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Let $f:[0,T]\rightarrow \mathbb{R}$ $$V(f) = \lim_{\|\Pi\|\rightarrow 0}\sum_{j=0}^{n-1}|f(t_{j+1})-f(t_j)|$$

where $\Pi$ is a partition of $[0,T]$ and define $\|\Pi\|$ as

$$\Pi=\{t_0,t_1,\cdots,t_n\}, \ \ 0=t_0<t_1<\ldots<t_n=T, \ \ \|\Pi\|=\max_i(t_{i+1}-t_i)$$

Now the solution manual says that

Suppose $V(f)$ is finite. Then for any $\epsilon>0$, there exists an $N\geq 1$, $$\sum_{j=0}^{n-1}|f(t_{j+1})-f(t_j)|<V(f) + \epsilon$$ for all $n\geq N$.

I am confused about the description of $+\epsilon$ and $\forall n \geq N$ parts. How do they and this description imply finite of $V(f)$?

I believe this is a trick widely used in real analysis; however, cannot still understand this.

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It just follows from the definition of $V(f)$ as a limit.

Given $\epsilon>0,$ there is a $\delta>0$ such that for $any$ paritition $P=\{0,x_1,\cdots ,x_{n-2},T\}$ of mesh less than $\delta$, we have

$\left|\sum_{j=0}^{n-1} |f(t_{j+1}) - f(t_j)|-V(f)\right|<\epsilon\tag1$

So, choose a partition $P$ with $N-1$ large enough so that the mesh of $P$ is less than $\delta.$ Then, $(1)$ holds, from which it follows that

$-\epsilon+V(f)<\sum_{j=0}^{N-1} |f(t_{j+1}) - f(t_j)|<V(f)+\epsilon\tag2$

But any partition with $n>N$ that refines $P$ will have mesh less than $\delta$ so in fact $(2)$ holds for $\textit{all}\ n\ge N$ and this is exactly the result you have in blockquotes.

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If $a_n$ is a sequence whose limit exists (call it $a := \lim_{n \to \infty} a_n$), then by the definition of a limit, for each $\epsilon > 0$ there exists $N$ such that $|a_n - a| < \epsilon$ for all $n \ge N$. Note that $|a_n - a| < \epsilon$ implies $a_n < a + \epsilon$.

Your situation is pretty much the same, except the limit is taken as $\|\Pi\| \to 0$. So for any $\epsilon > 0$, there exists a $\delta$ such that $$\sum_{j=0}^{n-1} |f(t_{j+1}) - f(t_j)| < V(f) + \epsilon$$ for any partition satisfying $\|\Pi\| < \delta$. Presumably you have omitted some condition (e.g., you are considering evenly-spaced partitions) that would imply that partitions $\Pi$ of size $\ge N$ will have small $\|\Pi\|$.

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What are you asking for the converse of what is sated in the manual. The manual says that if $V(f)<\infty$ the something happens. The stated property is immediate from the definition of limit. It is something like this: if $\lim a_n=l$ and $\epsilon >0$ then there exists $N $ such that $a_n <l+\epsilon$ for $n \geq N$. Do you agree with this statement?

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