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Mean value theorem

  1. Oct 9, 2006 #1
    Let be [tex] F(x)=\int dx f(x) [/tex] my question is if F(x) is continous (but not differentiable ) does the Mean-value theorem for integrals hold?..

    a<c<b and [tex] f(c)(b-a)=\int_{a}^{b} dx f(x) [/tex] ?... :rolleyes: :confused:
     
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  3. Oct 9, 2006 #2

    matt grime

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    Have you even bothered to try to think up some examples to see that this cannot hold? Just take some easy discontinuous f and construct counter examples.
     
  4. Oct 9, 2006 #3

    Office_Shredder

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    Try thinking about functions that can't achieve every value possible between f(a) and f(b) (which is basically the requirement needed to ensure f(c) can take on the right value)
     
  5. Oct 9, 2006 #4
    How can a function be continuous and not differentiable? The definition of continuity hold that the function has limits such as any value of y in the interval [a,b] could be represented as the funciton of a point (be it rational or irrational). This being said, it also holds that there are intervals where the function is strictly monotonously increasing, decreasing or stale. This last point, proves us that there is no bad behaviour after a certain point when evaluating the limit that the derivative is. Anyway have you checked a good proof of the mean value theorem? If the function is continuous, it certainly has a maximum and a minimum on a closed interval. Say Max {f(x)} = M
    and Min {f(x)} = m. If you we say the integral between a and b of the function is i, we have

    m(b-a) < i < M(b-a)

    Which needs no proof since it goes without saying. We then have,

    m < i/(b-a) < M

    By continuity of the function, i/b-a must be one of the value f(x) assumes between in the interval [a,b]. We can thus express it as f(e).

    f(e) = i/(b-a)
    i = f(e)(b-a)

    Which is the proof of the Mean Value Theorem. All you need is certainty of continuity.
     
  6. Oct 9, 2006 #5

    matt grime

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    f is not continuous. If it were F would be differentiable. So what was the point? It is trivial to find examples for Karlisbad/eljose/lokofer that demonstrate that it is false in general. Suppose that f is a function that only takes the value 0 and 1, then the condition states that the integral of f over any interval is the length of the interval, which can only happen if f is 0 on a set of measure zero, so just pick an f that is not zero on set of strictly positive measure. If that is too much measure thoery then just consider f to be the most elementary discontinuous function, the Heaviside step function.
     
    Last edited: Oct 9, 2006
  7. Oct 9, 2006 #6

    arildno

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    "How can a function be continuous and not differentiable?"
    try this one:
    [tex]f(x)=\sum_{n=0}^{\infty}\frac{\cos((n!)^{2}x)}{n!}[/tex]

    This is continuous, and non-differentiable at every point.
     
  8. Oct 9, 2006 #7

    HallsofIvy

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    No, it doesn't. Arildno's example is not monotone on any interval.
     
  9. Oct 9, 2006 #8

    arildno

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    Just to add another example:
    In general, one cannot conclude anything about the derivative from the function itself. This is basically what you could call the "un-smoothing" property of the derivative.

    Here's another example:
    Let a sequence of functions be defined as:
    [tex]f_{n}(x)=\frac{\sin(n^{2}x)}{n}[/tex]
    Note that we have [tex]f_{n}(x)\to{f}(x)\equiv{0}[/tex] UNIFORMLY.
    However, a limiting function of the derivatives of fn wrt. to x doesn't exist.

    Not even the strict requirement of uniform continuity is sufficient to say anything about the existence of derivatives.
     
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