Let $(X, d)$ be a metric space, and $\gamma: [a,b]\to X$ be a curve. For any partition $P=\{a=y_0 We denote: $\|P\|=\max_i|y_i-y_{i+1}|$ Now my question is the proof of the following statement:
$$\lim_{\|P\|\to 0}\Sigma(P)=L(\gamma)$$ The hard part for me to prove the statement is if $P$ and $Q$ are two partitions, with $\|P\|\le \|Q\|$, we only know $$\Sigma(P\cup Q)\ge\max(\Sigma(P), \Sigma(Q))$$ and this won't give me any contradiction when we assume there is a sequence of partitions say $P_i$ with $\|P_i\|\to 0$ and $\Sigma(P_i)\le L(\gamma)-\varepsilon_0$ for some fixed $\varepsilon_0>0$. Anybody can help? (btw. I thought that this may be similar with the proof of the Riemann sum for integrable function, but there one has the Osilation)
Length of curve in metric space
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real-analysis
geometry
differential-geometry
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0Is there a typo in the first formula? It should be $\sum_id(\gamma(y_i),\gamma(y_{i+1}))$. Are there particular assumptions on $\gamma$? – 2012-06-04
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0Thanks Davide, I've corrected the typo. We can assume the $\gamma$ is continuous and rectifible. – 2012-06-04
1 Answers
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- Take a partition $Q=\{y_0,\dots, y_n\}$ such that $\Sigma(Q)>L(\gamma)-\epsilon$.
- By the uniform continuity of $\gamma$, there exists $\delta>0$ such that $d(\gamma(t),\gamma(s))<\epsilon/n$ whenever $|t-s|<\delta$
- Let $P=\{x_0,\dots,x_m\}$ be any partition with $\|P\|<\delta$. For each $y_j$ there exists $x_{k(j)}$ such that $|x_{k(j)}-y_j|<\delta$.
- Use uniform continuity to estimate $\Sigma(\{x_{k(j)}\colon j=0,\dots,n\})$ from below.
There are some things to tidy up here, but this being homework, I'll leave the rest to you.