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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.

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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 $|l-\lambda|>|u-\lambda|$. plot
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.

EDIT
For the revised question, we get the following: plot 2
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.

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  • $\begingroup$ 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. $\endgroup$ – chp61 Jan 2 '14 at 15:58
  • $\begingroup$ That makes the problem more interesting :)\ $\endgroup$ – Ragnar Jan 2 '14 at 16:01
  • $\begingroup$ That makes sense to me. Thanks @Ragnar BTW can we always find a convex decreasing function in some interval that passes given three points?... $\endgroup$ – chp61 Jan 3 '14 at 14:36
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    $\begingroup$ 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. $\endgroup$ – Ragnar Jan 3 '14 at 15:33

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