MthT 430 Term Project 2006 Notes Revised

Problem 9 Remarks Revised December 6, 2006

This project should be completed by a group of not less than two nor more than four persons. The group should turn in one typeup of the project. The paper should include a description of each member's contributions to the project.

The project should be typed. For suggestions on typing, see

http://www2.math.uic.edu/~jlewis/mtht430/430type.pdf

Assignment due dates:

November 22, 2006 - 6 PM: Progress Report - A note to jlewis@uic.edu on your progress on the project - include the names of the members of your group.

November 29, 2006 - 5 PM: Completed typed project due.

I. Warmup - Inequalities Again and a Useful Fact

1.

Show that if x and y are numbers, then x £ y if and only if for every e > 0, x < y + e.

2.

Let A be a bounded set of numbers. Define

-A = {-x | x Î A }.

Show that

inf A = - sup (-A).

II. Understanding sup and inf - Equivalent Definitions

It is useful to note a working characterization of supA (A ¹ Æ):

If A ¹ Æ, supA is a number a such that ì ï í ï î For every x Î A, x £ a, and For every e > 0, there is an x Î A such that x > a- e.

The first condition means that a is an upper bound for A . The second condition means for every e > 0, a- e is not an upper bound for A .

3.

Let A be a nonempty set of numbers which is bounded above. Show that

b = sup A

if and only if for every e > 0

ì ï í ï î x < b + e for all x Î A, and x > b - e for some x Î A.

4.

Let A be a nonempty set of numbers which is bounded below. Show that

b = inf A

if and only if for every e > 0

ì ï í ï î x > b - e for all x Î A, and x < b + e for some x Î A.

III. Adding sup and inf

5.

(See Chapter 8 - Problem 13) Let A and B be two nonempty sets of numbers which are bounded (both above a nd below). Define

A+ B = {x | x = a + b,a Î A, b Î B }.

Show that

sup (A + B) = sup A + sup B.

Solution. If x = a + b, a Î A, b Î B, then

x = a + b £ sup A + sup B .

Thus supA + supB is an upper bound for A + B and

sup (A + B) £ sup A + sup B.

Given e > 0, there is an a Î A such that a > supA- e, and a b Î B such that b > supB - e. Then

sup (A + B) ³ a + b ³ sup A + sup B - 2e ,

so that

sup A + sup B < sup (A + B) + 2 e.

Thus, using Problem 1,

sup A + sup B £ sup (A + B).

Show that

inf (A + B) = inf A + inf B.

IV. More Adding sup and inf

If f is a bounded function on [0,1], we define

sup f = sup x Î [0,1]  f(x) inf f = inf x Î [0,1]  f(x)

6.

(Easy - see also Spivak Chapter 8 - Problem 13.) Show that if f and g are bounded functions on [0,1], then

sup (f + g) £ sup f + sup g.

Solution. The number supf +supg is an upper bound for the set {(f + g)(x) | x Î [0,1]}.

7.

Give an example of a pair of bounded functions f and g on [0,1] such that

sup (f + g) < sup f + sup g.

8.

Show that if f and g are bounded functions on [0,1], then

inf f + inf g £ inf (f + g).

9.

Show that if f and g are bounded functions on [0,1], then

inf f + sup g £ sup (f + g).

Solution 1. Fix x Î [0,1]. Then

g(x) = (f + g)(x) - f(x) £ sup (f + g) - inf f.

and

sup g £ sup (f + g) - inf f.

Solution 2. Given e > 0, there is an x Î [0,1] such that g(x) > supg - e. For this x,

inf f £ f(x) = (f + g)(x) - g(x) £ sup (f + g) - sup g + e.

It follows that: for every e > 0,

inf f + sup g £ sup (f + g) + e.

Remark. Several groups tried to use the two [in]equalities

inf f £ sup f, (* -true) sup f + sup g = sup (f + g). (** - false)

[In]equality (**) looks like Problem 5, but was shown not always true by the counterexample in Problem 7. I think the confusion arises from the interpretation of the notation

sup f = sup x Î [0,1]  f(x) = sup {f(x) |x Î [0,1]}.

Note that, as sets,

{(f + g)(x) |x Î [0,1]} Í {f(x) |x Î [0,1]} + {g(x) |x Î [0,1]},

but equality is not always true. The right hand side is

{f(x) |x Î [0,1]} + {g(x) |x Î [0,1] } = {f(x) + g(y) |x Î [0,1],y Î [0,1] }.

Remark on Language. Note that

{g(x) |x Î [0,1]} = {g(y) |y Î [0,1]} = {g(t) |t Î [0,1]} = {g(¨) |¨ Î [0,1]} = ¼.

Name an internal variable - x, y, t, ¨, ¼ - and then say what it means. Similarly, with a given function f,

lim x ® a  f(x) = lim y ® a  f(y) = lim t ® a  f(t) = lim ¨® a  f(¨) = ¼.

10.

(Easy - See the previous problem(s).) and Show that if f and g are bounded functions on [0,1], then

inf (f + g) £ inf f + sup g .

11.

(Monster Counterexample to Equality) Your group has shown: For two bounded functions f and g on [0,1],

inf f + inf g £ inf (f + g) £ inf f + sup g £ sup (f + g) £ sup f + sup g.

Give an example of a pair of bounded functions f and g on [0,1] such that

inf f + inf g < inf (f + g) < inf f + sup g < sup (f + g) < sup f + sup g.

Monster Counterexample. Let

f(x) = ì ï í ï î 2, 0 £ x < 0.5 , 0, 0.5 £ x £ 1.

g(x) = ì ï í ï î 0, 0 £ x < 0.25 , 1, 0.25 £ x £ 0.75, 2, 0.75 < x £ 1.

f(x) + g(x) = ì ï ï í ï ï î 2, 0 £ x < 0.25 , 3, 0.25 £ x < 0.5, 1, 0.5 £ x £ 0.75, 2, 0.75 < x £ 1.

inf f = inf g = 0, inf (f + g ) = 1, sup (f + g) = 3, sup f = sup g = 2.

N.B. The string of inequalities is related to similar inequalities regarding limsup and liminf in

http://www2.math.uic.edu/~jlewis/mtht430/chap8fproj.pdf

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