$$f(x) - f(a) = (x - a)f'(\xi)$$for some $\xi$ in $(a, b)$, and by the mean value theorem for integrals, the numerator is$$(x - a)\sqrt{1 + f'(\eta)^2}$$for some $\eta$ in $[a, b]$. So we are considering$${{(x - a)\sqrt{1 + f'(\eta)^2}}\over{\sqrt{(x - a)^2 + f'(\xi)^2(x - a)^2}}} = {{\sqrt{ 1 + f'(\eta)^2}}\over{\sqrt{1 + f'(\xi)^2}}},$$which approaches $1$ as $x \to a$. We need to assume that $f'$ is continuous at $a$.

Simply Beautiful Art
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user90184
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  • Yes I think this was the proof I found too, I guess it does depend on continuity of f'. I realize now that it's not true if you allow for any function with well defined length. But I'm curious whether it's true with only the constraint that $sqrt(1+f'(x)^2)$ is integrable ? – davik Jun 05 '16 at 07:58
  • @Danielle Doerr On my side, I'm not convinced by this proof. $\eta$ and $\xi$ can be such that ${{(x - a)\sqrt{1 + f'(\eta)^2}}\over{\sqrt{(x - a)^2 + f'(\xi)^2(x - a)^2}}} = {{\sqrt{ 1 + f'(\eta)^2}}\over{\sqrt{1 + f'(\xi)}}},$ is always equal to $1/2$ or whatever. – mathcounterexamples.net Jun 05 '16 at 20:58
  • @mathcounterexamples.net how so? –  Dec 09 '20 at 07:26
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I just wanted to share some of my progress let $$f(x)=\int_0^x {\cos{(1/u)} \, \mathrm{d}u}$$ where the integral is on $(0,x)$ so it is well-defined. Note $f$ is differentiable on $(0,\infty)$ by FTC. And $f'(x)=\cos(1/x)$. Furthermore, since $$f(x)=\int_0^x {\left(2 u \sin(\frac{1}{u}) - {d \over du} (u^2 \sin(\frac{1}{u}))\right) \, \mathrm{d}u}$$ we have $f'(0)=0$ if $f(0)$ is defined to be $0$.

Now, since $\sqrt{1+f'(x)^2}$ is integrable on $[0,1]$ $$L(x)=\int_0^x \left(\sqrt{1+\cos(1/u)^2}\right)du$$ $$d(x)=\sqrt{x^2+f(x)^2}$$ We just need to evaluate the limit $$\lim_{x\to 0^+} {\frac{L(x)}{d(x)}}$$ which evaluates to ${2E(-1) \over \pi} > 1$

(See Differentiability of a integral)

Thus this limit can fail to be 1 even if f is differentiable and f' is integrable, we really need the continuity at a.

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davik
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