Properties of Functions of Bounded Variation

[joypav]

Sorry for all the questions. Reviewing for my midterm next week. Fun fun.
If someone could take a look at my proof for (a) and help me out with (b) that'd be awesome!

(a) Let $\Delta$ be a partition of $[a, b]$ that is a refinement of partition $\Delta'$. For a real-value function $f$ on $[a, b]$, show that $V(f, \Delta') \leq V(f, \Delta)$.

Proof:
Consider
$\Delta' = \left\{x_0, x_1, ... , x_n\right\}$
and a refinement of $\Delta'$
$\Delta = \left\{x_0, x_1, ... , x_n, y\right\}$

Any refinement of $\Delta'$ can be created simply by adding points to $\Delta'$. So, it will suffice to consider the case when one point is added.
Assume, $x_j < y < x_{j+1}$ for some $0 \leq j \leq n-1$.
Then,
$\left| f(x_{j+1}) - f(x_j) \right| = \left| f(x_{j+1}) - f(y) + f(y) - f(x_j) \right| \leq \left| f(x_{j+1}) - f(y) \right| + \left| f(y) - f(x_j) \right|$
(By the triangle inequality)
Then,
$V(f, \Delta') = \sum_{k=0}^{n-1}\left| f(x_{k+1}) - f(x_k) \right| = \left| f(x_{j+1}) - f(x_j) \right| + \sum_{k=0, k \ne j}^{n-1}\left| f(x_{k+1}) - f(x_k) \right| \leq \left| f(x_{j+1}) - f(y) \right| + \left| f(y) - f(x_j) \right| + \sum_{k=0, k \ne j}^{n-1}\left| f(x_{k+1}) - f(x_k) \right| = V(f, \Delta) $

Then the proof is complete when adding one single point and we are done. (If we added more we would simply apply the triangle equality multiple times and repeat the same process.)

(b) Assume $f$ is of bounded variation on $[a, b]$. Show that there is a sequence of partitions $(\Delta_n)$ of $[a, b]$ for which the sequence $V(f, \Delta_n)$ is increasing and converges to $V(f, [a, b])$.

 

[GJA]

Hi joypav,

Your work for part (a) looks good, nice job.

Part (b) is essentially an induction argument. Here are a few hints to help you with it:

1. For $n\in\mathbb{N}$, by the definition of a function's variation on an interval as a supremum, there are partitions $\Delta_{n}'$ such that $$V(f,[a,b])-\dfrac{1}{n}< V(f,\Delta_{n}')\leq V(f,[a,b]).$$
2. Refine the $\Delta_{n}'$ to a new sequence $\Delta_{n}$ so that $\Delta_{n}\subseteq \Delta_{n+1}$ for all $n\in\mathbb{N}.$
3. Use your result from part (a) to conclude that the sequence of values $V(f,\Delta_{n})$ is increasing and satisfies $$\lim_{n} V(f,\Delta_{n})=V(f,[a,b]).$$

Let me know if anything is still unclear.

 

[joypav]

Hi joypav,

Your work for part (a) looks good, nice job.

Part (b) is essentially an induction argument. Here are a few hints to help you with it:

1. For $n\in\mathbb{N}$, by the definition of a function's variation on an interval as a supremum, there are partitions $\Delta_{n}'$ such that $$V(f,[a,b])-\dfrac{1}{n}< V(f,\Delta_{n}')\leq V(f,[a,b]).$$
2. Refine the $\Delta_{n}'$ to a new sequence $\Delta_{n}$ so that $\Delta_{n}\subseteq \Delta_{n+1}$ for all $n\in\mathbb{N}.$
3. Use your result from part (a) to conclude that the sequence of values $V(f,\Delta_{n})$ is increasing and satisfies $$\lim_{n} V(f,\Delta_{n})=V(f,[a,b]).$$

Let me know if anything is still unclear.

Let me see if I'm getting it...

