1. 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

2. Originally Posted by Mark84
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$} \qbezier(0,1.2)(0,1.2)(5,1.2) \qbezier(0,0)(0,0)(2,1.2) \qbezier(0,0)(0,0)(5,1.2) \qbezier(0,0)(1,0)(2,.90) \end{picture}$

3. Originally Posted by JakeD
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

4. Originally Posted by JakeD
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$} \qbezier(0,1.2)(0,1.2)(5,1.2) \qbezier(0,0)(0,0)(2,1.2) \qbezier(0,0)(0,0)(5,1.2) \qbezier(0,0)(1,0)(2,.90) \end{picture}$
Originally Posted by Mark84
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.

5. Originally Posted by JakeD
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

6. Originally Posted by Mark84
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.

7. Thanks for all the help

/Mark

8. 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.