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

[STEMucator]

Homework Statement

Suppose that f(x) is a bounded function on [a,b]

If M = sup(f) and m = inf(f), prove that -M = inf(-f) and -m = sup(-f). If m>0, show that 1/f is bounded and has 1/m as its supremum and 1/M as its infimum.

Let f be integrable on [a,b]. Prove that -f is integrable on [a,b] and if m>0, prove that 1/f is integrable on [a,b].

Homework Equations

So M is the least upper bound for f on [a,b] and m is the greatest lower bound for f on [a,b].

The Attempt at a Solution

So there's lots of stuff involved in this question. I'll start by trying to prove the first thing :

If m = inf(f) and M = sup(f), prove that -m = sup(-f) and -M = inf(-f).

So we know that : m ≤ f ≤ M

Forgive me if I'm wrong, but this seems like a one liner? Simply multiplying by (-1) gives :

-m ≥ -f ≥ -M so that -m is the least upper bound for -f and -M is the greatest lower bound for -f.

Before I go any further I'd like to check if I'm not jumping the gun a bit there.

 

[micromass]

Yes, I think you are jumping the gun. You correctly deduced that

-M\leq -f\leq -m

and this shows that -m is an upper bound of -f. But why is it a least upper bound?? How does that follow from the above inequality? Same for -M being the greatest lower bound.

 

[STEMucator]

Whoops I mistyped a few things in my original post. I fixed them now. I meant for M to be the sup and m to be the inf.

 

[micromass]

Whoops I mistyped a few things in my original post. I fixed them now. I meant for M to be the sup and m to be the inf.

Please don't edit your original posts, it destroys the flow of the topic and is hard to read

Anyway, my reply still holds, you have only proven that -m is an upper bound, but why is it the least upper bound? And why is -M the greatest lower bound?

 

[STEMucator]

Now as for what you were saying.

Since f is a bounded function on a nonempty set of real numbers, it has to have a least upper bound ( also a greatest lower bound ).

So if L₁ is any upper bound for -f, and L₂ is any lower bound for -f, then :

L_1 ≥ -m ≥ -f ≥ -M ≥ L_2

 

[micromass]

So if L₁ is any upper bound for f, and L₂ is any lower bound for f, then :

L_1 ≥ -m ≥ -f ≥ -M ≥ L_2

How did you obtain that inequality?? And why does that imply anything?

You need to prove: if a is another lower bound of -f, then a≤-M. And same for m.

 

[STEMucator]

Ah yes, so I want to show that -m is the least upper bound and -M is the greatest lower bound.

Right now we already have that -m is an upper bound for -f and -M is a lower bound for -f.

If L₁ is any upper bound bound for -f, we want to show L₁ ≥ -m. So :

L₁ ≥ -f
L₁ ≥ -m

This seems too obvious? Showing it for the lower bound is going to be exactly the same so I'll focus on this.

 

[micromass]

Ah yes, so I want to show that -m is the least upper bound and -M is the greatest lower bound.

Right now we already have that -m is an upper bound for -f and -M is a lower bound for -f.

If L₁ is any upper bound bound for -f, we want to show L₁ ≥ -m. So :

L₁ ≥ -f
L₁ ≥ -m

This seems too obvious? Showing it for the lower bound is going to be exactly the same so I'll focus on this.

So you say: if L_1\geq -f, then L_1\geq -m?? Can you clarify this? That doesn't seem obvious to me.

 

[STEMucator]

So you say: if L_1\geq -f, then L_1\geq -m?? Can you clarify this? That doesn't seem obvious to me.

Whoops, I had a brain hiccup there, forgot to do this :

L₁ ≥ -f
-L₁ ≤ f

The rest of the algebra is obvious if what I'm thinking is correct. I believe I can sub m in for f here?

 

[micromass]

Whoops, I had a brain hiccup there, forgot to do this :

L₁ ≥ -f
-L₁ ≤ f

The rest of the algebra is obvious if what I'm thinking is correct. I believe I can sub m in for f here?

