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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- Aug 22nd 2007, 12:49 AMMark84Upper 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 - Aug 22nd 2007, 02:49 AMJakeD
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}

$ - Aug 22nd 2007, 03:09 AMMark84
- Aug 22nd 2007, 03:21 AMJakeD
- Aug 22nd 2007, 04:06 AMMark84
- Aug 22nd 2007, 05:00 AMJakeD
- Aug 22nd 2007, 05:20 AMMark84
Thanks for all the help

/Mark - Aug 22nd 2007, 05:30 AMmathisfun1
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.