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I would like to have some hints on this exercise since I struggle to begin.

Let $c : [a,b] \rightarrow \mathbb{R}^n$ be a curve such that

$$ l(c) := \sup_{a<t_0<...<t_k<b} \left \{ \sum_{i=0}^{k-1} \| c(t_{i+1})-c(t_i) \| \right \} < \infty. $$

I need to show that: for all $\epsilon>0$, there exist a $\delta>0$ such that

$$ \Big| l(c)- \sum_{j=0}^{k-1}\| c(t_j)-c(t_{j+1}) \| \Big| < \epsilon$$

for all partition $P=\{ a=t_0<t_1<...<t_k=b \}$ with $\delta(P)<\delta$, where $\delta(P)= \max_{k=0,...,n-1} \{ t_{k+1}-t_k \}$.

Thank you for your help.

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  • 1
    $\begingroup$ There's lots of information packed in this problem, lots of little facts to exploit. Have you attempted to use of any of these facts? For example the function $c$ is uniformly continuous. Also, if $P'$ is a refinement of $P$ then $\Sigma_{P'} \ge \Sigma_P$ (I use $\Sigma_P$ as shorthand for the summation expression in your question); in other words, refining the partition moves the approximation closer to the supremum $l(c)$. $\endgroup$ – Lee Mosher Apr 30 at 15:46
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Let's proceed by "reductio ad absurdum" i.e. by contradiction: suppose that there exists an $\epsilon_o>0$ such that, for every $\delta>0$ the following inequality holds $$ \Big| l(c)- \sum_{j=0}^{k-1}\| c(t_j)-c(t_{j+1}) \| \Big| \ge \epsilon_0>0\label{1}\tag{1} $$ for all partitions $P=\{ a=t_0<t_1<\ldots<t_k=b \}$ of the interval $[a,b]$ (note that the can chose the unrestricted set of partitions since by the contradiction hypothesis, \eqref{1} holds for all $\delta >0$).
Now, since $$ l(c) := \sup_{a<t_0<...<t_k<b} \left \{ \sum_{i=0}^{k-1} \| c(t_{i+1})-c(t_i) \| \right \} < \infty. $$ we have that $$ \begin{split} l(c)\ge \sum_{i=0}^{k-1} \| c(t_{i+1})-c(t_i) \| &\: \text{ for all paritions $P$ of }[a,b]\\ &\Updownarrow\\ \end{split} $$ $$ \Big| l(c)- \sum_{j=0}^{k-1}\| c(t_j)-c(t_{j+1}) \| \Big|= l(c)- \sum_{j=0}^{k-1}\| c(t_j)-c(t_{j+1}) \|.\label{2} \tag{2} $$ Therefore for all paritions $P$ of $[a,b]$ we have that $$ \begin{split} l(c)\;- &\sum_{j=0}^{k-1}\| c(t_j)-c(t_{j+1}) \|\ge \epsilon_o \\ l(c)>l(c)-\epsilon_o\ge&\sum_{j=0}^{k-1}\| c(t_j)-c(t_{j+1}) \|>0 \end{split}\quad \text{ for all paritions $P$ of }[a,b]\label{3}\tag{3} $$ Thus, there exists an upper limit for all the sums $\sum_{j=0}^{k-1}\| c(t_j)-c(t_{j+1}) \|$ which is strictly less than $l(c)$ and this contradicts the fact that $l(c)$ is defined as the supremum i.e. the minimum of the set of all upper limits.

Final note
Note that the proof offered is non constructive as all proofs by contradiction. This means that, given an arbitrary $\epsilon>0$, we are not able to estimate the maximum partition dimension $\delta$ for which the inequality $$ \Big| l(c)- \sum_{j=0}^{k-1}\| c(t_j)-c(t_{j+1}) \| \Big| < \epsilon $$ holds by using the steps of the proof. Perhaps one could try an constructive proof by exploiting the fact that, as noted by Lee Mosher in his comment to the question, a partition finer than a given one produces a larger value of the associated sum: however, it seem to me that this conjectured constructive way would be quite hard to follow.

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