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An integral and function inequality proof

  1. Sep 26, 2007 #1
    1. The problem statement, all variables and given/known data
    Let f be a function such that |f(u)-f(v)|[tex]\le|u-v|[/tex] for all u and v in an interval [a,b]. Assume that f is integrable on [a,b]. Prove that

    |[tex]\int_a^bf(x)dx[/tex] - (b-a)f(a)|[tex]\leq\frac{(b-a)^{2}}{2}[/tex]

    That's the absolute value of the difference of the integral of f(x) from a to b and (b-a)f(a). The subtraction and absolute value signs aren't quite coming out right, but it should be good enough to interpret.

    Once I prove this, I have to prove the more general statement that for any c in [a,b],

    |[tex]\int_a^bf(x)dx[/tex] - (b-a)f(c)|[tex]\leq\frac{(b-a)^{2}}{2}[/tex]

    2. Relevant equations

    Standard integration formulas. I might be able to throw the triangle equality somewhere to break up those absolute value differences.

    3. The attempt at a solution

    I've realized that the given information implies that |m|[tex]\leq1[/tex] because
    m = [tex]\frac{|f(u)-f(v)|}{|u-v|}[/tex] and |u-v| is less than or equal to |f(u)-f(v)|.

    [tex]\frac{(b-a)^{2}}{2}[/tex] also looks like the formula for a triangle that has a height equal to its base length. So I think that if the slope were 1, then the integral would be equal to the area of the triangle. If the slope is less than 1, then I think the integral is less than the area of the triangle, because it won't reach the same height at point b that the triangle will reach. Using this, I started working with f(x) = (1)x + c, where c is the y intercept (I used c instead of b to avoid confusion). This hasn't gotten me very far, though, and I'm having trouble putting the logic into precise enough mathematical language to call it a proof. Any advice?
     
  2. jcsd
  3. Sep 26, 2007 #2

    NateTG

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    Well, let's say we have [itex]f(x)[/itex], and [itex][a,b][/itex] as described. If you can you find (nice) functions that are upper and lower bounds for [itex]f(x)[/itex], then you might be able to use those instead of this nebulous [itex]f(x)[/itex].
     
  4. Sep 26, 2007 #3
    I've noticed now that (b-a)f(a) removes the area under the rectangle generated from 0 to the y-intercept from a to b. Can I then rewrite the integral - (b-a)f(a) as just the integral of the function without a y-intercept; namely, as [tex]\int_a^b(mx)dx[/tex]?

    Then I get that [tex]\frac{(b^{2}-a^{2)}}{2}[/tex][tex]\leq\frac{(b-a)^{2}}{2}[/tex].

    I'm not sure if this is the right way to go, though.

    From here it follows that [tex]{(b+a)}[/tex] [tex]\leq[/tex] [tex]{(b-a)}[/tex], which isn't always true, so I seem to have either made a mistake or gone down the completely wrong path.
     
  5. Sep 27, 2007 #4

    NateTG

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    I'd try to attack this problem with:
    [tex]f(a)-(x-a) \leq f(x) \leq f(a)+(x-a)[/tex]
    [tex]\int f(a)-(x-a) dx \leq \int f(x) dx \leq \int f(a)+(x-a) dx[/tex]
     
  6. Sep 27, 2007 #5
    Thank you for your help. I managed to solve it by starting with the given information and integrating that to work toward the statement I wanted to prove.
     
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