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Prove that if a function $f:[0,1] \to R$ is twice differentiable, then its graph must be a rectifiable curve. (You may assume that the corresponding curve is $\lambda (t)=(t,f(t))$, where $t \in [0,1]$.)

We have defined a rectifiable curve as a mapping $ \gamma$ such that sup $ \Lambda (P,\gamma)< \infty$, where $P={(x_0, . . ., x_n)}$ is a partition of the interval $[a,b]$, in this case $[0,1]$, and $\Lambda (P,\gamma)$ is the length of the polygonal path from $\gamma(x_0)$ to $\gamma(x_n)$.

I have no clue where to begin with this problem. Any help is greatly appreciated.

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  • $\begingroup$ This would be true for a $C^1$ function actually $\endgroup$ – zhw. Mar 3 '17 at 20:33
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You have a partition $P = (x_0,\ldots,x_n)$ of $[0,1]$, a curve $\gamma(t) = (t,f(t))$, and $$\Lambda(P,\gamma) = \sum_{k=1}^n \sqrt{ |x_k - x_{k-1}|^2 + |f(x_k) - f(x_{k-1})|^2}.$$ You need to show this is bounded irrespective of the partition $P$. Note that $\sqrt{a^2 + b^2} \le |a| + |b|$ so that $$\Lambda(P,\gamma) \le \sum_{k=1}^n \left( |x_k - x_{k-1}| + |f(x_k) - f(x_{k-1})| \right) = 1 + \sum_{k=1}^n |f(x_k) - f(x_{k-1})|.$$ Since $f'$ is assumed to be continuous the fundamental theorem of calculus gives you $$|f(x_k) - f(x_{k-1})| = \left| \int_{x_{k-1}}^{x_k} f'(t) \, dt \right|\le \int_{x_{k-1}}^{x_k} |f'(t)| \, dt$$ so finally $$\Lambda(P,\gamma) \le 1 + \int_0^1 |f'(t)| \, dt.$$ Can you show that the last integral is finite?

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  • $\begingroup$ I don't understand why $\int_0^1 |f'(t)| \, dt$ must be finite. Clearly, if the function is bounded, this is the case. But I don't get why it is true for an unbounded function. $\endgroup$ – mathqueen459 Mar 3 '17 at 20:11
  • $\begingroup$ You are given that $f$ is twice differentiable on a closed interval. $\endgroup$ – Umberto P. Mar 3 '17 at 20:16
  • $\begingroup$ Ohhhh. I didn't recognize that. Got it now. Thanks! $\endgroup$ – mathqueen459 Mar 3 '17 at 20:20

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