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Math Help - Convex & bounded above

  1. #1
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    Convex & bounded above

    True or false: If a function f: \mathbb{R} \longrightarrow \mathbb{R} is convex and bounded above, then f is constant.
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    MHF Contributor chisigma's Avatar
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    According to...

    Convex function - Wikipedia, the free encyclopedia

    ... a constant can't be a convex function ...

    Kind regards

    \chi \sigma
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  3. #3
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    Quote Originally Posted by chisigma View Post
    According to...

    Convex function - Wikipedia, the free encyclopedia

    ... a constant can't be a convex function ...

    Kind regards

    \chi \sigma
    It can be a convex function, but not a strictly convex function.
    Last edited by mr fantastic; September 19th 2009 at 12:35 AM. Reason: Restored original reply
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    Quote Originally Posted by NonCommAlg View Post
    True or false: If a function f: \mathbb{R} \longrightarrow \mathbb{R} is convex and bounded above, then f is constant.
    I believe this statement is TRUE, and the squeeze theorem for limits may be helpful to prove it.

    Known the minimum point \left(x_1, f(x_1)\right), i.e. f(x_1) = \min_x f(x) . This minimum point is global and there is only one such point.

    Choose an arbitrary point \left(x_2, f(x_2)\right). Without loss of generality, assume x_2\geq x_1, i.e. on the right hand side of the minimum point.

    Construct a line segment connecting the arbitrary point \left(x_2, f(x_2)\right) to the minimum point \left(x_1, f(x_1)\right).

    The segment of function f(x), x \in [x_1, x_2] should lie between line y=\frac{f(x_2)-f(x_1)}{x_2-x_1}(x-x_1)+f(x_1) and line y=f(x_1).

    Therefore,

    \frac{f(x_2)-f(x_1)}{x_2-x_1}(x-x_1)\geq f(x)-f(x_1) \geq 0 (1)

    Inequality (1) holds for any x and x_2 as long as  x \in [x_1, x_2]

    For an arbitrary x, inequality (1) holds as x_2\rightarrow+\infty, but
    \lim_{x_2\to+\infty}\frac{f(x_2)-f(x_1)}{x_2-x_1}(x-x_1)=0 (2)
    Remembering both f(x_1) and f(x_2) are bounded.

    Thefore,
    0\geq f(x)-f(x_1) \geq 0\Rightarrow f(x)=f(x_1)
    Last edited by mr fantastic; September 19th 2009 at 12:35 AM. Reason: Restored original reply
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    Quote Originally Posted by NonCommAlg View Post
    True or false: If a function f: \mathbb{R} \longrightarrow \mathbb{R} is convex and bounded above, then f is constant.
    Here is another method to prove this statement. This method investigates intervals [-\infty,x_1],[x_2,+\infty], which differs from previous method which investigates interval [x_1,x_2].

    Choose two arbitrary points \left(x_1, f(x_1)\right) and \left(x_2, f(x_2)\right), where x_2 > x_1.

    (1) If f(x_2) > f(x_1), the slope of the straight line is positive. And f(x) is above this line in the interval  [x_2, +\infty), which implies  f(x)\rightarrow +\infty as x\rightarrow +\infty, and f(x) is unbounded.

    (2) If f(x_2) < f(x_1), the slope of the straight line is negative. And f(x) is above this line in the interval  [-\infty, x_1), which implies  f(x)\rightarrow +\infty as x\rightarrow -\infty, and f(x) is unbounded.

    Therefore, we must have f(x_1)=f(x_2), meaning f(x) is constant.
    Last edited by mr fantastic; September 19th 2009 at 12:37 AM. Reason: Restored original reply
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