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Math Help - Challenging proofs

  1. #1
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    Question Challenging proofs

    Hello!

    Can anyone help me solve the problems? It paralyzes my work

    It is obvious in the first two problems that f'(0)=0 and f is Lipschitz. But I can't go any further.

    1) A continuously differentiable function f is defined on \mathbb{R} such that f(0) = 0 and |f'(x)| \leq |f(x)| for all x\in \mathbb{R}. Show that f is constant.

    2) Suppose that for some positive constant M, |f'(x)|\leq M|f(x)|, x\in[0,1]. Show that if f(0)=0, then f(x) =0 for any x, 0\leq x\leq 1.

    3) Show that there exists no real-valued function f such that f(x)>0 and f'(x)=f(f(x)) for all x.

    Thank you very much!
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  2. #2
    Super Member Showcase_22's Avatar
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    Quote Originally Posted by Ivan View Post
    Hello!

    Can anyone help me solve the problems? It paralyzes my work

    It is obvious in the first two problems that f'(0)=0 and f is Lipschitz. But I can't go any further.

    1) A continuously differentiable function f is defined on \mathbb{R} such that f(0) = 0 and |f'(x)| \leq |f(x)| for all x\in \mathbb{R}. Show that f is constant.
    Here's my attempt:

    f(x) is continuous on \mathbb{R} since it's differentiable. Hence f(x) is continuous on some interval [0,a], and differentiable on (0,a) where a \in \mathbb{R} and a \neq 0.

    By applying the mean value theorem \exists \ x_0 \in (0,a) \ s.t \ f'(x_0)=\frac{f(a)-f(0)}{a-0}=\frac{f(a)-f(0)}{a}.

    We know that |f'(x)| \leq |f(x)| \Rightarrow \ |f'(x_0)| \leq |f(x_0)|.

    Therefore:

    |\frac{f(a)-f(0)}{a}|=|f'(x_0)| \leq |f(x_0)|

    \Rightarrow |f(a)-f(0)| \leq |a||f(x_0)|

    \Rightarrow |f(a)-f(0)| \leq |af(x_0)|

    Let \epsilon>|af(x_0)|

    This gives:

    |f(a)-f(0)| \leq |af(x_0)|<\epsilon so |f(a)-f(0)|< \epsilon \forall a \in \mathbb{R} where a \neq 0.

    Hence f(a)=f(0)=0 since |f(a)| < \epsilon as f(0)=0.
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  3. #3
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    ?

    Great, but the \epsilon in your proof isn't arbitrary, is it?
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  4. #4
    Super Member Showcase_22's Avatar
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    I don't think it is , shoot!

    hmm, this really is pretty tricky....
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