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10:22 AM
I have taken it :) yup!
@LittleRookie
 
It seems that your professor gave a correct proof, after all: Fixed point iteration exercise.
In our conversation here in chat I simply missed that it can be understood as proof by contradiction - as LutzL's answer suggest.
BTW since you accepted that answer, I suppose it was helpful to you. I will just remind that you can also upvote answers.
 
 
3 hours later…
1:41 PM
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6
Q: Bounded variation, difference of two increasing functions

AthiraProve that if $f$ is of bounded variation in $[a,b]$, it is the difference of two positive, monotonic increasing functions; and the difference of two bounded monotonic increasing functions is a function of bounded variation.

I was referring the answer here in this question ,BTW what is $var[a,x]$ ?
It may be notation for total variation.
 
If you are talking about Davide Giraudo answer, it is defined in the answer.
$\sup \sum_{j=1}^{n-1}|f(x_{j+1})-f(x_j)|=:\operatorname{Var}[a,x]$
I would say that something like $\operatorname{Var}_f[a,x]$ is more common. Or some other notation, which also includes the function $f$.
 
hi@MartinSleziak , can you help me the proof?
 
And calling it total variation seems to be correct, see Wikipedia: Total variation
 
Yes
 
@BAYMAX Proof of what?
 
1:47 PM
Proof of the above question, $G =F -f$
we don't know whether $f$ is increasing or not
 
ok, are there some parts of Davide Giraudo's answer which are unclear?
I'm going to have a closer look at the answer.
 
ok
okk...i think it's ok
that is $f = F - G$
 
Yes, in that answer he expressed $f$ as difference of those two functions.
 
yes
 
And the point is to shown that both of them are bounded and non-decreasing.
 
1:50 PM
ok as $f$ and $F$ are bounded so $G$ is bounded,right
 
Yes. It is difference of two bounded functions.
 
next I am looking at incresing part.
 
You meant the proof that $G$ is increasing?
If I am reading that answer correctly, he shows that $G(x_1)\le G(x_2)$ whenever $x_1\le x_2$.
 
Yes
 
In fact, I believe it should not be too difficult to show $\operatorname{Var}[a,x_2] = \operatorname{Var}[a,x_1]+ \operatorname{Var}[x_1,x_2]$. Which is slightly stronger than the inequality in the answer.
In the other words, total variation is additive.
 
1:55 PM
I am trying to show.
It is like Supremum property?
 
We will definitely need to use some properties of supremum.
What exactly are you trying to show now? Maybe we could start with the inequality from the answer, which is easier.
 
a --------------$x_{1}$---------------$x_{2}$ be the numberline
Yes I was thinking about how t oget that inequality?
 
This one?
$\operatorname{Var}[a,x_1]+|f(x_2)-f(x_1)|\leq \operatorname{Var}[a,x_2]$
 
Yes
first one is absolute or modulus property
 
If you do not mind, I'd like to change notation to $$\operatorname{Var}[a,t_1]+|f(t_2)-f(t_1)|\leq \operatorname{Var}[a,t_2]$$
 
1:59 PM
Yes sure
 
Simply because we have already used the $x_1,x_2,\dots$ in the definition of variation.
I do not want to use the same variable in two meanings.
 
I think it must be equalt to
wait
Can I answer
?
 
Ok, I'm listening. (Or reading.)
 
In $Var[a,x_{2}]$ there is a term $sup{|f(x_{2}) - f(x_{1})|}$
but in the L.H.S
there is only $|f_{x_{2}} - f_{x1}|$
I think it's wrong!
 
@BAYMAX Did you mean to say L.H.S. here?
 
2:03 PM
yes
L.H.S
 
Well, what you wrote as your argument is either incorrect, or I do not understand it.
One of the problems might be that you are confusing two meanings in which the variable $x_2$ is used.
 
Ok..I am listening.
 
So let us stick to $\operatorname{Var}[a,t_2]$ to avoid the problem.
 
yes
 
The value $\operatorname{Var}[a,t_2]$ is a supremum taken over all partitions of the interval $[a,t_2]$.
 
