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If integral equals zero, then function equals zero

  1. Mar 11, 2009 #1
    1. The problem statement, all variables and given/known data

    Let f : [a,b] [tex]\rightarrow[/tex] [tex]\Re[/tex] be continuous and assume f [tex]\geq[/tex] 0. Prove that if [tex]\int_{[a,b]}[/tex]f = 0 then f = 0.

    2. Relevant equations

    Nothing really. If relevant, mean value theorem was discussed in earlier problems, so I'm not sure if it fits though.

    3. The attempt at a solution

    I tried using the MVT and stating that if all that was mentioned above holds true then you can say f(c)(b-a) = 0, then solving b = a. From there I couldn't go anywhere. Pretty sure I wasn't on the right track to begin with in the first place anyway.
     
  2. jcsd
  3. Mar 11, 2009 #2
    Intuitively the problem is clear i would guess. SInce f>=0 then what it menas is that from a to b there is no area accumulated at all, so the only possibility in this case is for f to be the zero function(f=0).

    Now, to formally prove it i would suggest starting from the definition of the definite integral in terms of Riman sums would help.
     
  4. Mar 12, 2009 #3

    matt grime

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    Suppose that f is not identically zero, then a lower bound for the integral is...
     
  5. Mar 12, 2009 #4
    Well the upper bound, sup(f) = b , and the lower bound, inf(f) = a

    And with Riemann Sums, I don't see how you can add up the partitions and take the Upper Sum and Lower Sum of a function that you don't know.

    I just can't see where they fit into the proof.
     
  6. Mar 12, 2009 #5

    matt grime

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    No, sup(f) is not b, nor is inf(f) equal to a. In fact you don't know what inf and sup are.

    You have also failed to use or mention the fact that f is continuous.

    Suppose that f is not identically 0, i.e. there exists a c in (a,b) with f(c)=d>0.

    f is continuous. What does that say about f in small enough neighbourhood of c?
     
  7. Mar 12, 2009 #6
    f converges to c??
     
  8. Mar 12, 2009 #7

    Dick

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    No, no, no. If f(c)=d>0 and f is continuous, then there is an interval N=(c-delta,c+delta) around x=c such that for x in N, f(x)>d/2, for example. If you can tell me why, then I'll tell you why the integral must be greater than zero.
     
  9. Mar 12, 2009 #8
    With your example f(x)>d/2, since f(c)=d where you have x=c so then f(x)=d. And d/2<d. But I still can't see the relation to my original problem.
     
  10. Mar 12, 2009 #9

    Dick

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    The relation to your original problem is to show that if f(x)>0 at any point then the integral must be greater than zero. Because if f is continuous and positive at any point, then it is positive on an interval around that point. That forces the integral to be nonzero. That's what people have been trying to tell you. Hence, f(x) is equal to zero everywhere. Now can you add the details?
     
  11. Mar 12, 2009 #10
    Haha, I must've just stared at your statement for about 2 minutes, then it clicked. So f would have to be zero everywhere because if it was greater than zero than it's integral would have to be greater than zero as well. Thanks. Now I just have to put everything together.
     
  12. Mar 12, 2009 #11
    You still haven't fully(formaly) justified, as Dick asked you to do, as to why f(x)>0 on some interval N, if the function f(x) is nonzero at any point whatsoever........
     
  13. Mar 12, 2009 #12
    Well if the function is non-zero at every point and continuous, then when taking the value of the function at some point along a given interval, that point will be non-zero as well since every other point along the function is non-zero. Is that what you were asking??
     
  14. Mar 12, 2009 #13
    No, not really. What i am saying is that, you haven't shown why f(x)>0, if say,(like Dick already explained it really well) at some point c,
    f(c)=d>0. In other words, if at some single point f is greater than zero, then you have to show, why will there exist an interval
    N=(c-delta,c+delta), such that for every point x in this interval N, f(x)>0?
     
  15. Oct 31, 2009 #14
    Apologies for bringing up an old thread but I am attempting to solve a similar problem...

    I can understand the workings of the proof that has been discussed above, but am having difficulties in actually writing it formally.

    In particular, I am having trouble trying to prove why if at some single point f is greater than zero, then there must exist an interval N=(c-delta,c+delta), such that for every point x in this interval N, f(x)>0?

    Clearly, by the epsilon-delta definition of continuity, if there is a point f(c)=d>0 then there must be points within epsilon either side of f(c)=d. Can we then just increase our value epsilon (by definition of continuity, we can take any value for epsilon) until all x such that f(x) is non-zero is within delta either side of c? Hence, there would then be an interval N=(c-delta,c+delta) such that for every point x in this interval N, f(x)>0.

    Does this work? How do I then show that the integral of f(x) between c-delta and c+delta is greater than 0?

    Thanks very much for your help!
     
  16. Oct 31, 2009 #15
    You don't have to get that involved with it. Just choose [tex]\epsilon < d[/tex] to begin with.
     
  17. Oct 31, 2009 #16
    I don't know how I didn't see that before...thanks!

    How do I then go on to show that the integral of f(x) between c-delta and c+delta (and therefore the whole function between a and b) is greater than 0?
     
  18. Oct 31, 2009 #17
    I messed up, you don't want to take [tex]\epsilon < d[/tex] but some fraction of d. 1/2 is convenient.

    then [tex]f > d/2[/tex]. But [tex]c - \delta < x < c + \delta[/tex] gives you an interval for which [tex]f(x) > d/2[/tex]
     
  19. Oct 31, 2009 #18
    THe idea is relatively simple. If f is a continuous function on an interval (a,b), and if there is a point c in that interval such that f(c)=d>0, then we have to show that there must be some neighborhood Q of c such that for all x in Q f(x)>0.

    It is not difficult to construct a sequence {x_n} of members of (a,b) each distinct from c, and converging to c, hence c is an accumulation point of (a,b).

    Now, given that f is continuous, for every epsilon greater than zero and also for e=d/2 there should exist some delta such that anytime |x-c|<delta |f(x)-d|<d/2.

    so: -d/2<f-d<d/2=> 0<d/2<f<3d/2, for all x in (c-delta, c+delta).


    Then to show that the integral is not zero, go to the definition of the integral using rieman sums. when you divide the interval [a,b] into n subintervals take (c-delta, c+delta) as one of your intervals.
     
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