Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Take two weakly convex, weakly increasing non-negative functions, $g(x)$ and $h(x)$, domain of $x$ is $[a,d]$, $g(x)=0$ for $x \in [a,b]$, $h(x)=0$ for $x \in [a,c]$, and $g(d)=h(d)$. So $g$ and $h$ start at zero, stay at zero over some (possibly different) domain(s), and intersect again at $d$. Can we make a general statement on the relative convexities and slopes of $g$ and $h$ that achieves a weak ranking of $g$ and $h$ over all $x$ in $[a,d]$? i.e. $g(x)\geq h(x)$?

Clearly if $b<c$ and $g'(x)\leq h'(x)$ for $x \in [c,d]$ that will do it, but i'm more interested in situations where the first derivatives can't necessarily be ranked in this way (and, preferably, where we don't have to invoke a condition on $b$ or $c$). Intuitively it seems like there must be some condition on the scaled second derivative? like on $g''(x)/g'(x)$ vs $h''(x)/h'(x)$?

share|improve this question
    
For future reference, please use the LaTeX capabilities of this website for displayed mathematics. Also please double check your question after you posted it: a stray, unescaped less than sign (&lt;) can "chop off" the tail end of your post. –  Willie Wong Mar 3 '11 at 11:41
    
A pointwise condition on the scaled second derivative is actually not really weaker than a condition on the first derivative. If you have a comparison result $g''/g' \leq h''/h'$, note that $f''/f' = (\log f')'$, you can integrate and get $g' \leq C h'$ using monotonicity of $\log$. In any case, your question feels a bit open-ended, it would help if you give some motivation or some more detailed restrictions on what properties of $g$ and $h$ you are allowed to use. –  Willie Wong Mar 3 '11 at 11:52
    
Also, the $b<c$ condition is unnecessary: if $b > c$ and $g'(x) \leq h'(x)$ for $x \in [c,d]$, then you get a contradiction to the assumption that $h(y) \neq 0$ for some $y\in (c,b]$ and $g(d) = h(d)$. So you can't really object to that condition on the basis of it depending on ranking of $b$ versus $c$. –  Willie Wong Mar 3 '11 at 11:57
    
Thanks very much. Sorry, I can't really give a detailed motivation without being very long-winded. I take your point that the scaled 2nd derivative is equivalent to a condition on the first. But the latter is not constructive since $C$ depends on $g$ and $h$. Is it true that the condition on the scaled second derivative achieves a ranking of g and h? If not, does it help that $g'(x), h'(x)\in[0,1]\forall x$? –  user7728 Mar 7 '11 at 10:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.