Upper bounded convex funtion

**Mark84** · August 22nd 2007, 12:49 AM

Hi

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

Thanks
Mark

**JakeD** · August 22nd 2007, 02:49 AM

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} $

**Mark84** · August 22nd 2007, 03:09 AM

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

**JakeD** · August 22nd 2007, 03:21 AM

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} $

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.

**Mark84** · August 22nd 2007, 04:06 AM

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

**JakeD** · August 22nd 2007, 05:00 AM

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.

**Mark84** · August 22nd 2007, 05:20 AM

Thanks for all the help

/Mark

**mathisfun1** · August 22nd 2007, 05:30 AM

Here's another approach which I find more intuitive. Let M be an upper bound and suppose there exists a 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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