Can anybody explain this to me? (Analysis)

1. Nov 22, 2012

Artusartos

For question 32.2 in this link:

http://people.ischool.berkeley.edu/~johnsonb/Welcome_files/104/104hw11sum06.pdf [Broken]

I did not understand how $b^2/2 \leq U(f)$. We know that we have strict inequality in $t_{k+1} > \frac{t_k + t_{k+1}}{2}$...so don't we need to have $b^2/2 < U(f)$ instead of $b^2/2 \leq U(f)$?

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2. Nov 22, 2012

Ray Vickson

If a strict inequality folds for any finite partitions, all you can conclude is that the non-strict inequality holds in the limit, or when taking the sup, etc. In other words, we cannot have U(f) < b^2/2, because in order to have that we would have to have U_P(f) < b^2/2 for some finite partition P.

To put it another way: we have 1/n > 0 for all positive integers n, but inf{1/n: n is a positive integer} = 0.

RGV

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3. Nov 22, 2012

Artusartos

Thanks, but...

"we cannot have U(f) < b^2/2, because in order to have that we would have to have U_P(f) < b^2/2 for some finite partition P."...but if you look at the last part of the solution, it actually shows that U(f,P) < b^2/2 for all partitions P...

4. Nov 22, 2012

Ray Vickson

OK, so I got the inequalities reversed. The argument is still the same.

RGV

5. Nov 22, 2012

Artusartos

No I wasn't saying that you got the inequalities reversed...

I was just saying that (at the very end of the solution), it is shown that U(f,P)> b^2/2 for all P.

6. Nov 22, 2012

haruspex

No, you had it right the first time.
Artusartos, Ray is saying the argument runs like this.
We know that for any finite partition U_P(f) > b2/2. Suppose in the limit U(f) < b2/2. That would imply U_P(f) < b2/2 for some finite partition P, contradicting what we know. Therefore U(f) ≥ b2/2.
The OP says, in part:
don't we need to have b2/2<U(f) ?​

Therein lies a misunderstanding. The relationship between b2/2 and U(f) is not a condition that's needed, it's a relationship we are trying to deduce. The point is that we cannot deduce U(f) > b2/2. This is because it is the limit of a sequence, and a sequence in which every term is > x can be equal to x in the limit. Instead, we can deduce the weaker relationship U(f) ≥ b2/2.

7. Nov 23, 2012

Artusartos

Thanks a lot, I think I understand it more now...but...

The other poster says "In other words, we cannot have U(f) < b^2/2, because in order to have that we would have to have U_P(f) < b^2/2 for some finite partition P." So (according to this statement) if we know that U_P(f) < b^2/2 for "some finite" partition P, then we can say that U(f) < b^2/2. In the text that I attached, it says that U_P(f) < b^2/2 for "all" finite partitions P...so why can't we say the same thing? Since, if we have U(f)<b^2/2 for "all" partitions, then we definitely have it for "some" partitions...

8. Nov 23, 2012

haruspex

No, that doesn't follow. You're turning A $\Rightarrow$ B into B $\Rightarrow$ A:
If {U(f) < b^2/2} then {$\exists$ P s.t. U_P(f) < b^2/2}​
is not the same as
If {$\exists$ P s.t. U_P(f) < b^2/2} then {U(f) < b^2/2}​

9. Nov 23, 2012

Artusartos

Oh, ok...I get it. Thanks :)