2:05 PM
so $a = x_{1}<x_{2}<...<x_{n} = t_{2}$ ?
 
yes
 
To compare it with the LHS, you should consider only special kind of partitions. Namely the ones where the last subinterval is $[t_1,t_2]$.
 
yeah..that is the point
 
I.e., partitions of the form $a=x_1<x_2<\dots<x_{n-1}=t_1<x_n=t_2$.
 
2:06 PM
Yes
 
If you take supremum only over thesepartitions, you get $\operatorname{Var}[a,x_1]+|f(x_2)-f(x_1)|$.
 
yes
 
And since the RHS is a supremum of a superset, you get the inequality.
 
$Sup(|f(x2 )-f(x1)|) = |f(x2)-f(x1))|$
 
And of course, I should have written $\operatorname{Var}[a,t_1]+|f(t_2)-f(t_1)$.
 
2:09 PM
yes
how inequality
We get equality
 
No we do not get equality.
 
chat.stackexchange.com/transcript/message/36122922#36122922
4 mins ago, by Martin Sleziak
If you take supremum only over thesepartitions, you get $\operatorname{Var}[a,x_1]+|f(x_2)-f(x_1)|$.
We calculated that for $a = x1 <....<x_{n-1} = t1 < x_{n} = t2$
 
Oh, I thought that you meant the equality between $\operatorname{Var}[a,t_1]+|f(t_2)-f(t_1)$ and $\operatorname{Var}[a,t_2]$.
 
Yes i mean that only!
 
Well, $\operatorname{Var}[a,t_1]+|f(t_2)-f(t_1)|=\operatorname{Var}[a,t_2]$ is not true in general.
We can only show $\operatorname{Var}[a,t_1]+|f(t_2)-f(t_1)| \le \operatorname{Var}[a,t_2]$.
 
2:13 PM
5 mins ago, by BAYMAX
$Sup(|f(x2 )-f(x1)|) = |f(x2)-f(x1))|$
is this correct!
 
But let's step back a little.
 
ok
 
What you wrote is correct, but it does not have anything to do what we are trying to prove.
 
I am saying that in the summation of $Var[a,t_{2}]$
 
The basic idea to show that the supremum over those special partitions is indeed $\operatorname{Var}[a,t_1]+|f(t_2)-f(t_1)|$ is noticing that $$\sup \left(|f(t_2)-f(t_1)| +\sum_{i=1}^{n-1} |f(x_n)-f(x_{n-1})|\right)= |f(t_2)-f(t_1)| + \sup \left(\sum_{i=1}^{n-1} |f(x_n)-f(x_{n-1})|\right)$$
 
2:15 PM
We have the last term as $Sup|f(t2)- f(t1)|$
yes
hence the equality
 
Is anyone willing to look at a proof I am working on and offer some help?
1
Q: Three Distinct Points and Their Normal Lines

Nick Pavini Suppose That three points on the graph of $y=x^2$ have the property that their normal lines intersect at a common point. Show that the sum of their $x$-coordinates is $0$. I have a lot going but can not finish it. Proof: Let $(a,a^2)$, $(b,b^2)$, and $(c,c^2)$ be three distinct points on $...

 
@BAYMAX So you are claiming that you can prove $\operatorname{Var}[a,t_1]+|f(t_2)-f(t_1)|=\operatorname{Var}[a,t_2]$?
 
Yes, what you wrote above states it right?
 
It should be relatively easy to find a counterexample. In fact, I think any function which is not monotone on the interval $[t_1,t_2]$ will give you a counterexample.
@BAYMAX No, it does not. What I am trying to prove is to get in the end this inequality: $$\operatorname{Var}[a,t_1]+|f(t_2)-f(t_1)|\le\operatorname{Var}[a,t_2].$$
 
$Var[a,t2] = \sup \left(|f(t_2)-f(t_1)| +\sum_{i=1}^{n-1} |f(x_n)-f(x_{n-1})|\right) $
 
2:19 PM
And what I wrote above basically says that supremum taken only over some special partitions is equal to $\operatorname{Var}[a,t_1]+|f(t_2)-f(t_1)|$.
@BAYMAX This is not true. Because you only counted the partitions which contain the interval $[t_1,t_2]$.
But the definition of $\operatorname{Var}[a,t_2]$ takes all partitions of $[a,t_2]$.
In some of them, the interval $[t_1,t_2]$ can be subdivided into smaller subintervals.
In your last message, you did not count those partitions, where $[t_1,t_2$ is divided into smaller parts.
 
