# Geometry question with convexity

Assume that a function $h(\lambda)$ is decreasing and convex given interval $[l,u]$ and has an unique root $\lambda^*\in (l,u)$. Also, assume $|l-\lambda^*| > |\lambda^*-u|$. Consider any $z\in (l,\lambda^*)$. By connecting points $(z,h(z))$ and $(u,h(u))$, we obtain $(w,0)$. Then, I want to show $|z-\lambda^*|>|w-\lambda^*|$.

Any suggestions or comments are welcome.

In the following plot, we have $(l,u)=(-4,0)$, $\lambda=-\sqrt 2$ (I omit the $\ ^*$) and $w=-\frac 12$. (The function is $x^2-2$.) We can see that $|l-\lambda|>|u-\lambda|$. When $z=-\sqrt 2-\epsilon$ for a small $\epsilon>0$, we have $|z-\lambda|=\epsilon$ and $|w-\lambda|=\sqrt 2-\frac 12$, so you inequality isn't true.
For the revised question, we get the following: But now, the function (not this one) could go from $(-3,6)$ to $(-2,0)$ and then to $(0,-2)$, and still be convex. In that case, we would have $|z-\lambda|<|w-\lambda|$, so the theorem still can't be true.
• You are right. There was my mistake in the problem. Sorry. Actually, $(w,0)$ is an intersection point of the secant line (which connects points $(z,h(z))$ and $(u,h(u))$) and horizontal axis. – chp61 Jan 2 '14 at 15:58
• Of course you need that $f(a)>f(b)>f(c)$ if $a<b<c$. Also, you need $\frac{f(a)-f(b)}{a-b}<\frac{f(b)-f(c)}{b-c}$, because the slope can't decrease. If these conditions are satisfied, you can always find a convex decreasing function. – Ragnar Jan 3 '14 at 15:33