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Homework Help: Riemann integrable proof

  1. Mar 17, 2010 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    Right now, I'm still trying to understand why the hint is true. This is what I've got so far...
    Let ||f||= sup{|f(x)|: x E [a,b]}
    [tex]M_i(f,P)[/tex] = sup{f(x): [tex]x_{i - 1}[/tex] ≤ x ≤ [tex]x_i[/tex]}
    [tex]m_i(f,P)[/tex] = inf{f(x): [tex]x_{i - 1}[/tex] ≤ x ≤ [tex]x_i[/tex]} where P is a partition of [a,b]

    Let x,t E [[tex]x_{i - 1}, x_i[/tex]]
    Then |f(x)g(x)-f(t)g(t)| ≤ |f(x)| |g(x)-g(t)| + |f(x)-f(t)| |g(t)|
    ≤ ||f|| [[tex]M_i(g,P) - m_i(g,P)[/tex]] + [[tex]M_i(f,P) - m_i(f,P)[/tex]] ||g||

    How can we finish proving the hint from here? I have no idea how to get Mi(fg, P) - mi(fg, P) on the LHS of the inequality...

    I hope somebody can help me!
    Any help is much appreciated!
  2. jcsd
  3. Mar 17, 2010 #2


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    OK, so your inequality holds for EVERY [itex]x,t \in [x_{i-1},x_i][/itex]. Therefore it also remains true if you replace the LHS with

    [tex]\sup_{x,t \in [x_{i-1},x_i]} |f(x)g(x) - f(t)g(t)|[/tex]

    where the RHS remains unchanged because it does not depend on [itex]x[/itex] or [itex]t[/itex].

    How does this relate to

    [tex]|M_i(fg,P) - m_i(fg,P)|[/tex]?
  4. Mar 17, 2010 #3
    Yes, I agree that [tex]\sup_{x,t \in [x_{i-1},x_i]} |f(x)g(x) - f(t)g(t)|[/tex], but I really have no idea how I can possibly get [tex]M_i(fg,P) - m_i(fg,P)[/tex] on the LHS (also I think we need [tex]M_i(fg,P) - m_i(fg,P)[/tex] without the absolute values, i.e. [tex]M_i(fg,P) - m_i(fg,P)[/tex], not |[tex]M_i(fg,P) - m_i(fg,P)[/tex]| ). This part is exactly where I'm having trouble. Can you explain this step, please?
  5. Mar 17, 2010 #4


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    First, [itex]M_i(fg,P)[/itex] is the supremum of a set, and [itex]m_i(fg,P)[/itex] is the infimum of the same set, so [itex]M_i(fg,P) \geq m_i(fg,P)[/itex] and therefore

    [tex]|M_i(fg,P) - m_i(fg,P)| = M_i(fg,P) - m_i(fg,P)[/tex]

    and the absolute values are irrelevant. Furthermore, by definition boundedness means "bounded above and below," so that you would need to check that

    [tex]|M_i(fg,P) - m_i(fg,P)| \leq \ldots[/tex]

    i.e., in general to prove boundedness, you DO need the absolute values. And in any case,

    [tex]x \leq |x|[/tex]

    for any real number [itex]x[/itex], so if you can prove the inequality with the absolute values, you get the inequality without the absolute values as a side effect.

    Now onto the heart of the question. I claim that

    [tex]|M_i(fg,P) - m_i(fg,P)| = \sup_{x,t \in [x_{i-1},x_i]} |f(x)g(x) - f(t)g(t)|[/tex]

    and if this claim is true then, because of what I wrote previously, that proves that the hint is true.

    So let's prove that the claim is true. First,

    [tex]|M_i(fg,P) - m_i(fg,P)| = \left|\sup_{x \in [x_{i-1},x_i]} f(x)g(x) - \inf_{t \in [x_{i-1},x_i]} f(t)g(t) \right|[/tex]

    by definition. But

    [tex]\inf_{t} f(t)g(t) = - \sup_{t} (-f(t)g(t))[/tex]


    [tex]|M_i(fg,P) - m_i(fg,P)| = \left|\sup_{x} f(x)g(x) + \sup_{t} (-f(t)g(t)) \right|[/tex]

    The suprema are over different variables, so in particular the supremum over t doesn't have any effect on the first term. Thus I can write

    [tex]\sup_x f(x)g(x) + \sup_t(-f(t)g(t)) = \sup_t \left[\left(\sup_x f(x)g(x)\right) - f(t)g(t)\right][/tex]

    Similarly, the supremum over x doesn't have any effect on the second term, so it should be clear what to do next.

