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Show that the integral tends to zero

  1. Sep 29, 2013 #1
    1. The problem statement, all variables and given/known data

    How do I show that this integral tends to zero as b tends to a ?

    2. Relevant equations

    [itex]\int_a^b \int_0^a (f(a)-f(y))(x-y)dy(b-x)dx[/itex]

    3. The attempt at a solution

    I'm not sure how to start.
    Any hints or ideas will be very much appreciated.
    We are given that function f is continuous over the domain.

    Thank you.
     
  2. jcsd
  3. Sep 29, 2013 #2

    jbunniii

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    The inner integral ##\int_{0}^{a} (f(a) - f(y)) (x-y) dy## is a function of ##x##. Can you show that it is bounded? In other words, there is some ##M## such that for all ##x \in [a,b]##,
    $$\left|\int_{0}^{a} (f(a) - f(y)) (x-y) dy\right| \leq M$$
    If so, then you will be able to argue as follows:
    $$\left|\int_{a}^{b} \int_{0}^{a} (f(a) - f(y)) (x-y) dy (b-x) dx\right| \leq
    M \int_{a}^{b} |b-x| dx$$
    and the result should follow easily.
     
  4. Sep 29, 2013 #3
    Thank you very much.

    To show that the inner integral is bounded, I first simplified
    [itex]|\int_0^a (f(a)-f(b))(x-y)dy |[/itex] since f(a) is independent of x.

    So I split the integral into 2 parts:
    [itex]\int_0^a f(a)(x-y)dy-\int_0^a f(b)(x-y)dy[/itex]

    Evaluating the first part:
    [itex]\int_0^a f(a)(x-y)dy=\frac{1}{2}f(a)(2xa+a^2)[/itex]

    So I have to show that there is an [itex]M[/itex] such that for [itex]x\in (a,b)[/itex]
    [itex]|\frac{1}{2}f(a)(2xa+a^2)-\int_0^a f(b)(x-y)dy|\le M[/itex]

    Would M be a constant or function of x? I'm a bit confused.
     
  5. Sep 29, 2013 #4

    jbunniii

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    It needs to be independent of ##x##. So far your work looks good. You can now apply the triangle inequality to get
    $$\left|\frac{1}{2}f(a)(2xa+a^2)-\int_0^a f(b)(x-y)dy\right| \leq
    \left|\frac{1}{2}f(a)(2xa+a^2)\right| + \left|\int_0^a f(b)(x-y)dy\right|$$
    The first term should be easy to bound since you know ##a \leq x \leq b##. The second term should be similarly easy if you note that ##f(b)## is a constant and that ##|\int \ldots| \leq \int |\ldots|##.
     
  6. Sep 29, 2013 #5
    Thank you.

    I see, so I have:
    [itex]|\frac{1}{2}f(a)(2xa+a^2)-\int_0^a f(y)(x-y)dy|\le M[/itex]
    Using the triangle inequality, we have:
    [itex]|\frac{1}{2}f(a)(2xa+a^2)|+|\int_0^a f(y)(x-y)dy|\le M[/itex]
    The first term is bounded by
    [itex]\frac{1}{2}f(a)(2ba+a^2)[/itex]
    So, we can write:
    [itex]|\frac{1}{2}f(a)(2ab+a^2)|+\int_0^a |f(y)(x-y)|dy\le M[/itex]

    Now, for the second term, I'm still unsure, if we can't integrate it, is it still easy to see what it is bounded by?

    I just noticed that in my second post, all f(b) should actually read f(y).
     
  7. Sep 29, 2013 #6

    jbunniii

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    Well, you shouldn't write ##\leq M## yet because you haven't shown that. Instead, you can write
    $$\begin{align}
    \left|\frac{1}{2}f(a)(2xa + a^2) - \int_{0}^{a} f(y)(x-y)dy \right|
    &\leq \left| \frac{1}{2}f(a)(2xa + a^2)\right| + \left|\int_{0}^{a}f(y)(x-y)dy \right|\\
    \end{align}$$
    Now, you said the first term is bounded by ##\frac{1}{2}f(a)(2ba + a^2)##, but even after adding the missing absolute value signs, that isn't necessarily true if ##a## and ##b## are allowed to be negative. Consider for example ##a=-1##, ##b=0##. However, you can get a bound by applying the triangle inequality again:
    $$\begin{align}
    \left| \frac{1}{2}f(a)(2xa + a^2)\right|
    &\leq \frac{1}{2}|f(a)|(|2xa| + |a^2|)\\
    &= \frac{1}{2}|f(a)|(2|a|\cdot|x| + |a|^2)\\
    \end{align}$$
    Now find a bound for ##|x|## and that takes care of the first term.

    For the second term, you need to use the fact that ##f## is continuous, hence bounded on ##[0,a]## (why?), i.e. there exists some bound ##B## such that ##|f(y)| \leq B## for all ##y \in [0,a]##. Now, can you find a bound for ##|x-y|##? Remember that ##y \in [0,a]## and ##x \in [a,b]##.
     