I am little confused! will see your argument and think over it deep.
will message after that
sorry
 
Ok, see you later!
 
ok!
 
2:47 PM
Just in case somebody wonders, the above conversation is related to details of this answer.
 
3:05 PM
Oh@MartinSleziak I got it I think
There are two partitions involved here
one is normal partition from $a$ to $t2$
another is from $a$ to $t1$ and a single interval of $[t1,t2]$
in $[t1,t2]$ there are no partitions .
Now I am thinking about how many no. of partitions , does that matter
in both the cases of partitions?
Now as I scroll above I see that you mentioned superset term , I was thinking which is the superset here?
 
@BAYMAX I think it is two kinds of partitions, rather than just two partitions.
 
yes two kinds of partitions
 
@BAYMAX The set of partitions containing interval $[t_1,t_2]$ is clearly a subset of the set of all partitions of $[a,t_2]$.
The LHS and the RHS are suprema taken over these two sets of partitions.
 
yes@MartinSleziak
i got firstline
so $Var[t1,t2] \leq Var[a,t2]$
 
Ok, let $\mathcal P_1$ be the set of all partitions containing $[t_1,t_2]$ as the last interval.
And let $\mathcal P_2$ be the set of all partitions of $[a,t_2]$.
Clearly $\mathcal P_1 \subseteq \mathcal P_2$.
 
3:18 PM
The set of partitions containing interval [t1,t2] , if I take $Var$ of the latter then I would get $Var[t1,t2]$
 
For each partition we have the corresponding sum $S(P)$.
Because of the inclusion, we have $\sup\{S(P); P\in\mathcal P_1\} \le \sup\{S(P); P\in\mathcal P_2\}$.
 
yes i got it $P1 \subset P2$
 
We also have $\operatorname{Var}[a,t_2]=\sup\{S(P); P\in\mathcal P_2\}$. (This is precisely the definition.)
I claim that $\sup\{S(P); P\in\mathcal P_1\} \overset{(*)}= \operatorname{Var}[a,t_1]+|f(t_2)-f(t_1)|$.
 
yes
Yes
 
The equation $(*)$ in combination with what I wrote before gives use the desired inequality: $$\operatorname{Var}[a,t_1]+|f(t_2)-f(t_1)| \le \operatorname{Var}[a,t_2]$$
 
3:22 PM
we are left to prove $\sup\{S(P); P\in\mathcal P_1\} \overset{(*)}= \operatorname{Var}[a,t_1]+|f(t_2)-f(t_1)|$
?
we still have to prove this ?
 
Well, unless you agree that we have already proved it.
We have discussed about this.
1 hour ago, by Martin Sleziak
The basic idea to show that the supremum over those special partitions is indeed $\operatorname{Var}[a,t_1]+|f(t_2)-f(t_1)|$ is noticing that $$\sup \left(|f(t_2)-f(t_1)| +\sum_{i=1}^{n-1} |f(x_n)-f(x_{n-1})|\right)= |f(t_2)-f(t_1)| + \sup \left(\sum_{i=1}^{n-1} |f(x_n)-f(x_{n-1})|\right)$$
 
here we have used $sup(|f(t1) -f(t2)|) = |f(t1) -f(t2)|$ ?
 
I do not think we have used this anywhere. (And I am not really sure what you mean by this.)
BTW there is an obvious typo in my message above.
It should have been $$\sup \left(|f(t_2)-f(t_1)| +\sum_{i=1}^{n-1} |f(x_i)-f(x_{i-1})|\right)= |f(t_2)-f(t_1)| + \sup \left(\sum_{i=1}^{n-1} |f(x_i)-f(x_{i-1})|\right)$$
 
sorry I am not alert enough!
Yes in the above equation how you carried over the supremum over the two additive terms
 
And perhaps I should add that supremum is taken over all partitions from $\mathcal P_1$. (The ones where $x_{n-1}=t_1$ and $x_n=t_2$.
@BAYMAX I am just using that $\sup\{x+C; x\in A\} = \sup\{x; x\in A\}+C$.
Or to write it more succinctly, $\sup\{x+C; x\in A\}=C+\sup A$.
Here $C$ is a constant (does not depend on $x$).
It should be clear that if we add a constant $C$ to each element of some set, the supremum changes in the same way.
 
