Ratio of convex functions with dominating derivatives is convex?

Mathematics Asked by chlee on October 30, 2020

Let $f,g:mathbb [0,infty)rightarrow (0,infty)$ satisfy $f^{(n)}(x)geq g^{(n)}(x)>0$ for all $n=0,1,2,ldots$ and $xin [0,infty)$. In particular, $fgeq g> 0$ are increasing and convex (from the case when $n=1,2$). Further, assume $f^{(n)}(x)geq f^{(n-1)}(x)$ and $g^{(n)}(x)geq g^{(n-1)}(x)$ for all $ngeq1$.

Question: Is $frac{f}{g}$ convex?

This seems true, once I draw a picture, but I couldn’t convince myself thoroughly. Can anybody help? Thanks very much.

One Answer

I think the answer is no necessarily. Note that

$$ f=g+ h mbox{ where } h^{(n)}(x)ge0 ;; forall n ge 0 $$ And $frac{f}{g}$ is convex iff $frac{h}{g}$ is convex.

Now suppose $g(x)=e^x$. Then we need $h$ such that $h^{(n)}(x)ge0 ; forall n$ and

$$ left(frac{h}{e^x}right)''=frac{h''(x)-2h'(x)+h(x)}{e^x}<0 mbox{ somewhere } $$ Set $h(x)=ax+b$, where $a$ and $b$ are positive constants. Then $$ h''(x)-2h'(x)+h(x)= 0-2a+ax+b=a(x-2)+b $$ which is negative in $x=2-frac{b}{a}-varepsilon$.


I think that given any $g$, we can choose $h$ such that

$$ left(frac{h}{g}right)''<0 $$ but surely it's a little more difficult to prove.

Answered by Pocho la pantera on October 30, 2020

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