OK, but can you write this out with some explanations instead of only writing down the inequalities??

For example, what you could writes:
We want to prove that -m is the greatest upper bound of -f. So let L be another upper bound of -f, this means: L≥-f. Multiplying by -1 yields f≤-L. Now ...

Can you complete the above? Try to write using sentences instead of just symbols.

 

[STEMucator]

Yessir, so.

Now ... since we are given that m ≤ f ≤ M and that m is the infimum of f, we know f attains the value m on the interval [a,b]. That is, f = m at some point. So :

-L₁ ≤ m
L₁ ≥ -m

Hence L₁ is an upper bound for -f and -m is a least upper bound for -f.

 

[micromass]

Yessir, so.

Now ... since we are given that m ≤ f ≤ M and that m is the infimum of f, we know f attains the value m on the interval [a,b].

Why should f attain the value m somewhere on the interval [a,b]??
If you're using the extreme value theorem, then think again: it only holds for continuous functions.

 

[STEMucator]

Why should f attain the value m somewhere on the interval [a,b]??
If you're using the extreme value theorem, then think again: it only holds for continuous functions.

I just re-read the theorem, you're right. I haven't used it in so long I had forgotten. Continuity would've made things nice :(.

So I need a different explanation, but the rest of what I said looks good.

Now I argued that L₁ was an upper bound for -f, so it would be a lower bound for f. Since m is the greatest lower bound for f, we get :

-L₁ ≤ m
L₁ ≥ -m

Hence L₁ is an upper bound for -f and -m is a least upper bound for -f.

 

[micromass]

I just re-read the theorem, you're right. I haven't used it in so long I had forgotten. Continuity would've made things nice :(.

So I need a different explanation, but the rest of what I said looks good.

Now I argued that L₁ was an upper bound for -f, so it would be a lower bound for f. Since m is the greatest lower bound for f, we get :

-L₁ ≤ m
L₁ ≥ -m

Hence L₁ is an upper bound for -f and -m is a least upper bound for -f.

OK, that is good! (although the "Hence L₁ is an upper bound of -f" is unnecessary since that was the hypothesis)

 

[STEMucator]

OK, that is good! (although the "Hence L₁ is an upper bound of -f" is unnecessary since that was the hypothesis)

Sweet, okay. I'll write everything nice and clean with a good explanation here then.

So we know that m ≤ f ≤ M which implies that -m ≥ -f ≥ -M. What we want to show that -m is the supremum of -f and that -M is the infimum of -f. Right now we have that -m is an upper bound for -f and -M is an lower bound for -f.

So suppose that L₁ is any upper bound for -f. So we get L₁ ≥ -f which yields -L₁ ≤ f. Now, since we argued that L₁ is any upper bound for -f, so it must be a lower bound for f. Since m is the greatest lower bound for f, we get -L₁ ≤ m which implies that L₁ ≥ -m. Hence -m is the least upper bound for -f.

Suppose now that L₂ is any lower bound for -f. So we get L₂ ≤ -f which yields -L₂ ≥ f. Now, since we argued that L₂ is any lower bound for -f, it must be an upper bound for f. Since M is the least upper bound for f, we get -L₂ ≥ M which implies that L₂ ≤ -M. Hence -M is the greatest lower bound for -f.

 

[STEMucator]

Now I'm going to attempt the second part of the question.

If m>0, show that 1/f is bounded and has 1/m as its supremum and 1/M as its infimum.

So we know that m ≤ f ≤ M which implies that 1/m ≥ 1/f ≥ 1/M.

This looks like an exact copy and paste of my post above except with a few different numbers if I'm not mistaken?

 

[micromass]

Now I'm going to attempt the second part of the question.

If m>0, show that 1/f is bounded and has 1/m as its supremum and 1/M as its infimum.

So we know that m ≤ f ≤ M which implies that 1/m ≥ 1/f ≥ 1/M.

This looks like an exact copy and paste of my post above except with a few different numbers if I'm not mistaken?