  8. Sep 30, 2013 #7
    We have: [itex]|x|\le max(|a|,|b|)[/itex].

    For the second term:
    We can say that f is continuous and bounded on [0,a] since we can take

    [itex]B=max(|f(a)|,f(0)|[/itex]
    so that
    [itex]|f(y)|\le B\,\,\,\, \forall\,\, y\in[0,a][/itex]

    I think |x-y| must be bounded by b. is that right?
    It's the biggest difference it can take for [itex]x\in[a,b]; y\in[0,a][/itex].

    So does that mean that the initial double integral is bounded by M*b ?
     
    Last edited: Sep 30, 2013
  9. Sep 30, 2013 #8
    I think you guys are making a mistake you fill in f(b) while it should be f(y) in the remainder of your calculations.
     
  10. Sep 30, 2013 #9
    Thank you, yes you're right.
    The f(b) should be f(y) in post numbers 3 and 4.
    We have fixed this in the following posts.
     
  11. Sep 30, 2013 #10

    jbunniii

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    No, the max won't necessarily occur at an endpoint of the interval. You have to invoke the theorem that a function which is continuous on a closed bounded interval achieves a maximum and a minimum on the interval, hence it is bounded. The theorem doesn't say where the max or min occurs, but it does guarantee that there is some ##C## (to use a different letter from the other bounds we already found) such that ##|f(x)| \leq C## for all ##x \in [0,a]##.
    Not necessarily. Is ##b## even guaranteed to be positive? It's not stated in the problem anywhere. However, you can use the good old triangle inequality again to get a bound:
    ##|x-y| \leq |x| + |y| \leq ???##
     
  12. Sep 30, 2013 #11
    Oh, I see now why f is bounded. thank you!

    I've just checked and we have that [itex]a[/itex] and [itex]b[/itex] are positive.
    I should have stated that in the problem.
    Also, [itex]b \ge a[/itex].

    So, can I say that
    [itex]|x-y|\le |x|+|y| \le a+b[/itex] ?
     
  13. Sep 30, 2013 #12

    jbunniii

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    Yes, that works fine. Now what happens when you apply all the bounds that you've found to the original inequality?
     
  14. Sep 30, 2013 #13
    [itex]
    \left|\frac{1}{2}f(a)(2xa + a^2) - \int_{0}^{a} f(y)(x-y)dy \right|
    \leq \left| \frac{1}{2}f(a)(2xa + a^2)\right| + \left|\int_{0}^{a}f(y)(x-y)dy \right| [/itex]

    [itex]\le \frac{1}{2}|f(a)|(2a^2 + a^2) + B(a+b)[/itex]
    (with a>0)

    What then happens with the original inequality with the double integral?
    [itex]\int_{a}^{b} \int_{0}^{a} (f(a) - f(y)) (x-y) dy (b-x) dx[/itex] ?
     
  15. Sep 30, 2013 #14

    jbunniii

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    Let's give your bound a name: ##M = \frac{1}{2}|f(a)|(2a^2 + a^2) + B(a+b)##. Then you have shown that
    $$\left|\int_{0}^{a} (f(a) - f(y)) (x-y) dy\right| \leq \left|\frac{1}{2}f(a)(2xa + a^2)\right| + \left|\int_{0}^{a} f(y)(x-y)dy \right| \leq M$$
    So apply that to your original integral:
    $$\left|\int_{a}^{b} \int_{0}^{a} (f(a) - f(y)) (x-y) dy (b-x) dx\right| \leq
    \int_{a}^{b} \left|\int_{0}^{a} (f(a) - f(y)) (x-y) dy\right| |b-x| dx \leq ???$$
     
  16. Sep 30, 2013 #15
    We need to find a bound for
    [itex]|b-x|[/itex] for [itex]x[/itex] between [itex]a[/itex] and [itex]b[/itex]

    So,
    [itex]\left|\int_{a}^{b} \int_{0}^{a} (f(a) - f(y)) (x-y) dy (b-x) dx\right| \leq
    \int_{a}^{b} \left|\int_{0}^{a} (f(a) - f(y)) (x-y) dy\right| |b-x| dx \leq M(b-a)[/itex]
    Is that right?

    (Thank you by the way, very much appreciated.)
     
  17. Sep 30, 2013 #16

    jbunniii

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    Yes, that looks good. Now what can you conclude as ##b \rightarrow a##?
     
  18. Sep 30, 2013 #17
    As [itex]b \rightarrow a[/itex], we have that the upper bound tends to zero... ?
     
  19. Sep 30, 2013 #18

    jbunniii

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    Right, so what does that imply about ##\left|\int_{a}^{b} \int_{0}^{a} (f(a) - f(y)) (x-y) dy (b-x) dx\right|## as ##b \rightarrow a##?
     
  20. Sep 30, 2013 #19
    This implies that it tends to zero...?
     
  21. Sep 30, 2013 #20

    jbunniii

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    Are you asking me, or telling me?

    You have an inequality of this type:
    $$0 \leq |\text{ function of }a\text{ and }b| \leq M(b-a)$$
    If ##b \rightarrow a##, then the right hand side tends toward zero. The left side is already zero. So the guy in the middle is squeezed between zero and something which approaches zero. What can you conclude?
     
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