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Math Help - Continuity of Function on Intervals

  1. #1
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    Continuity of Function on Intervals

    If we let I := [a,b] and let f:I\rightarrow\mathbb{R} be a continuous function on I such that for each x in I there exists y in I such that |f(y)| \leq \frac{1}{2}|f(x)|. Prove there exists a point c in I such that f(c) = 0.
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  2. #2
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    Assume that  \forall x \in I , 0 < |f(x)| . You know that f takes on a minimum on  I = [a,b] . Say it takes on its minimum at  x = m . Note that  \frac{|f(m)|}{2} < |f(m)| since  |f(m)| \neq 0 .

     \exists y \in I : |f(y)| \le \frac{|f(m)|}{2} < |f(m)| , contradicting the fact that  f(m) is the minimum of f on I.
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  3. #3
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    Additionally, you could suppose that \sup_I|f|=1 on the interval (if you can't do this then f is the zero function and there is nothing to prove).
    Then choose a point x_0, so you have a new point x_1 such that |f(x_1)|\le\frac{|f(x_0)|}{2}\le\frac{1}{2}.
    The by induction you have points \{x_n\}_{n=0}^\infty which satisfy |f(x_n)|\le\frac{1}{2^n}

    Now some subsequence x_{n_k} converges to a point x. And since f is continuous

    \lim_k|f(x_{n_k})|=\lim_k\frac{1}{2^{n_k}}=0=|f(x)  |.

    Really, the \frac{1}{2} wasn't important. You just needed a constant 0<\alpha<1 and the same proof will work.
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  4. #4
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    Quote Originally Posted by JG89 View Post
    Assume that  \forall x \in I , 0 < |f(x)| . You know that f takes on a minimum on  I = [a,b] . Say it takes on its minimum at  x = m . Note that  \frac{|f(m)|}{2} < |f(m)| since  |f(m)| \neq 0 .

     \exists y \in I : |f(y)| \le \frac{|f(m)|}{2} < |f(m)| , contradicting the fact that  f(m) is the minimum of f on I.
    Sorry but I need a little extra help on this one... I follow everything you're saying but can't see how to use that to show there is a c where f(c) = 0....
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  5. #5
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    f has a root in [a,b] if and only if |f| has a root in [a,b], right?

    If |f| doesn't have a root in [a,b], then 0 < |f(x)| for all x in [a,b], right? From that assumption, I derived a contradiction. Which must mean that there indeed does exist a point x = c such that 0 = |f(c)| = f(c)
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