OK, so that proves that 1/m is an upper bound of 1/f and that 1/M is a lower bound. So that implies indeed that 1/f is bounded. (where did you use that m>0 anyway?)

But you still have more work to do if you wand to show that 1/M is the greatest lower bound and that 1/m is the least upper bound. The work is very analogous to above though.

 

[STEMucator]

I'm assuming that M can't be zero either otherwise we would have a problem here, but here goes my attempt :

So we know that m ≤ f ≤ M which implies that 1/m ≥ 1/f ≥ 1/M hence 1/f is bounded above by 1/m and bounded below by 1/M. We want to show that 1/m is the supremum of 1/f and 1/M is the infimum of 1/f. Right now we have that 1/m is an upper bound for 1/f and 1/M is a lower bound for 1/f.

So suppose that Q₁ is any upper bound for 1/f. Then we have Q₁ ≥ 1/f which implies that 1/Q₁ ≤ f. Now since Q₁ is any upper bound for 1/f, 1/Q₁ must be a lower bound for f. Now, since m is the greatest lower bound for f, we have that 1/Q₁ ≤ m which yields Q₁ ≥ 1/m. Hence 1/m is the least upper bound for 1/f.

Now suppose that Q₂ is any lower bound for 1/f. Then we have Q₂ ≤ 1/f which implies that 1/Q₂ ≥ f. Now since Q₂ is any lower bound for 1/f, 1/Q₂ must be an upper bound for f. Now, since M is the least upper bound for f, we have that 1/Q₂ ≥ M which yields Q₂ ≤ 1/M. Hence 1/M is the greatest lower bound for 1/f.

I believe that should do it.

 

[micromass]

That's correct! And you have written that perfectly!

 

[STEMucator]

That's correct! And you have written that perfectly!

Whoop :D, so I only have one part left here : Let f be integrable on [a,b]. Prove that -f is integrable on [a,b] and if m>0, prove that 1/f is integrable on [a,b].

So I'll be using some notation here, but I'll try to make it clear what is what.

I have a definition here : If I = sup(s_p) = inf(S_p) = J, then f is integrable.

Now, s_p = \sum_{i=1}^{n} m_i Δx_i and S_p = \sum_{i=1}^{n} M_i Δx_i.

To fill in the last few blanks, m_i = inf \left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\} and M_i = sup \left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\}

I'll start writing my proof in the next post since this is pretty cluttered as is.

 

[STEMucator]

I just want to show why f is integrable as it will make showing -f and 1/f incredibly easy to show as it will be a reproduction of the proof.

Now, we are given that f is integrable on [a,b]. We also know that sup(f) = M and inf(f) = m. Which saves us writing some latex stuff out.

So we get our lower and upper sums :

s_p = \sum_{i=1}^{n} m Δx_i = m(b-a) and S_p = \sum_{i=1}^{n} M Δx_i = M(b-a).

So we get I = sup(s_p) = m(b-a) and J = inf(S_p) = M(b-a).

Hence we get that I = m ≤ M = J.

Hmm have I done something wrong here?

 

[micromass]

Huh? But you are given that f is integrable. Why do you need to prove it then??

 

[STEMucator]

Huh? But you are given that f is integrable. Why do you need to prove it then??

I'm simply going to re-produce the proof for -f and 1/f. Since I already did the work of finding bounds for -f and 1/f I figured if I wrote out the proof for f, then it's going to be another copy and pasta for both of them.

I'm just curious to know why the proof I wrote didn't come out as expected.

 

[micromass]

The difference is that you are given that f is integrable. And you need to prove that 1/f and -f are integrable. You can't just prove it for f (a proof which is just the hypothesis) and say that 1/f and -f are analogous.

 

[STEMucator]

The difference is that you are given that f is integrable. And you need to prove that 1/f and -f are integrable. You can't just prove it for f (a proof which is just the hypothesis) and say that 1/f and -f are analogous.

I suppose I got a bit lazy there then. I'll write out the proofs formally then.

