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Math Help - Proving properties of bumb function

  1. #1
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    Proving properties of bump function

    Let be f(x):=g(1-g(x)); x \in \mathbb{R}, where g: \mathbb{R} \rightarrow \mathbb{R}\text{; }  x \mapsto g(x):= \begin{cases}e^{1-\frac{1}{x^2} &  \text{if }x>0\\0 &  \text{if }x\leq 0\end{cases}
    Let be a, b \in \mathbb{R} and h(x)=f(a-x)f(x-b).
    Prove that: h \in C^\infty;\: h\geq0; h(x)=1 \text{ if } x\in [a,b] \text{ and } h(x)=0 \text{ if } x \not\in ]a-1,b+1[ .

    I know that g \in C^\infty but not analyitic in 0. I also know that f is a monotonically decreasing function f(x)=1 \text{ if } x \leq 0 \text{ and } f(x)=0 \text{ if } x>0.
    Thank you very much in advance!
    Last edited by zadir; September 11th 2011 at 02:30 AM. Reason: typo
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Proving properties of bumb function

    T
    Quote Originally Posted by zadir View Post
    Let be f(x):=g(1-f(x)); x \in \mathbb{R}, where g: \mathbb{R} \rightarrow \mathbb{R}\text{; }  x \mapsto g(x):= \begin{cases}e^{1-\frac{1}{x^2} &  \text{if }x>0\\0 &  \text{if }x\leq 0\end{cases}
    Hey, this doesn't make sense. Are you sure you wrote right?
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  3. #3
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    Re: Proving properties of bump function

    Quote Originally Posted by Drexel28 View Post
    T

    Hey, this doesn't make sense. Are you sure you wrote right?
    Thank you so much that you warned me. You are right, f(x)=g(1-g(x)), I have corrected it. Thanks!

    I have made another mistake that I have just noticed. In the title it is not bumb, but of course bump function. Sorry!
    Last edited by zadir; September 11th 2011 at 02:31 AM. Reason: other mistake
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  4. #4
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    Re: Proving properties of bump function

    Quote Originally Posted by zadir View Post
    Thank you so much that you warned me. You are right, f(x)=g(1-g(x)), I have corrected it. Thanks!

    I have made another mistake that I have just noticed. In the title it is not bumb, but of course bump function. Sorry!
    The problem is right and clear after correction, isn"t it?
    Thank you very much in advance!
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