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Math Help - Analysis Help

  1. #1
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    Analysis Help

    Suppose that  g:[0,1] \rightarrow \mathbb{R} is continuous, g(0) = g(1) = 0, and for any  c \in (0,1) there is some  k > 0 such that:

     0 < c-k < c < c+k < 1 and  g(c) = \frac{1}{2}(g(c+k) + g(c-k)) .

    Prove that g(x) = 0 for any x in [0,1].

    My work:

    Since g is a continuous mapping of a closed interval, it must achieve its maximum, say M, on [0,1]. So, let A = {  x \in [0,1] : g(x) = M }. Since g achieves its maximum, A is nonempty. A is clearly bounded above by 1, so  sup(A) = x_0 exists.

    Suppose that  x_0 \ne 0 . Take  x_0 > 0. Then,  x_0 \in (0,1) (since g(0) = 0 = g(1)).

    Then, there is a k > 0 such that  (x_0 - k) and  (x_0 +k) \in (0,1) and that  M = g(x_0) = \frac{1}{2} (g(x_0 + k) + g(x_0 -k)) .

    But, since M is the maximum of g, this implies that  g(x_0 + k) = g(x_0 -k) = g(x_0) = M .

    This is where I'm getting stuck. Can anyone point me in the right direction?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Math Major View Post
    Suppose that  g:[0,1] \rightarrow \mathbb{R} is continuous, g(0) = g(1) = 0, and for any  c \in (0,1) there is some  k > 0 such that:

     0 < c-k < c < c+k < 1 and  g(c) = \frac{1}{2}(g(c+k) + g(c-k)) .

    Prove that g(x) = 0 for any x in [0,1].

    My work:

    Since g is a continuous mapping of a closed interval, it must achieve its maximum, say M, on [0,1]. So, let A = {  x \in [0,1] : g(x) = M }. Since g achieves its maximum, A is nonempty. A is clearly bounded above by 1, so  sup(A) = x_0 exists.

    Suppose that  x_0 \ne 0 . Take  x_0 > 0. Then,  x_0 \in (0,1) (since g(0) = 0 = g(1)).

    Then, there is a k > 0 such that  (x_0 - k) and  (x_0 +k) \in (0,1) and that  M = g(x_0) = \frac{1}{2} (g(x_0 + k) + g(x_0 -k)) .

    But, since M is the maximum of g, this implies that  g(x_0 + k) = g(x_0 -k) = g(x_0) = M .

    This is where I'm getting stuck. Can anyone point me in the right direction?
    You almost got it.

    Let \displaystyle \sup_{x\in[0,1]}f(x)=\alpha and \beta=\sup f^{-1}\left(\{\alpha\}\right). We claim that either \beta=1 or \beta=0. To see this suppose not, then \beta\in(0,1). Thus, by assumption we have some k>0 such that 0<\beta-k<\beta<\beta+k<1 such that f(\beta)=\frac{1}{2}\left(f(\beta-k)+f(\beta+k)\right). Note that both f(\beta+k),f(\beta-k)\leqslant f(\beta) and so in particular if f(\beta+k)\ne f(\beta) then f(\beta)=\frac{1}{2}\left(f(\beta-k)+f(\beta+k)\right)<\frac{1}{2}\left(f(\beta)+f(\  beta)\right)=f(\beta). But, this is evidently a contradiction. It follows that f(\beta+k)=f(\beta)=\alpha and so \beta+k\in f^{-1}\left(\{\alpha\}\right) which contradicts that \beta=\sup f^{-1}\left(\{\alpha\}\right). Thus, it follows that \beta=0,1 but either way this means that \displaystyle \sup_{x\in[0,1]}f(x)=0. Doing a similar analysis shows that \displaystyle \inf_{x\in[0,1]}f(x)=0. Thus, f=0
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  3. #3
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    Ah, thank you so much!

    I was being too dumb to realize that  g(x_0 + k) \ne g(x_0) since  x_0 is the supremum of all points that map to the maximum of g on [0,1]. Sorry for the stupid question -_-;

    But thanks again. Your proof made everything much clearer!
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