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Math Help - Upper bounded convex funtion

  1. #1
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    Upper bounded convex funtion

    Hi

    I need to show that a real valued (defined on R) upper bounded convex function must be a constant.

    Thanks
    Mark
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  2. #2
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    Quote Originally Posted by Mark84 View Post
    Hi

    I need to show that a real valued (defined on R) upper bounded convex function must be a constant.

    Thanks
    Mark
    Let y be an upper bound for the convex function f. Suppose a < b < c and f(a) < f(b). By convexity, the line connecting the points (a,f(a)) and (c,y) must lie above the point (b,f(b)). Show for c large enough, that cannot be true, contradicting the assumption that f(a) < f(b).

    \setlength{\unitlength}{2.5cm}<br />
\begin{picture}(1,2)<br /> <br />
\put(0,-.2){$f(a)$}<br />
\put(1,.4){$f(b)$}<br />
\put(2,.8){$f(c)$}<br />
\put(5.1,1.1){$y$}<br />
\qbezier(0,1.2)(0,1.2)(5,1.2)<br />
\qbezier(0,0)(0,0)(2,1.2)<br />
\qbezier(0,0)(0,0)(5,1.2)<br />
\qbezier(0,0)(1,0)(2,.90)<br /> <br />
\end{picture}<br />
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  3. #3
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    Quote Originally Posted by JakeD View Post
    Let y be an upper bound for the convex function f. Suppose a < b < c and f(a) < f(b). By convexity, the line connecting the points (a,f(a)) and (c,y) must lie above the point (b,f(b)). Show for c large enough, that cannot be true, contradicting the assumption that f(a) < f(b).

    \setlength{\unitlength}{2.5cm}<br />
\begin{picture}(1,2)<br /> <br />
\put(0,-.2){$f(a)$}<br />
\put(1,.4){$f(b)$}<br />
\put(2,.8){$f(c)$}<br />
\put(5.1,1.1){$y$}<br />
\qbezier(0,1.2)(0,1.2)(5,1.2)<br />
\qbezier(0,0)(0,0)(2,1.2)<br />
\qbezier(0,0)(0,0)(5,1.2)<br />
\qbezier(0,0)(1,0)(2,.90)<br /> <br />
\end{picture}<br />

    Thanks

    I have already convinced myself that is true but I do not think this is a proof?

    /Mark
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  4. #4
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    Quote Originally Posted by JakeD View Post
    Let y be an upper bound for the convex function f. Suppose a < b < c and f(a) < f(b). By convexity, the line connecting the points (a,f(a)) and (c,y) must lie above the point (b,f(b)). Show for c large enough, that cannot be true, contradicting the assumption that f(a) < f(b).

    \setlength{\unitlength}{2.5cm}<br />
\begin{picture}(1,2)<br /> <br />
\put(0,-.2){$f(a)$}<br />
\put(1,.4){$f(b)$}<br />
\put(2,.8){$f(c)$}<br />
\put(5.1,1.1){$y$}<br />
\qbezier(0,1.2)(0,1.2)(5,1.2)<br />
\qbezier(0,0)(0,0)(2,1.2)<br />
\qbezier(0,0)(0,0)(5,1.2)<br />
\qbezier(0,0)(1,0)(2,.90)<br /> <br />
\end{picture}<br />
    Quote Originally Posted by Mark84 View Post
    Thanks

    I have already convinced myself that is true but I do not think this is a proof?

    /Mark
    It is not a proof. It is a sketch of a proof that can be turned into an actual proof by justifying the last two sentences with mathematical expressions.
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  5. #5
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    Quote Originally Posted by JakeD View Post
    It is not a proof. It is a sketch of a proof that can be turned into an actual proof by justifying the last two sentences with mathematical expressions.
    Thanks

    Now I have proved the last part when c -> oo.
    I need some help with why (b, f(b)) must be under the line?
    I understand it but I can not get i right on paper.

    Thanks

    /Mark
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  6. #6
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    Quote Originally Posted by Mark84 View Post
    Thanks

    Now I have proved the last part when c -> oo.
    I need some help with why (b, f(b)) must be under the line?
    I understand it but I can not get i right on paper.

    Thanks

    /Mark
    t = (c-b)/(c-a) implies b = ta + (1-t)c and by definition of convexity that

    \begin{aligned}f(b) &\le tf(a) + (1-t)f(c) \\ &\le tf(a) + (1-t)y \end{aligned}

    since y is an upper bound.
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  7. #7
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    Thanks for all the help

    /Mark
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  8. #8
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    Here's another approach which I find more intuitive. Let M be an upper bound and suppose there exists a<b such that f(a)<f(b)<M. Convexity tells us that f(ta+(1-t)b) <tf(a)+(1-t)f(b). There is some t_0 < 0 such that t_0f(a)+(1-t_0)f(b) > M. Since for some t< 0 we have f(ta+(1-t)b) > tf(a)+(1-t)f(b), by convexity we must have f(ta+(1-t)b)> tf(a)+(1-t)f(b) for all t<0. For t<t_0, f(ta+(1-t)b) > tf(a)+(1-t)f(b)> M - contradiction.
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