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Math Help - Darboux Theorem

  1. #1
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    Darboux Theorem

    This is my lecturer's Darboux Theorem, however I have a problem with one of the steps there, I signed it wih ***. He said xmin is not equal to a and b. Buf if it is equal then what? I mean intuitively when I think about it if it is equal then I cant get c such that f ' (c)=0 but still can I just say xmin is not equal to a and b? Isn't that specific case? Thank you.
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  2. #2
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    Re: Darboux Theorem

    Quote Originally Posted by davidciprut View Post
    He said xmin is not equal to a and b. Buf if it is equal then what?
    He explains in the folloing two lines why $x_{text{min}}$ cannot equal $a$ or $b$, though his explanation leaves a lot to be desired. A function cannot be decreasing in one point, only on a set. However, it is easy to show from the definition of derivative that if $f'(a)<0$, then there exists a $t>a$ such that $f(t)<f(a)$.
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    Re: Darboux Theorem

    So he had to say that if f'(a)<0 then there exists h>0 such that a-h<x<a+h where the function is decreasing. so there exsist t such that a<t<a+h with f(t)<f(a). And since it's an interval you can find a point that is smaller than a and therefore it can't be xmin I get it ! Thanks
    Last edited by davidciprut; April 7th 2014 at 04:00 AM.
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    Re: Darboux Theorem

    Quote Originally Posted by davidciprut View Post
    So he had to say that if f'(a)<0 then there exists h>0 such that a-h<x<a+h where the function is decreasing.
    If $f'(a)<0$, it does not follow that there exists a neighborhood of $a$ where $f$ is decreasing. That would be true if $f'(x)$ were continuous at $a$: then there would be a neighborhood where $f'(x)<0$ and we could use the mean value theorem to show that in $f$ is decreasing in that neighborhood. But $f'(x)$ does not have to be continuous.

    The following example is taken from the book "Counterexamples in Analysis" (chapter 3, example 5). Let
    $$
    f(x)=
    \begin{cases}
    x+2x^2\sin\frac{1}{x}&\text{if }x\ne0\\
    0&\text{if }x=0
    \end{cases}
    $$
    Then
    $$
    f'(x)=
    \begin{cases}
    1+4x\sin\frac{1}{x}-2\cos\frac{1}{x}&\text{if }x\ne0\\
    1&\text{if }x=0
    \end{cases}
    $$
    Thus $f'(1)=1>0$, but the term $2\cos(1/x)$ causes oscillations around $x=0$. The result is that $f'(x)$ takes both positive and negative values, and so $f(x)$ increases and decreases, in each neighborhood of 0. As expected, $f'(x)$ is not continuous at 0.

    What I was saying is that if $f'(a)<0$, then using the definition of derivative it is easy to show that there exists (at least) one point $t>a$ such that $f(t)<f(a)$.
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    Re: Darboux Theorem

    If h(x) is continuous on [a,b], it takes on all values between h(a) and h(b). (standard theorem)

    let h(x) =f'(x)
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    Re: Darboux Theorem

    Quote Originally Posted by emakarov View Post
    If $f'(a)<0$, it does not follow that there exists a neighborhood of $a$ where $f$ is decreasing.
    Yes, you are right, however according to the derivative definition, L=f'(a)<0 exists , which means lim(f(x)-f(a))/((x-a)) when x goes to a exists, and according to the definition of limit
    For every epsilon there exists delta such that, |x-a|<delta then |f(x)-f(a))/((x-a))-L|<epsilion, and this means that there is a neighbourhood, otherwise we couldn't talk about the limit. Since f'(a) exists so does the limit and therefore according to the definition of limits there has to be a neighbourhood. Is my logic wrong?
    Last edited by davidciprut; April 7th 2014 at 11:37 AM.
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    Re: Darboux Theorem

    If $f'(a)<0$, then there does indeed exist a neighborhood of $a$ where $f(x)>f(a)$ for $x<a$ and $f(a)>f(x)$ for $x>a$. However, there may not be a neighborhood of $a$ where $f(x)$ is decreasing. Do you see the difference between the two claims?
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    Re: Darboux Theorem

    Hmmm, I can't really tell the difference. I am going to think about it for a while and if I can't tell the difference I am going to post a message again. Thank you for giving feedbacks, I really appreciate it.
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    Re: Darboux Theorem

    My previous post was wrong. Continuity of f(x) is sufficient but not necessary to prove Darbouxs theorem by intermediate value theorem.
    So I took OP proof seriously. Having done so, the OP attachment seems quite straight-forward, and I dont see all the fuss about ***:

    The OP begins:
    Let f(a)<0<f(b).
    f(x) continuous on [a,b], therefore f(x) achieves a min on [a,b], and
    a<xmin<b by definition of right and left hand derivative.*

    *f(x)<f(a) for points near a to the right of a (f(a)<0) and f(x)<f(b) for points near b to the left of b (f(b)>0).

    Continuity of derivative is irrelevant to this proof.
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    Re: Darboux Theorem

    Quote Originally Posted by Hartlw View Post
    *f(x)<f(a) for points near a to the right of a (f(a)<0)
    The question was how to show that $f'(a)<0$ implies that there exists a $\delta$ such that $f(x)<f(a)$ for $a<x<a+\delta$. The OP said that $f$ decreases at $a$, which does not make sense. It also does not follow that there exists a $\delta$ such that $f$ is decreasing on $[a,a+\delta]$.
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    Re: Darboux Theorem

    Quote Originally Posted by emakarov View Post
    The question was how to show that $f'(a)<0$ implies that there exists a $\delta$ such that $f(x)<f(a)$ for $a<x<a+\delta$. The OP said that $f$ decreases at $a$, which does not make sense. It also does not follow that there exists a $\delta$ such that $f$ is decreasing on $[a,a+\delta]$.
    f(a)=lim (f(a+h)-f(a))/h, h→0, h+
    |(f(a+h)-f(a))/h f(a)|< ε if h<δ
    -ε < (f(a+h)-f(a))/h f(a) < ε
    f(a+h) f(a) < h[f(a)+ε]

    f(a)= -|f(a)| and let ε =|f(a)|/2
    then
    f(a+h) < f(a) and f(a) is not a minimum.

    A similar argument applies at b.
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