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I have encountered the following problem:

Let $f$ be continuous on $[a,b]$. Define the length of $f$ on $[a,b]$ by $$l=\sup_P[\lambda_P(f)],$$ where $$\lambda_P(f)=\sum_{k=1}^N\sqrt{(x_k-x_{k-1})^2+(f(x_k)-f(x_{k-1}))^2},$$ and the supremum is taken over all partitions $P=\{a=x_0<x_1<\cdots<x_N=b\}$ of $[a,b]$.

Show that $\lambda_P(f)\leqslant\lambda_Q(f)$ for any refinement $Q$ of $P$. Then, show that there is a sequence $(P_n)_{n=1}^\infty$ such that $$l=\lim_{n\to\infty}\lambda_{P_n}(f).$$

This is what I have done:

Let $Q\supseteq P$. If $Q=P$, then $\lambda_Q(f)=\lambda_P(f)$. If $Q\neq P$, then it must be the case that there is at least one $c\in Q$ such that $c\notin P$. This implies that $\lambda_Q(f)$ will have at least one more sum than $\lambda_P(f)$, and because distance is a non-negative value, we must have that $\lambda_P(f)\leqslant\lambda_Q(f)$.

Moreover, take the sequence $(P_n)_{n=1}^\infty=P_1\supset P_2\supset\cdots$. Then, from above, $\lambda_{P_1}(f)\leqslant\lambda_{P_2}(f)\leqslant\cdots$. Hence, if the limit exists, we must have that $$l=\lim_{n\to\infty}\lambda_{P_n}(f).$$

Does this seem reasonable?

wjmolina
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    There are more terms in the sum associated to $Q$ than the one associated to $P$, but they might be smaller. To prove $\lambda_P(f) \le \lambda_Q(f)$ you need to use the triangle inequality. As for the second part, your sequence is increasing and bounded, so it does indeed converge, but you can't be sure it converges to $l$. Instead of taking any sequence $(P_n)$, you should try to construct one (recall that $l$ is defined as a sup so try to use that) to make sure you get the right limit. – Joel Cohen Apr 27 '12 at 01:57
  • You just shed a ton of light on this problem for me! However, I have a few questions: for the first question, how could the sums of $Q$ be smaller than those of $P$, when we have already assumed that $Q\supseteq P$? And for the second question, would taking the sequence $P_1=[a,b]$, $P_2=[a,\frac{a+b}{2},b]$, $P_3=[a,\frac{3a+b}{4},\frac{a+b}{2},\frac{a+3b}{4},b]$, $\dots$ be a good start? – wjmolina Apr 27 '12 at 02:42
  • @Josué: I think that sequence would work, but only because $f$ is assumed to be continuous. That assumption actually isn't necessary to derive the result, and I think it would be a lot more work to do it using that than just using the properties of the supremum as Joel suggested. The only potential benefit would be that your proof would be constructive whereas the other one would require the axiom of choice. – joriki Apr 29 '12 at 17:11
  • @JoelCohen: perhaps you should turn your comment to an answer? – leonbloy Apr 29 '12 at 17:14

1 Answers1

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Ad $(1)$: The quantity $\lambda_P(f)$ is a sum of lengths of segments $[z_{k-1},z_k]$ where $z_k=\bigl(x_k,f(x_k)\bigr)$. Introducing a new data point $\zeta:=\bigl(\xi,f(\xi)\bigr)$ with $x_{k-1}<\xi<x_k$ makes $|\zeta-z_{k-1}|+|z_k-\zeta|\geq|z_k-z_{k-1}|$. This proves $\lambda_Q\geq \lambda_P$ when $Q\supset P$.

Ad $(2)$: Whatever $\ell:=\sup_P\lambda_P(f)\leq\infty$ there is a sequence $(Q_n)_{n\geq0}$ of partitions with $$\lim_{n\to\infty}\lambda_{Q_n}(f)=\ell\ .\qquad(3)$$ Put $P_n:=\bigcup_{k\leq n} Q_k$. Then by $(1)$ the $\lambda_n:=\lambda_{P_n}$ form an increasing sequence with $\lambda_n\leq \ell$ for all $n\geq1$. On the other hand, given any $\ell'<\ell$, because of $(3)$ there is an $m$ with $\lambda_{Q_m}\geq\ell'$. Therefore one has $$\lambda_n\geq \lambda_m\geq \lambda_{Q_m}\geq\ell'$$ for all $n\geq m$. This proves $\lim_{n\to\infty}\lambda_n=\ell$.

Christian Blatter
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