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Thread: Convex & bounded above

  1. #1
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    Convex & bounded above

    True or false: If a function $\displaystyle f: \mathbb{R} \longrightarrow \mathbb{R}$ is convex and bounded above, then $\displaystyle f$ 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

    $\displaystyle \chi$ $\displaystyle \sigma$
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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

    $\displaystyle \chi$ $\displaystyle \sigma$
    It can be a convex function, but not a strictly convex function.
    Last edited by mr fantastic; Sep 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 $\displaystyle f: \mathbb{R} \longrightarrow \mathbb{R}$ is convex and bounded above, then $\displaystyle f$ is constant.
    I believe this statement is TRUE, and the squeeze theorem for limits may be helpful to prove it.

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

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

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

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

    Therefore,

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

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

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

    Thefore,
    $\displaystyle 0\geq f(x)-f(x_1) \geq 0\Rightarrow f(x)=f(x_1)$
    Last edited by mr fantastic; Sep 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 $\displaystyle f: \mathbb{R} \longrightarrow \mathbb{R}$ is convex and bounded above, then $\displaystyle f$ is constant.
    Here is another method to prove this statement. This method investigates intervals $\displaystyle [-\infty,x_1],[x_2,+\infty]$, which differs from previous method which investigates interval $\displaystyle [x_1,x_2]$.

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

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

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

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