    Also keep in mind that the order in which you take the suprema makes no difference, so

    [tex]\sup_{t} \sup_{x} = \sup_{x} \sup_{t} = \sup_{x,t}[/tex]
    Last edited: Mar 17, 2010
  6. Mar 17, 2010 #5
    I've never seen this way of proving it before. The way I remember it is you prove that the square of an integrable function is integrable. Then you prove that the sum of any two integrable functions is integrable. Then you use the fact that 2fg = f2 + g2 - (f+g)2.
  7. Mar 17, 2010 #6


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    Yeah, same here. I guess that way is slightly easier, but it always seemed like a clever trick to me ("very sneaky, Mr. Rudin, but how the heck would I have proved it if I hadn't had that insight?") so actually I prefer this proof. It's more of a "direct attack" on the problem, and showcases some good inf/sup techniques that show up a lot in analysis.
  8. Mar 18, 2010 #7
    I understand everything up to here. But we may also say that [tex]|M_i(fg,P) - m_i(fg,P)| = \left|\sup_{x \in [x_{i-1},x_i]} f(x)g(x) - \inf_{x \in [x_{i-1},x_i]} f(x)g(x) \right|[/tex] since t is just a dummy variable. It's also natural to write both terms with "x". Did you purposely replace the second one by t? Why are we doing this?

    But I really don't follow what's going on here. This doesn't seem obvious to me at all. How can we rigorously prove it?

    Thank you for your detailed response. :)
  9. Mar 18, 2010 #8


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    Yes, I intentionally used two different variables, because doing so allows me to do the manipulations that follow, namely handling first one supremum and then the other one. That would not have been legitimate if both suprema were over the same variable.

    Can you tell me which line is the first one you don't understand? If I do much more, I will have done the whole problem for you, and then my post gets yanked because that's against PF policies.
  10. Mar 18, 2010 #9
    Assuming the hint and moving on, I think I actually finished the proof by using the integrability criterion. Now I'm only left with proving the hint.
    I think I may have an easier way to prove the hint, but I can't prove the needed sup-inf proof. I am pretty confused about this type of proofs in general :( I hope someone can explain this.


    Thanks for helping!
  11. Mar 18, 2010 #10


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    OK, you're almost there. Tell me which of the following steps you aren't sure about and I'll try to explain in more detail. All the sups and infs are taken over the interval [itex][x_{i-1},x_i][/tex] so I won't bother writing that part.


    [tex]\inf_t f(t)g(t) = -\sup_t (-f(t)g(t) )[/tex]


    [tex]\sup_x f(x)g(x) - \inf_t f(t)g(t) = \sup_x f(x)g(x) + \sup_t (-f(t)g(t) )[/tex]

    Next, note that the term

    [tex]\sup_x f(x)g(x)[/tex]

    does not depend on t. Therefore, with regard to the supremum over t, this term is a CONSTANT. I can freely move it inside or outside of the supremum over t, just as I can do so with any constant:

    [tex]c + \sup_t (-f(t)g(t)) = \sup_t \left( c - f(t)g(t) \right)[/tex]

    I choose to move it inside:

    [tex]\left(\sup_x f(x)g(x)\right) + \sup_t (-f(t)g(t) ) = \sup_t \left[ \left(\sup_x f(x)g(x)\right) - f(t)g(t) \right][/tex]

    And now similarly, the term [itex]f(t)g(t)[/itex] does not depend on x, so I can include it or exclude it in the supremum over x. I choose to include it:

    [tex]\sup_t \left[ \left(\sup_x f(x)g(x)\right) - f(t)g(t) \right] = \sup_t \left[ \sup_x \left( f(x)g(x) - f(t)g(t) \right) \right][/tex]

    Finally, it doesn't matter in which order I take the suprema. All three of the following expressions are equal:

    [tex]\sup_t \left[ \sup_x \left( f(x)g(x) - f(t)g(t) \right) \right][/tex]

    [tex]\sup_x \left[ \sup_t \left( f(x)g(x) - f(t)g(t) \right) \right][/tex]

    [tex]\sup_{x,t} \left( f(x)g(x) - f(t)g(t) \right) [/tex]

    This shows that

    [tex]\sup_x f(x)g(x) - \inf_t f(t)g(t) = \sup_{x,t} \left( f(x)g(x) - f(t)g(t) \right)[/tex]

    Which of the above manipulations are you uncomfortable with?
    Last edited: Mar 18, 2010
  12. Mar 18, 2010 #11
    I don't understand the last step.

    Why is [tex]\sup_t \left[ \sup_x \left( f(x)g(x) - f(t)g(t) \right) \right][/tex]
    equal to [tex]\sup_{x,t} \left( f(x)g(x) - f(t)g(t) \right) [/tex] ? How can we formally prove this?

    Thank you very much!
  13. Mar 18, 2010 #12
    Is there an easier way to prove that
    sup{A(x)-B(t): x,t in a specified range}=sup {A{x}}+sup{-B(t)} ??
  14. Mar 18, 2010 #13


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    OK, let's introduce some notation:

    [tex]h(x,t) = f(x)g(x) - f(t)g(t)[/tex]

    [tex]X(t) = \sup_x h(x,t)[/tex]

    [tex]M = \sup_t X(t)[/tex]

    [tex]K = \sup_{x,t} h(x,t)[/tex]

    I claim that

    [tex]M = K[/tex]

    Proof of claim:

    For any choice of [itex]x,t[/itex],

    [tex]h(x,t) \leq K[/tex]

    because [itex]K[/itex] is an upper bound for [itex]h[/itex].

