# Upper bounded convex funtion

• August 22nd 2007, 12:49 AM
Mark84
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
• August 22nd 2007, 02:49 AM
JakeD
Quote:

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 $y$ 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}
\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}
$
• August 22nd 2007, 03:09 AM
Mark84
Quote:

Originally Posted by JakeD
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}
\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
• August 22nd 2007, 03:21 AM
JakeD
Quote:

Originally Posted by JakeD
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}
\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
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.
• August 22nd 2007, 04:06 AM
Mark84
Quote:

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
• August 22nd 2007, 05:00 AM
JakeD
Quote:

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

$t = (c-b)/(c-a)$ 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.
• August 22nd 2007, 05:20 AM
Mark84
Thanks for all the help

/Mark
• August 22nd 2007, 05:30 AM
mathisfun1
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) . 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, $f(ta+(1-t)b) > tf(a)+(1-t)f(b)> M$ - contradiction.