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2 True/False Questions -- Integral and convergence

  1. Apr 4, 2016 #1
    1. The problem statement, all variables and given/known data
    a) If ##f: [0,1] \rightarrow \mathbb{R}## is continous and ##\int^{b}_{a} f(x)dx = 0## for every interval ##[a,b] \subset [0,1]##, then ##f(x)=0 \forall x \in [0,1]##

    b) Let ##f: [0,\infty) \rightarrow [0,\infty)## be continous. If ##\int^{\infty}_{0} f(x)dx## converges, then ##f(x) \rightarrow 0## when ##x \rightarrow \infty ##

    2. Relevant equations

    3. The attempt at a solution
    For the a) part, my guess is that it is true ( I know that guesses don't mean much). I've tried to come up with a counter example such as

    ## f(x)=
    \left\{
    \begin{array}{ll}
    1 & \mbox{if } x \in Q \\
    0 & \mbox{if } x \in I
    \end{array}
    \right.
    ##


    , which will satisfy the latter conditions, but the ones that say that the function is continuous and that ##f: [0,1] \rightarrow \mathbb{R}## (which would prevent me to use symmetric intervals around zero, and hence use sin(x) for example) make me think that it's true. However I'm not sure how to prove it.

    b) Like before, the ##f: [0,\infty) \rightarrow [0,\infty)## condition makes me think it is true, since I cannot use bounded functions like sin(x) which would give me a counter example (here's a similar problem, but without the interval and continous conditions: http://math.stackexchange.com/questions/538750/if-the-improper-integral-int-infty-a-fx-dx-converges-then-lim-x→∞fx ). Also I have in my notes, before Cauchy's integral criterion (which states, for using it, that the function must be a decreasing one, something which is not specified here), that if

    ##S_{n}= \sum_{k=0}^{N} A_{k}## converges, then ##A_{k} \rightarrow 0##, but I cannot find a proof of this.

    Any hint would be much appreciated.
     
  2. jcsd
  3. Apr 4, 2016 #2

    HallsofIvy

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    Suppose f were not identically 0 on the interval. That is, suppose that, for some a in the interval f(a)= b> 0. Then, since f is continuous, there would be an interval around a where f(x)> 0 and the integral over that interval could not be 0.
     
  4. Apr 5, 2016 #3
    Thanks for your reply Hallsoflvy, but I don't quite understand what you are saying. The statement has the condition that the integral is zero on every interval ##[a,b] \subset [0,1]##, therefore how could I suppose that for some ##a## the conclusion is that its integral is not zero?
    I thought of a counterexample (let me know if you think this is correct): Let ##F(x)'=f(x)##, and I define this F(x) to be constant on every interval in ## [a,b] \subset [0,1]##, therefore, since f(x) is continous

    ##\int^{b}_{a} f(x) dx = F(b) - F(a)=0## and it doesn't assume for this to happen that it must true that that ##f(x)=0##, therefore the statement is false. Is this right?
     
  5. Apr 5, 2016 #4

    HallsofIvy

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    You can't and I did not say that. I said assume the conclusion, that the function is not identically 0, is false and then show a contradiction.

    No, that is not correct since, if F(x) is constant on every interval, its derivative, f, is identically 0.
     
  6. Apr 5, 2016 #5
    Thank you for your patience, I misread you, I see what you mean now, I'll work on that.

    Meanwhile I thought something for the b) part. If could treat the problem (since f is continous) from a series point of view (maybe justifying this with the archimedean property of real numbers that states that for every ##x \in \mathbb{R}## there's an ##n \in \mathbb{N}## such, that ##n>x## ?):
    The series ##\sum_{n=0}^{\infty} a_{n}## converges if the sequence of partial sums ##S_{k}=\sum_{n=0}^{k} a_{n}## also converges, and the sum of the series is ##\lim_{k \to \infty} S_{k}=L##

    Now ##a_{n}= S_{n} - S_{n-1}= L - L = 0## as ##n \rightarrow \infty##. Therefore the statement is true.
     
  7. Apr 5, 2016 #6

    HallsofIvy

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    What statement is true? What you have proved is "If a series converges then the individual terms must go to 0." I don't see how that has anything to do with the original statement "If the integral of continuous function, f, over any subinterval of an interval is 0, then f is 0 on the interval". For one thing, "continuous" is crucially important in this statement but has no part in your "series" variation.
     
  8. Apr 5, 2016 #7
    The b) is a different problem, that says if the integral from zero to infinity converges then its integrand must go to zero when x extends to infinity. I was refering to that problem when I brought up the series.

    For the a) problem, following your suggestion: suppose that for some ##a \subset [0,1]##, ##f(a)>0##, then, since f(x) is continuous there is an interval ##(a-\epsilon, a+\epsilon)## around a, where ##f(a)>0##, therefore

    ##\int^{a+\epsilon}_{a-\epsilon} f(x) = F(a+\epsilon) - F(a-\epsilon) \neq 0## and so the original statement is true.
     
  9. Apr 5, 2016 #8

    pasmith

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    Homework Helper

    Cauchy's integral test will not assist you: it says that a series converges if the integral of a related decreasing function converges. You are asked to prove or disprove that if an integral of an arbitrary continuous non-negative function converges then that function tends to zero.

    Intuitively it seems like statement (b) should be true. But consider a continuous, non-negative function which is zero except for triangular peaks centered on the positive integers, such that the height of the peaks increase without limit but the widths decrease sufficiently fast that the integral over the peak at [itex]n[/itex] is a strictly decreasing function of [itex]n[/itex]. Does that not suggest how one might construct a counterexample?
     
  10. Apr 8, 2016 #9
    Thanks pasmith, indeed that is a counterexample, but I have no idea how to define that function (except for the x=0, when x<0, part, of course). Nevertheless, describing it as you did would not count as correct answer (whatever the explicit expression for its formula might be) in replying that the statement is false?
     
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