    Therefore for every [itex]t[/itex],

    [tex]X(t) = \sup_x h(x,t) \leq K[/tex]

    and therefore

    [tex]M = \sup_t X(t) \leq K[/tex]

    Now, can we have strict inequality? Suppose that [itex]M < K[/itex]. Since [itex]K[/itex] is the LEAST upper bound for [itex]h[/itex], that means there must exist particular [itex]x_0,t_0[/itex] such that

    [tex]M < h(x_0,t_0)[/tex]


    [tex]M = \sup_t X(t)[/tex]

    so the following must be true for all [itex]t[/itex]:

    [tex]X(t) \leq M < h(x_0,t_0)[/tex]

    Substituting the definition of [itex]X(t)[/itex], the following is true for all [itex]t[/itex]:

    [tex]\sup_x h(x,t) < h(x_0,t_0)[/tex]

    But then this inequality must hold for all [itex]x,t[/itex]:

    [tex]h(x,t) \leq \sup_x h(x,t) < h(x_0,t_0)[/tex]

    which is a contradiction.

    Therefore my assumption that [itex]M < K[/itex] was incorrect, and since we already showed [itex]M \leq K[/itex], the only possibility is [itex]M = K[/itex].

    The above can be shortened a LOT, but I wanted each step to be very clear. I hope I succeeded!
    Last edited: Mar 18, 2010
  15. Mar 18, 2010 #14


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    I'm not sure what would qualify as "easier" - if you look at my post #10 and strip out the chatter, it only takes three steps, starting with "Next, note that the term..."

    1) Move term inside the "sup x"
    2) Move term inside the "sup t"
    3) Recognize that "sup x sup t" is the same as "sup x,t"

    Once you are more comfortable with sups and infs you don't need to provide all the other details (such as all of post #13) because they will seem pretty obvious.

    In fact, "sup{A(x)-B(t): x,t in a specified range}=sup {A{x}}+sup{-B(t)}" itself may eventually seem obvious enough to you that you will not see the need to warrant further justification.
  16. Mar 18, 2010 #15
    I'm not sure if the following would work...

    Claim: Sup (A -B) = Sup A - inf B

    Let a in A, b in B.
    a - b <= Sup A - b
    <= Sup A - inf B ( b>= inf B)
    So, Sup A - inf B is an upper bound of the set {a in A, b in B : a - b}

    For all e > 0
    Sup A - inf B - e = (Sup A - e/2) - (inf B + e/2)
    By definition of Sup, there is an element a' in A s.t. a' > Sup A - e/2
    By definition of inf, there is an element b' in B s.t. b' < inf B + e/2
    hence, Sup A - inf B - e < a' - b'
    i.e. Sup A - inf B is the least upper bound of the set {a in A, b in B : a - b}

    Is this correct? But in our case we have functions f(x),g(t), etc., can this proof be modified to show that sup{A(x)-B(t): x,t in range}=sup{A(x)}-inf{B(t)}?
    Is this going to work?

    Because I saw you online, I am posting this before reading the previous 2 posts. I will read your posts now. Thanks for all your time and patience! :)
    Last edited: Mar 18, 2010
  17. Mar 18, 2010 #16


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    Yes, that looks right.

    Sure, just define

    [tex]A = B = \{f(x)g(x) | x \in [x_{i-1},x_i]\}[/tex]

    and then your proof applies without any modification.

    It doesn't matter whether you call the variable x or t.
  18. Mar 18, 2010 #17
    I learned a lot from your post. Thanks! :smile:

    Just one small question at the end.
    I understand that for all x, [tex]h(x,t) \leq \sup_x h(x,t) [/tex] (because the sup is taken over all x)
    But why is it true that for all x and t, [tex]h(x,t) \leq \sup_x h(x,t) [/tex]?
    Last edited: Mar 18, 2010
  19. Mar 18, 2010 #18


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    Because it's the same [itex]t[/itex] on both sides of the inequality.

    If I pick any specific [itex]t'[/itex], then

    [tex]h(x,t') \leq \sup_x h(x,t')[/tex]

    i.e. hold one variable fixed, and take the sup over the other variable. But this is true no matter which [itex]t'[/itex] I choose, so it's true for all [itex]t[/itex].
  20. Mar 18, 2010 #19
    "It doesn't matter whether you call the variable x or t."
    Why? But then we'll always have f(x)g(x) - f(x)g(x) =0 ?

    If A is as you defined (in terms of x), shouldn't we define B = {f(t)g(t): t E [xi-1,xi]}??
  21. Mar 18, 2010 #20


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    Yes, you can call it that if you like, but the name of the variable used to define the set is irrelevant:

    [tex]B = \{f(t)g(t) : t \in [x_{i-1},x_i]\}[/tex]

    [tex]B' = \{f(x)g(x) : x \in [x_{i-1},x_i]\}[/tex]

    These are exactly the same two sets of numbers. [itex]B = B'[/itex].
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