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Thread: derivatives 2/analysis

  1. #1
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    derivatives 2/analysis


    problem 3:


    Prove that if f : [0,1)--- IR is nonnegative, integrable, and uniformly
    continuous, then lim f(x) =0 as x tends to 00.

    Problem 4:

    Suppose that a differentiable function f : IR ---IR and its derivative f'
    have no common zeros. Prove that f has only finitely many zeros in [0, 1].
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  2. #2
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    Re: derivatives 2/analysis

    In problem 3, where f is supposed to be defined? You deal with the limit at infinity, but f seems to be only defined on [0,1).
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  3. #3
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    Re: derivatives 2/analysis

    Sorry, f is defined on the interval : [0, 00) where 00 is infinity
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  4. #4
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    Re: derivatives 2/analysis

    Quote Originally Posted by skybluesea2010 View Post

    problem 3:


    Prove that if f : [0,1)--- IR is nonnegative, integrable, and uniformly
    continuous, then lim f(x) =0 as x tends to 00.

    Problem 4:

    Suppose that a differentiable function f : IR ---IR and its derivative f'
    have no common zeros. Prove that f has only finitely many zeros in [0, 1].
    For the first one, assume that it doesn't, then there exists $\displaystyle a>0$ and a sequence such that $\displaystyle x_n \rightarrow \infty$ and $\displaystyle f(x_n)>a$. By uniform continuity there is a $\displaystyle \delta >0$ such that if $\displaystyle |x_n-y|<\delta$ then $\displaystyle |f(x_n)-f(y)|<a/2$ then if $\displaystyle y$ satisifies this we have, by the triangle inequality $\displaystyle f(y)>a/2$ but then

    $\displaystyle \int_0^{\infty} f \geq \sum_{n=1}^{m} \int_{x_n-\delta}^{x_n+\delta} f = 2\delta a m$

    for all $\displaystyle m$. This is a contradiction.

    For the second one, since $\displaystyle [0,1]$ is compact, assuming there is a sequence of zeroes of $\displaystyle f$, we get that there is a subsequence that converges to something in $\displaystyle [0,1]$ and this limit is also a zero. Using Rolle's theorem we can construct a sequence of zeroes of $\displaystyle f'$ that converges to the same limit point. If $\displaystyle f'$ were continous this would be enough, but at the moment I don't see how to finish the argument in general.
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