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I have been trying to solve a certain exercise which i found in my Real Analysis book but have not been able to solve it.

Context/Relevant definitions:

Given $f:[a, b] \rightarrow \mathbb{R}$, for each partition $P=\left\{t_{0}, \ldots, t_{n}\right\}$ in $[a, b]$ let: $$V(f ; P)=\sum_{i=1}^{n}\left|f\left(t_{i}\right)-f\left(t_{i-1}\right)\right|$$

When the set $\{V(f ; P) ; P=$ partition of $[a, b]\}$ is bounded, one says that f is a function of bounded variation and writes that: $$ V_{a}^{b}(f) = \sup _{p} V(f ; P) $$

Problem: Let $f$ be a continuous function of bounded variation. Show that, for each $\epsilon > 0$, there exists $\delta > 0$ such that: $$|P|<\delta \Rightarrow\left|V(f ; P)-V_{a}^{b}(f)\right|<\varepsilon$$ where $|P|$ stands for the length of the partition.

(The book does not say it, but I am thinking that $|P| = t_{n} - t_{0}$ does that make sense?)

What have I tried so far?

Given $f$ a continuous function of bounded variation and $\epsilon > 0$ we observe that:

$$\left|V(f ; P)-V_{a}^{b}(f)\right| = \left|V(f ; P)-\sup _{p} V(f ; P)\right|$$

Now, by using the triangle inequality, we obtain that: $$ \begin{split} \left|V(f ; P)-V_{a}^{b}(f)\right| & = \left|V(f ; P)-\sup _{p} V(f ; P)\right| \\ & \leq \left |V(f ; P)\right|+\left| \sup _{p} V(f ; P) \right| \end{split}$$ Now, since $f$ is of bounded variation, the set $V(f ; P)$ is bounded and we can write that:

$$\left|V(f ; P)\right|+\left| \sup _{p} V(f ; P) \right| \leq $$

$$\leq M + \left| \sup _{p} V(f ; P) \right|$$

Here I got stuck. I just don't know how to proceed.

Second attempt:

I tried writing what $\sup _{p} V(f ; P)$ means to see if i could do something meaningful with it, but i was only able to write that if:

$$ V_{a}^{b}(f) = \sup _{p} V(f ; P)$$

Then, given $\epsilon > 0$ there exists $\Phi \in V(f ; P)$ such that: $$V_{a}^{b}(f) - \epsilon < \Phi \leq V_{a}^{b}(f)$$ which means that:

$$V_{a}^{b}(f) - \epsilon < \sum_{i=1}^{n}\left|f\left(t_{i}\right)-f\left(t_{i-1}\right)\right| \leq V_{a}^{b}(f)$$

But got nothing good out of it. I also tried reverse engineering the inequality, but did not go so well.

Can someone help? I would greatly appreciate it. Thanks in advance, Lucas

Daniele Tampieri
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Lucas
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    I think you mean $|P|$ is the length of the largest subinterval of the partition $P$. The length of the partition is $b-a$. This is typically called the norm of $P$. – C Squared Jan 09 '21 at 03:04
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    See this nice answer: https://math.stackexchange.com/a/3130591/72031 Continuity is essential for the conclusion to hold. – Paramanand Singh Jan 09 '21 at 03:33
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    $|P|$ (norm or mesh of $P$) denotes the length of largest subinterval created by the partition. Thus if $P={x_0,x_1,x_2,\dots,x_n}$ then $|P|=\max_{i=1}^{n}(x_i-x_{i-1})$. – Paramanand Singh Jan 09 '21 at 03:35
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    I have upvoted your question which should serve as an example of how users should ask questions. Clear context, well written attempts (even though they don't lead to a solution). This is strong indication of a sincere effort on part of asker. – Paramanand Singh Jan 09 '21 at 03:48
  • @ParamanandSingh I really try very hard to write clear and precise questions! I am very happy about that! Regarding the answer you mentioned in the aforementioned link, i went there and wrote some questions about the solution. Is that the correct procedure? – Lucas Jan 09 '21 at 04:16
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    @Lucas: yes it is the accepted practice to ask clarification in comments. You should get a response soon from that user as they are very active on the site. – Paramanand Singh Jan 09 '21 at 04:17

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