3:30 PM
Yes now I got it!
I was thinking sup to be distributive over
still knowing that when we add
 
To related it to the above inequality, $|f(t_2)-f(t_1)$ is the constant. (It does not depend on which partition belonging to $\mathcal P_1$ we choose.)
 
each element of a set a number $C$ then the sup gets added with $C$ too!
Yes yes
but wait
yes it is ok!
I will write a proof on this again and post it here.
 
3:52 PM
@BAYMAX cool, then can i ask you about proofs on showing a set is countable?
 
actually I am busy writing a proof! but still you should post your question so that other potential users can see and answer your question!
@LittleRookie
 
Its more on the concept of countable sets
I'll ask you when u're free okay? :)
Because im kind of confused about the "counting algorithm/proof"
 
but still you can try it now ... just even it is one line
 
So there are instances where i can find an explicit expression for the function mapping natural numbers to a set which we would like to show that it is countable.
And there are also instances where i cannot find an explicit expression for the function, then i would suggest a method/algorithm to map the natural numbers to my set
sometimes, i show it for first 10 natural numbers, then i would write repeating the process for the other natural numbers
Now, the big issue i have is to show that the mapping/function is indeed bijective.
In cases, where i do not have the explicit formula for the function, how do i show that it is a bijection?
For example, the well known case for Rational numbers, the mapping has no explicit form, instead a counting algorithm is proposed.
 
4:15 PM
may be the second one is blur , sorry for that!
Yes @LittleRookie this is a nice question!
 
Yep, can u help me with it. Until now, im still confused on showing the mapping is a bijection when it does not have an explicit expression. :(
 
@LittleRookie I have copied your messages into set theory chat room: chat.stackexchange.com/transcript/2318/2017/3/18
Asking about countable sets is probably closer to the topic of that room.
However, I have to go off-line at the moment. If I have time, I might leave some comments there later.
 
4:33 PM
Sure thanks
 
4:46 PM
@LittleRookie I left some comments on rationals in the other chatroom. chat.stackexchange.com/transcript/2318/2017/3/18
 
4:59 PM
@BAYMAX I think you are missing right bracket in $\sup \left(\sum_{i=1}^{n-1} |f(x_i)-f(x_{i-1})|\right)$. (The second line from the end of proof.)
But other than some typos, it seems ok to me. (And nice handwriting.)
 
Ha ha.. thank you very much @MartinSleziak for the proof...
what should I study next in functions of bounded variations?
 
I am not sure. Don't you follow some kind of textbook?
 
$\int_{0}^{x} f$ is absolutely continuous
Yes
but I saw in some textbooks as an appendix of 3 pages!
will look into other books
ohh
Royden pg116
ohhh
wow
I
It was Jordan's theorem!
Yes my next objective is "Absolutely continuous functions" !
 
I'd guess good books in real analysis should have something on bounded variation and absolutely continuous functions.
BTW the word absolutely continuous is used also for measures, not only for functions: Absolute continuity
So the question whether integral is absolutely continuous can be interpreted in two possible ways.
 
yes!
I was wondering how you can do that in finger tips! ... still I am interested to be like that one day or some day!
 
5:08 PM
@BAYMAX Do what?
 
Mathematics , nice illustrations , explanations...@MartinSleziak.
 
I am not entirely sure how to interpret that thing from Royden - it might be about absolute continuity w.r.t Lebesgue measure.
 
hmm...will read about it and will inform you
hey @MartinSleziak , I am able to find very less users on my MSc thesis
like on the topic of Dynamical systems
but still it was interesting that in those too measure theory was used!
like Lebesgue measures
 
I would guess that there are a few people around this site who know enough about dynamical systems.
 
but pinging them might not be a good idea!
 
5:17 PM
On the other hand, if you have a question and ask on the main site, somebody might respond.
 
Yes
 
Ok, I'll have to leave for a bit.
 
Bye
see you later!
 

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