Finding upper sum and lower sums

Question

Question: Let a, b ∈ R with a < b. Suppose that f is increasing on [a, b] (i.e., ∀x, y ∈ [a, b], f(x) < f(y) whenever x < y). Notice that f is automatically bounded on [a, b] because f(a) < f(x) < f(b) for all x ∈ [a, b]. If P = {x0 = a, x1, x2, . . . , xn = b} is a partition of [a, b], find L(f, P) and U(f, P).

My attempt:

For $i = 1, \cdots, n$ Define: $$M_i = sup\{f(x)|x \in [x_{i-1}, x_{i}]\} \text{ By definition} = sup\{x,y\} = y \text{ because Q and II are densed in R}$$

$$m_i = inf\{f(x)|x \in [x_{i-1}, x_{i}]\} \text{ By definition} = inf\{x,y\} = x \text{ because Q and II are densed in R}$$

U(f,P) = $\sum_{i=1}^{n} M_i(x_i - x_{i-1})$ by def

= $M_i (x_n - x_0)$ given info / telescoping

= $f(x_i)(b-a)$ (given info)

L(f,P) = $\sum_{i=1}^{n} m_i(x_i - x_{i-1})$ by def

= $f(x_{i-1}) (x_n - x_0)$ given info / telescoping

= $f(x_{i-1})(b-a)$ (given info)

Is this right, if not how do I do this. Thx

Answer 1

No, this is not correct. First, let's look at this step:

$$\sum_{i=1}^n M_i(x_i - x_{i-1}) = M_i(x_n - x_0)$$

Something you might notice about it: the value of the left-hand side does not depend on $i$ at all. $i$ occurs in it as a "dummy variable". It is only there for the sake of the notation. If you expand it out, there is no $i$ in the expression at all: $$M_1(x_1 - x_0) + M_2(x_2 - x_1) + \dots + M_n(x_n - x_{n-1})$$

But on the right-hand side, $i$ appears! Which value of $i$ is it? The only way this expression could be true is if it held for all $i$, which since the rest of the expression does not depend on $i$ would mean that all $M_i$ have to be the same value. However, it is trivial to come up with example functions $f$ that satify the conditions without the $M_i$ being the same. $f(x) = x$ is one.

Second, the only way this could "telescope" is again if all the $M_i$ have the same value, which is something that we know is usually not going to be true. Telescoping occurs when the exact same value is both added in and subtracted out of a sum. That is not the case here. When you drop those parentheses, the sum becomes: $$\color{red}{M_1x_1} - M_1x_0 + \color{blue}{M_2x_2} \color{red}{- M_2x_1} + M_3x_3 \color{blue}{- M_3x_2} +\dots + M_nx_n - M_nx_{n-1}$$

Note that is $M_1x_1$ that is added in, but it is $M_2x_1$ that is subtracted out. It is $M_2x_2$ that is added in, but $M_3x_2$ that is subtracted out. If all the $M_i$ were the same, these terms would cancel out and you would indeed be left with only the lowest and highest. But when the $M_i$ are different, this does not hold.

Third, you define $M_i = \sup\{f(x) \mid x \in [x_{i-1}, x_i]\}$, which is fine. But then you say $M_i = \sup\{x,y\} = y$ Wait a moment! What is $x$, what is $y$? There is no fixed value of $x$ or $y$ defined in this problem or your work. Even in the starting equation, $x$ only occurs as a dummy variable again. The scope of its definition is just within that set notation. And then you justify this with "because $\Bbb Q$ and $\Pi$ are dense in $\Bbb R$". Huh?? What does $\Bbb Q$ have to do with this problem? There is no mention of rational numbers anywhere in it! And what is $\Pi$? That isn't a standard notation, and you haven't given a definition for it either. And anyway what would the density of any set have to do with the supremum of $f$ on the interval $[x_{i-1}, x_i]$?

So how do you solve it? NOT by waving your hands and muttering mathemagical mumbo-jumbo when the going gets rough.

As I said, you were fine with $M_i = \sup\{f(x) \mid x \in [x_{i-1}, x_i]\}$. But what do we know about $f$? We were told that it is increasing, so the larger a value $x$ is, the larger $f(x)$ will be. What is the largest value of $x$ possible when $x \in [x_{i-1}, x_i]$? It will be $x_i$. So as user247327 pointed out $M_i = f(x_i)$, and we can drop the unknown $M_i$ for an explicit value. Thus $$U(f, P) = \sum_{i=1}^n f(x_i)(x_i - x_{i-1})$$

That is it. With only the information given, there is no further simplification possible for this expression.

I'll leave you to figure out $L(f, p)$, which just requires figuring out where the lowest value of $f$ occurs on each interval $[x_{i-1}, x_i]$ I'll give you the hint that it does not occur at the same place as the maximum.