Results 1 to 8 of 8

Thread: Upper bounded convex funtion

  1. #1
    Newbie
    Joined
    Aug 2007
    Posts
    5

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Apr 2006
    Posts
    399
    Awards
    1
    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 $\displaystyle y$ be an upper bound for the convex function $\displaystyle f$. Suppose $\displaystyle a < b < c$ and $\displaystyle f(a) < f(b)$. By convexity, the line connecting the points $\displaystyle (a,f(a))$ and $\displaystyle (c,y)$ must lie above the point $\displaystyle (b,f(b))$. Show for $\displaystyle c$ large enough, that cannot be true, contradicting the assumption that $\displaystyle f(a) < f(b)$.

    $\displaystyle \setlength{\unitlength}{2.5cm}
    \begin{picture}(1,2)

    \put(0,-.2){$f(a)$}
    \put(1,.4){$f(b)$}
    \put(2,.8){$f(c)$}
    \put(5.1,1.1){$y$}
    \qbezier(0,1.2)(0,1.2)(5,1.2)
    \qbezier(0,0)(0,0)(2,1.2)
    \qbezier(0,0)(0,0)(5,1.2)
    \qbezier(0,0)(1,0)(2,.90)

    \end{picture}
    $
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2007
    Posts
    5
    Quote Originally Posted by JakeD View Post
    Let $\displaystyle y$ be an upper bound for the convex function $\displaystyle f$. Suppose $\displaystyle a < b < c$ and $\displaystyle f(a) < f(b)$. By convexity, the line connecting the points $\displaystyle (a,f(a))$ and $\displaystyle (c,y)$ must lie above the point $\displaystyle (b,f(b))$. Show for $\displaystyle c$ large enough, that cannot be true, contradicting the assumption that $\displaystyle f(a) < f(b)$.

    $\displaystyle \setlength{\unitlength}{2.5cm}
    \begin{picture}(1,2)

    \put(0,-.2){$f(a)$}
    \put(1,.4){$f(b)$}
    \put(2,.8){$f(c)$}
    \put(5.1,1.1){$y$}
    \qbezier(0,1.2)(0,1.2)(5,1.2)
    \qbezier(0,0)(0,0)(2,1.2)
    \qbezier(0,0)(0,0)(5,1.2)
    \qbezier(0,0)(1,0)(2,.90)

    \end{picture}
    $

    Thanks

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

    /Mark
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Apr 2006
    Posts
    399
    Awards
    1
    Quote Originally Posted by JakeD View Post
    Let $\displaystyle y$ be an upper bound for the convex function $\displaystyle f$. Suppose $\displaystyle a < b < c$ and $\displaystyle f(a) < f(b)$. By convexity, the line connecting the points $\displaystyle (a,f(a))$ and $\displaystyle (c,y)$ must lie above the point $\displaystyle (b,f(b))$. Show for $\displaystyle c$ large enough, that cannot be true, contradicting the assumption that $\displaystyle f(a) < f(b)$.

    $\displaystyle \setlength{\unitlength}{2.5cm}
    \begin{picture}(1,2)

    \put(0,-.2){$f(a)$}
    \put(1,.4){$f(b)$}
    \put(2,.8){$f(c)$}
    \put(5.1,1.1){$y$}
    \qbezier(0,1.2)(0,1.2)(5,1.2)
    \qbezier(0,0)(0,0)(2,1.2)
    \qbezier(0,0)(0,0)(5,1.2)
    \qbezier(0,0)(1,0)(2,.90)

    \end{picture}
    $
    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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Aug 2007
    Posts
    5
    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
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Apr 2006
    Posts
    399
    Awards
    1
    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
    $\displaystyle t = (c-b)/(c-a)$ implies $\displaystyle b = ta + (1-t)c$ and by definition of convexity that

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

    since $\displaystyle y$ is an upper bound.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Aug 2007
    Posts
    5
    Thanks for all the help

    /Mark
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Jun 2007
    Posts
    18
    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 $\displaystyle f(ta+(1-t)b) <tf(a)+(1-t)f(b)$. There is some $\displaystyle t_0 < 0$ such that $\displaystyle t_0f(a)+(1-t_0)f(b) > M$. Since for some $\displaystyle t< 0$ we have $\displaystyle f(ta+(1-t)b) > tf(a)+(1-t)f(b)$, by convexity we must have $\displaystyle f(ta+(1-t)b)> tf(a)+(1-t)f(b)$ for all $\displaystyle t<0$. For $\displaystyle t<t_0$, $\displaystyle f(ta+(1-t)b) > tf(a)+(1-t)f(b)> M$ - contradiction.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. upper bounded taylor
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: Sep 6th 2011, 01:22 AM
  2. Replies: 2
    Last Post: Apr 24th 2010, 08:35 AM
  3. The union of two convex sets is not convex
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: Jan 30th 2010, 03:23 PM
  4. Proving that max{0,f(x)} is convex if f is convex
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Nov 5th 2009, 06:16 AM
  5. Convex & bounded above
    Posted in the Math Challenge Problems Forum
    Replies: 4
    Last Post: Aug 11th 2009, 08:05 PM

Search Tags


/mathhelpforum @mathhelpforum