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If F'(x) is bounded so is F(x)

  1. Oct 1, 2011 #1
    Hello everybody,
    A few years ago i tried to join a mathematics department and in the relevant exams
    i came up against the following problem. I apologise beforehand if the statement of the problem is a little bit ambiguous because i do not remember it exactly. However, I am sure you will get the point.

    1. The problem statement, all variables and given/known data

    if the first derivative of the real function F(x) is continuous and bounded over the interval [a,b] (or (a,b) ?) , prove that F(x) also is bounded on the interval (a,b) (or [a,b] ?) and the vice versa.

    2. Relevant equations
    So we can see that m =< F'(x) =< M.
    How can we get from this into the boundness of the F(x) without falling into pitfalls ?
    What about the vice versa ?
    Should we use the defintion or something else ?

    3. The attempt at a solution
    I will not attempt to publish the solution I proposed because many of you may laugh.:smile:
  2. jcsd
  3. Oct 1, 2011 #2
    For starters, for the "vice-versa" thing to have a fighting chance, it would have to be on a closed interval. For example, x^1/2 is bounded on (0,1] but its derivative is not.
  4. Oct 1, 2011 #3
    Thanks for the reply,

    Indeed the "vice-versa" does not hold for every function F(x) as it has also been discussed https://www.physicsforums.com/showthread.php?t=515616&highlight=bounded+function+derivative".

    How can we argue that F(x) is bounded ? :confused: By intuition the statement is obvious ...

    Is integrating over (a,t) where t a variable among a and b the inequality m =< F'(x) <= M a good argument ? We will get rid off the derivative by doing so ... :smile:

    I have an engineering background so make allowances ...
    Last edited by a moderator: Apr 26, 2017
  5. Oct 1, 2011 #4


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    That sounds like an excellent place to start
  6. Oct 1, 2011 #5
    Hmmm, you have a function F and you know that is derivative, F' is bounded on [a,b]. But nothing says that F' is continuous, which you would need to prove to be able to use FTC. For example, FTC1 says that if f is continuous on [a,x] and if F(x) is the integral of F(x) from a to x, then F'(X) = f(x). The proof 100% relies on the continuity of f at x. FTC2 (which is the one you are wanting to use) requires the continuity of f, as well.

    Now of course, you could still use FTC. However, I would suggest this approach: you know that F'(x) is bounded and you know that it exists at every x in [a,b]. This implies that F is continuous, but you are dealing with a compact (closed and bounded) interval and therefore F is uniformly continuous. Now, say that F is unbounded at some point c. Let epsilon = 1 and show that no matter how close you bring x to c, |F(x) - F(c)| > 1. Now, this is pretty clear since every F(x), x not equal c, is going to be finite, but F(c) isn't finite. Now, there is some care you will have to take to ensure that F(x) is, infact finite for x not equal c.
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