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It is a part of the proof of lower semi-continuity of mapping $x \mapsto \|x\|_{bv}$, where $x\colon[0,T]\to E$, $(E,d)$ is a metric space, and the norm is defined as $\|x\|_{bv}:=\sup_{t_{i}\in\mathcal{P}}\sum d(x_{t_{i}},x_{t_{i+1}})$, $\mathcal{P}$ is a partition of $[0,T]$.

The theory states:

If $\{x^{n}\}$ is a sequence of mapping from $[0,T]$ to $E$ of bounded variation (with $\|x^{n}\|_{bv}<\infty$) and $x^{n}\to x$ pointwise on $[0,T]$, then for all $0\leq s<t\leq T$: $$\|x\|_{bv}\leq \lim_{n\rightarrow \infty} \inf\|x^{n}\|_{bv}.$$

The proof is given as:

Let $P = \{0=t_{0}<t_{1}<\cdots<t_{K}=T\}$ be a partition.

$$\sum_{i=0}^{K-1}d(x_{t_{i}},x_{t_{i+1}})=\lim_{n\rightarrow \infty} \sum_{i=0}^{K-1}d(x_{t_{i}}^{n},x_{t_{i+1}}^{n})=\lim_{n\rightarrow \infty} \inf\sum_{i=0}^{K-1}d(x_{t_{i}}^{n},x_{t_{i+1}}^{n})\leq\lim_{n\rightarrow \infty} \inf\|x^{n}\|_{bv}.$$

I am stuck at the first equality, how to prove

$$\left|\sum_{i=0}^{K-1}d(x_{t_{i}},x_{t_{i+1}})-\sum_{i=0}^{K-1}d(x_{t_{i}}^{n},x_{t_{i+1}}^{n})\right| \rightarrow 0$$ as $n\rightarrow \infty$.

P.S: if my statement was not so clear, please feel free to point out.

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$K$ is a fixed integer, i.e. it doesn't depend on $n$. You can write $$\left|\sum_{i=0}^{K-1}d(x_{t_i},x_{t_{i+1}})-\sum_{i=0}^{K-1}d(x_{t_i}^n,x_{t_{i+1}}^n)\right|\leq \sum_{i=0}^{K-1}\left|d(x_{t_i},x_{t_{i+1}})-d(x_{t_i}^n,x_{t_{i+1}}^n)\right|,$$ and \begin{align*}\left|d(x_{t_i},x_{t_{i+1}})-d(x_{t_i}^n,x_{t_{i+1}}^n)\right|&\leq |d(x_{t_i},x_{t_{i+1}})-d(x_{t_{i+1}},x_{t_i}^n)|+|d(x_{t_{i+1}},x_{t_i}^n)-d(x_{t_i}^n,x_{t_{i+1}}^n)|\\ &\leq d(x_{t_i},x_{t_i}^n)+d(x_{t_{i+1}},x_{t_{i+1}}^n). \end{align*}

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hahaha...what an idiot I am. Thanks! –  newbie Nov 24 '11 at 18:47
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