How Do You Prove Inequality for Bounded Functions in Real Analysis?

[choirgurlio]

Homework Statement

Let f and g be bounded functions on [a,b].

1. Prove that U(f+g)</=U(f)+U(g).

2. Find an example to show that a strict inequality may hold in part 1.

Homework Equations

Definition of absolute value?

The Attempt at a Solution

I know that a function f is bounded if its range f(D) is a bounded subset of R; that is, f is bounded if there exists M in R such that absvalue(f(x))</=M for all x in D.

The best I can think to solve the first part is to use some type of proof similar to the Triangle Inequality Proof used in Calculus. With this particular problem, I am having trouble understanding what U really is. I am thinking it is a set, but does it function like a composite i.e. U(f(x)) (U circle f)?

The second part, I would say just to start plugging in different equations for f and g. It seems this would be easier to show as opposed to part 1, which asks for how to prove it in general.

Does this sound right? What other ways can you solve this problem?

Thank you very much for your help!

 

[╔(σ_σ)╝]

I think U means union but it could be another function.
And I think
$$\bigcup \left (f +g \right) = \left\{y \left| y \in f+g \right\}$$Basically, f is considered the range of the function f(x).If could very well mean U(x)... some function.

The problem doesn't seem to be constructed properly. First of all, how do you know the domain of f and g are the same ? I guess you have to assume it is. Secondly, the U seems to mean union but I am not sure. It might mean another function.

What is your intepretation of U?

EDIT

After re-reading it seems that U means a function...

If it was then ...Let
f(x) = -2
g(x) = 6

Both on R

Let U(x) = 1/x

U(f+g) = U(4) = 1/4

U(f) + U(g) = -1/2 + 1/6 < 0

So U(f+g) > U(f) + U(g)

But if U(x) = x^2

And f(x) = -2
g(x) = 4

U( f+g) = U(2) = 4

U(f) + U(g) = 20

In this case

U(f+g) < U(f) + U(g) .So it seem that if U is a function the equality is not even correct.

 

[hgfalling]

Is U mu, by chance: $$\mu$$. Then this makes sense in the measure theory sense.

 

[arestes]

I believe we are dealing here with the Supremum of a function, that is, the LEAST UPPER BOUND.

This is absolutely trivial but I think the OP is a newbie in the subject. So let's start from the beginning.
It is correct that the range (the image of the domain D of f, which you denote as D) is a bounded set, by the definition of bounded function.

According to a crucial property of the real numbers (which?), any bounded subset of the reals has a supremum.
You can also show, using the definition of supremum, that the supremum is always UNIQUE, that is, there can only be one supremum for any given set, if there is any at all. There are subsets of the reals numbers (including the whole real line itself, considered as a subset of itself) that do NOT have an upper bound, let alone a least upper bound. The aforementioned property of the reals numbers is quite important because it serves to distinguish the reals from the rational numbers, for example.

Since f and g have each a UNIQUE supremum, it makes sense to define $$U \left (f \right) = \sup_{x\in D}\{f(x)\}= \sup\{f(x) : x\in D\}$$
for f and g.

Note that if you "define" something you have to prove that it is WELL-DEFINED, that is, that it exists, that it doesn't depend on which way you represent the initial object you will associate it with (in this case the function f, which no matter if you write it as f-1 +1 or 2f-f, it still has the same supremum...this is something obvious in this case but in other areas of mathematics you will work with equivalence classes and this will not be so obvious) and that it is UNIQUE.Now, recall what a supremum is: as its name states it is THE SMALLEST UPPER BOUND OF ALL for that given set. Therefore, if you find any upper bound of that set, it must be true that this is bigger or at worst equal to the supremum.
From the mere fact that a supremum is, by definition, an upper bound we get
$$f(x) \leq U(f)$$

and also that

$$g(x) \leq U(g)$$

This implies that

$$(f+g)(x) \leq U(f) + U(g)$$
for all $$x\in D$$. This means that $$U(f) + U(g)$$ is an upper bound for the set $$(f+g)(D)$$. What does this imply according to the definition of the supremum of this last set?

For the second part, you need to prove that the the supremum of the sum is NOT the sum of supremums. This means that ocasionally we can get the equality to hold but in GENERAL this won't happen. Take the functions f(x)=x and g(x)=1-x defined in the interval [0, 1] and therefore, bounded in that domain. Check that the equality does NOT hold. This single COUNTEREXAMPLE (in which we specialize in a particular case using allowed functions and a specific interval) proves the assertion in the second part of your problem.
Since you only need ONE counterexample to show that something is not true, This part of the problem can be rephrased as saying "Show that the equality cannot hold identically". (as opposed to "ocassionally" because it CAN hold sometimes).

 

[╔(σ_σ)╝]

U been the suprema makes sense. I am suprised that OP didn't even know this; at least he should have verified what section of the book the question came from :-(.