$V(f, [a,b]) = sup_\Delta V(f,\Delta)$
(by definition of supremum) $\implies \forall \epsilon > 0, \exists V(f,\Delta_\epsilon)$ such that,
$V(f, [a,b]) - \epsilon < V(f,\Delta_\epsilon) \leq V(f, [a,b])$

To create our desired sequence, we will let $\epsilon = 1/n$.
Then,
$\forall \epsilon = 1/n > 0, \exists V(f,\Delta_n)$ such that,
$V(f, [a,b]) - 1/n < V(f,\Delta_n) \leq V(f, [a,b])$

Consider the sequence of partitions $(\Delta_n)_{n \in \Bbb{N}}$ of the smallest partition $\Delta_n$ that satisfy this inequality.
As $n$ increases, $1/n$ decreases, $\implies \forall n \in \Bbb{N}, V(f,\Delta_n) \leq V(f,\Delta_{n+1})$.
$\implies$ the sequence $(V(f,\Delta_n))_{n \in \Bbb{N}}$ is increasing.

Then,
$\forall \epsilon > 0$, if $n > 1/\epsilon$,
$\implies \left| V(f,\Delta_n) - V(f,[a,b]) \right| < \left| V(f,[a,b]) - 1/n - V(f,[a,b]) \right| = \left| -1/n \right| = 1/n < \epsilon$

$\implies (V(f,\Delta_n))_{n \in \Bbb{N}} \rightarrow V(f,[a,b])$I'm not sure when I need to refine my partitions $\Delta_n$? Let me know what needs fixing!

 

[GJA]

Hi joypav,

To create our desired sequence, we will let $\epsilon = 1/n$.
Then,
$\forall \epsilon = 1/n > 0, \exists V(f,\Delta_n)$ such that,
$V(f, [a,b]) - 1/n < V(f,\Delta_n) \leq V(f, [a,b])$

This is correct. In order to match the notation of part (a), I would recommend calling these $\Delta_{n}'$ because they are not the sequence of partitions we ultimately want to use.

Consider the sequence of partitions $(\Delta_n)_{n \in \Bbb{N}}$ of the smallest partition $\Delta_n$ that satisfy this inequality.
As $n$ increases, $1/n$ decreases, $\implies \forall n \in \Bbb{N}, V(f,\Delta_n) \leq V(f,\Delta_{n+1})$.
$\implies$ the sequence $(V(f,\Delta_n))_{n \in \Bbb{N}}$ is increasing.

This is not correct: $1-\dfrac{1}{n}<V(f,\Delta_{n})\leq V(f,[a,b])$ for all $n$ does not imply $V(f,\Delta_{n})$ is an increasing sequence. For example, $3.5<3.9\leq 4$ could correspond to $n=2$ and $3.\bar{6}<3.75\leq 4$ could correspond to $n=3$ (assuming $V(f,[a,b])=4$, say). But $3.75<3.9,$ so we can't say the sequence is necessarily increasing. This is why you need to refine the partitions and rely on part (a).

Edit. Put another way, the choices you're making for the $\Delta_{n}$ at this point are not related to each other in any way; i.e., what you choose at the $n^{\text{th}}$ step has no bearing on what choice you're making at the $(n+1)^{\text{st}}$ step. This is why we want to proceed inductively rather than randomly. That is, we want to use our knowledge of what occurs at the the $n^{\text{th}}$ step to guide our choice at the $(n+1)^{\text{st}}$ step. Doing so will allow us to meet the extra constraint of creating an increasing sequence, rather than one that simply converges to $V(f,[a,b]).$

Then,
$\forall \epsilon > 0$, if $n > 1/\epsilon$,
$\implies \left| V(f,\Delta_n) - V(f,[a,b]) \right| < \left| V(f,[a,b]) - 1/n - V(f,[a,b]) \right| = \left| -1/n \right| = 1/n < \epsilon$

$\implies (V(f,\Delta_n))_{n \in \Bbb{N}} \rightarrow V(f,[a,b])$

This is true because the $V(f,\Delta_{n})$ satisfy $V(f,[a,b])-\dfrac{1}{n}<V(f,\Delta_{n})\leq V(f,[a,b])$ for all $n$. Note that this step does not require the sequence $V(f,\Delta_{n})$ to be increasing.