How Does Negation and Reciprocal Affect Bounds and Integrability of Functions?

[micromass]

OK, so you know that f is integrable. So you know that $sup\{s_p\}=inf\{S_p\}$.

For -f, we define

$$m_i^\prime=inf\{-f(x)~\vert~x_{i-1}<x<x_i\}~\text{and}~M_i^\prime=sup\{-f(x)~\vert~x_{i-1}<x<x_i\}$$

and

$$s_p^\prime=\sum m_i\Delta x_i~\text{and}~S_p^\prime=\sum M_i^\prime \Delta x_i$$

You need to prove that $\sup\{s_p^\prime\}=inf\{S_p^\prime\}$ using the hypothesis $sup\{s_p\}=inf\{S_p\}$.

So, do you know a relation between $m_i^\prime,~M^\prime_i,s_p^\prime,S_p^\prime$ and $m_i,~M_i,~s_p,~S_p$??

 

[STEMucator]

OK, so you know that f is integrable. So you know that $sup\{s_p\}=inf\{S_p\}$.

For -f, we define

$$m_i^\prime=inf\{-f(x)~\vert~x_{i-1}<x<x_i\}~\text{and}~M_i^\prime=sup\{-f(x)~\vert~x_{i-1}<x<x_i\}$$

and

$$s_p^\prime=\sum m_i\Delta x_i~\text{and}~S_p^\prime=\sum M_i^\prime \Delta x_i$$

You need to prove that $\sup\{s_p^\prime\}=inf\{S_p^\prime\}$ using the hypothesis $sup\{s_p\}=inf\{S_p\}$.

So, do you know a relation between $m_i^\prime,~M^\prime_i,s_p^\prime,S_p^\prime$ and $m_i,~M_i,~s_p,~S_p$??

Indeed I do see a relationship between them.

$m_i ≤ m_{i}^{'} ≤ M_{i}^{'} ≤ M_i$

and

$s_p ≤ s_{p'} ≤ S_{p'} ≤ S_p$

 

[micromass]

Indeed I do see a relationship between them.

$m_i ≤ m_{i}^{'} ≤ M_{i}^{'} ≤ M_i$

and

$s_p ≤ s_{p'} ≤ S_{p'} ≤ S_p$

Do you have a proof for this relationship?

 

[STEMucator]

Do you have a proof for this relationship?

Well, ill start with : $m_i ≤ m_{i}^{'} ≤ M_{i}^{'} ≤ M_i$

First we start by partitioning [a,b] into sub-intervals where the ith sub interval is [x_i-1, x_i]. Then we get m_i ≤ M_i ( Clearly from how they're defined ).

Now we form a new partition p' from p ( a refinement ) by inserting a point, say x' into (x_i-1, x_i).

Then we get m_i' ≤ M_i'.

Now, m_i is the greatest lower bound over the entire interval while m_i' is the greatest lower bound on the refinement of the interval, so m_i ≤ m_i'.

M_i is the least upper bound over the entire interval while M_i' is the least upper bound on the refinement p', so M_i' ≤ M_i

Yielding the inequality. A similar argument will occur for the sums.

 

[micromass]

What do refinements have to do with this?? You seem to define $m_i^\prime$ as the infimum over a refinement. That's not how I defined it in my previous post. I defined it as the infimum of $-f$ over $(x_{i-1},x_i)$. I didn't say anything about refinements.

 

[STEMucator]

Oh boy, late night sloppiness kicking in here.

So knowing that sup{s_p} = inf{S_p}, I want to show that sup{s_p'} = inf{S_p'} with how you've defined it.

I know that s_p ≤ S_p and s_p' ≤ S_p' from the way you've constructed things.

So I want to get to the point that : s_p ≤ s_p' ≤ S_p' ≤ S_p

Is this what you were getting at?

 

[micromass]

So I want to get to the point that : s_p ≤ s_p' ≤ S_p' ≤ S_p

These inequalities won't even be true, so don't bother with trying to prove them.

Are there other relationships you see?? For example, by using $sup(-f)=-inf(f)$?

 

[STEMucator]

These inequalities won't even be true, so don't bother with trying to prove them.

Are there other relationships you see?? For example, by using $sup(-f)=-inf(f)$?

Yeah of course :

sup(-f) = -inf(f)
inf(-f) = -sup(f)

 

[micromass]

Yeah of course :

sup(-f) = -inf(f)
inf(-f) = -sup(f)

What does that imply in terms of the numbers $m_i,m^\prime_i,M_i,M_i^\prime$??

 

[STEMucator]

What does that imply in terms of the numbers $m_i,m^\prime_i,M_i,M_i^\prime$??

$m_i = inf \left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\}~\text{and}~M_i = sup \left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\}$

$m_i^\prime=inf\{-f(x)~\vert~x_{i-1}<x<x_i\}~\text{and}~M_i^\prime=sup\{-f(x)~\vert~x_{i-1}<x<x_i\}$

We also have :

sup(-f) = -inf(f)
inf(-f) = -sup(f)

Thus :

M_i' = -m_i
m_i' = -M_i

 

[micromass]

Right. So what does that imply in terms of $s_p,~s^\prime_p,~S_p,~S^\prime_p$?

 

[STEMucator]

Right. So what does that imply in terms of $s_p,~s^\prime_p,~S_p,~S^\prime_p$?

So :

S_p' = -s_p so S_p' + s_p = 0
s_p' = -S_p so s_p' + S_p = 0

Thus :

S_p' + s_p = s_p' + S_p
S_p' - s_p' = S_p - s_p

 

[micromass]

Right, so $S_p^\prime=-s_p$ and $s_p^\prime=-S_p$.

Now, try to prove that if $sup\{s_p\}=inf\{S_p\}$, then $sup\{s_p^\prime\}=inf\{S_p^\prime\}$.

 

[STEMucator]

Right, so $S_p^\prime=-s_p$ and $s_p^\prime=-S_p$.

Now, try to prove that if $sup\{s_p\}=inf\{S_p\}$, then $sup\{s_p^\prime\}=inf\{S_p^\prime\}$.

Well. We know : sup{s_p} = inf{S_p}, so :

sup{-S_p'} = inf{-s_p'}

So the least upper bound of -S_p' is the same thing as the greatest lower bound of -s_p'. So it must be the case that the least upper bound of s_p' is the same thing as the greatest lower bound of S_p' hence :

sup{s_p'} = inf{S_p'}

 

[micromass]

That's it!

 

[STEMucator]

That's it!

Hence -f is integrable on [a,b] o.o...

You sir have the patience of a saint for helping me with that one. The 1/f case is just a reproduction and I'm pretty sure I can handle it. Thanks so much for your help though.

I'm going to go pass out now lol...

 

[micromass]

I'm going to go pass out now lol...

Haha! You deserve some rest!