We want to show that -f is integrable on [a,b] and we know that -m = sup(-f) and -M = inf(-f). So we define our lower and upper sums :

s_p = - \sum_{i=1}^{n} MΔx_i = -M(b-a) and S_p = - \sum_{i=1}^{n} mΔx_i = -m(b-a).

Is this better?

 

[micromass]

I suppose I got a bit lazy there then. I'll write out the proofs formally then.

We want to show that -f is integrable on [a,b] and we know that -m = sup(-f) and -M = inf(-f). So we define our lower and upper sums :

s_p = - \sum_{i=1}^{n} MΔx_i = -M(b-a) and S_p = - \sum_{i=1}^{n} mΔx_i = -m(b-a).

Is this better?

Those lower and upper sums are not right. You seem to have a term MΔx_i in each sum. But shouldn't the supremum actually depend on each i??

 

[STEMucator]

Those lower and upper sums are not right. You seem to have a term MΔx_i in each sum. But shouldn't the supremum actually depend on each i??

Yes, this is true. How would I fix this actually given the definitions? I know that -m = sup(-f) and -M = inf(-f) from the prior work done in the question, but I'm guessing I'm overlooking something?

 

[micromass]

The terms in the sum should be M_i \Delta x_i. Where M_i is the supemum of f over [x_i,x_{i+1}]. Using what you have proven, you know that -M_i is the supremum of -f over [x_i,x_{i+1}]. Use this.

 

[STEMucator]

The terms in the sum should be M_i \Delta x_i. Where M_i is the supemum of f over [x_i,x_{i+1}]. Using what you have proven, you know that -M_i is the supremum of -f over [x_i,x_{i+1}]. Use this.

Ahhh I see now. So we have that -m = sup(-f) and -M = inf(-f).

So : m_i = inf\left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\} = -M and M_i = sup\left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\} = -m

So for the lower sum we get s_p = - \sum_{i=1}^{n} m_iΔx_i = -M(b-a) and for the upper sum we get S_p = - \sum_{i=1}^{n} M_iΔx_i = -m(b-a).

 

[micromass]

So : m_i = inf\left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\} = -M and M_i = sup\left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\} = -m

No, how did you conclude this?? You imply that each m_i are equal which is clearly not so.

 

[STEMucator]

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

We also have -m = sup(-f) and -M = inf(-f).

Hmm is it that -M_i ≤ -m and -m_i ≤ -M? Not quite sure what you're getting at.

 

[micromass]

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

We also have -m = sup(-f) and -M = inf(-f).

Hmm is it that -M_i ≤ -m and -m_i ≤ -M? Not quite sure what you're getting at.

Forget about m and M. They're not necessary here.

 

[STEMucator]

Forget about m and M. They're not necessary here.

Okay, I see. So. -M_i is the least upper bound for -f on the interval. -m_i is the greatest lower bound for -f on the interval.

So summing all of the M_i's over all the sub intervals I should get some number M_n if I'm not mistaken?

 

[micromass]

You need to prove that -f is integrable. Can you write out what you need to prove?? What is the definition of integrable?

 

[STEMucator]

You need to prove that -f is integrable. Can you write out what you need to prove?? What is the definition of integrable?

We partition [a,b] into sub-intervals.

For each i, we let : m_i = inf \left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\} and M_i = sup \left\{{f(x)|x_{i-1} ≤ x ≤ x_i}\right\}

Now, we define s_p = \sum_{i=1}^{n} m_i Δx_i as the lower sum and S_p = \sum_{i=1}^{n} M_i Δx_i as the upper sum.

Some more info in my notes :

Let M = sup \left\{{f(x)|x \in [a,b]}\right\} and m = inf \left\{{f(x)|x \in [a,b]}\right\}. Then we get s_p ≤ M(b-a) so the set of all possible s_p is bounded above.

Let I = sup{s_p} and J = inf{S_p}

Definition : if I = J, then f(x) is integrable.

 

[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!