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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 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?

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    @JoelCohen: perhaps you should turn your comment to an answer?2012-04-29

1 Answers 1

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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 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 $\lim_{n\to\infty}\lambda_n=\ell$.