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Consider a continuous function $u(q)\geq 0$ whose domain is $[0,\infty)$ that satisfies the following conditions:

  1. $u^{\prime }>0$ for all $q \in [0,\infty)$,

  2. $u^{\prime \prime }<0$ for all $q \in [0,\infty)$,

  3. $u(0)=0$,

  4. $u^{\prime }(0)<\infty $

Now I am wondering whether $\frac{u^{\prime \prime }(q)}{u^{\prime }(q)}$ is weakly increasing in $q$ always hold ? Is there any function $u(q)$ making $\frac{u^{\prime \prime }(q)}{u^{\prime }(q)}$decreasing in $q$ or non-monotonic in $q$?

Here is what I have done: $\frac{d\Big(\frac{u^{\prime \prime }(q)}{u^{\prime }(q)}\Big)}{d q}=\frac{u^{\prime \prime \prime}(q)u^{\prime}(q)-[u^{\prime \prime}(q)]^{2}}{[u^{\prime}(q)]^{2}}$. So if $u^{\prime \prime \prime}(q)<0$ or $u^{\prime \prime \prime}(q)u^{\prime}(q)<[u^{\prime \prime}(q)]^{2}$ holds for some $q$, I can conclude the above conjecture is not true. But I cannot figure out a such a function. Can you help to find out such a function? thank you !

congmingniao
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  • It's not possible that $u'''(q)<0$ for all $q$, as this would imply $u'(q)\leq u'(1)+u''(1)(q-1)$ for $q>1$, so $u'(q)\rightarrow-\infty$ as $q\rightarrow\infty$. – stewbasic Aug 05 '16 at 03:31
  • thank you. But is it possible that $u^{\prime \prime \prime}(q)<0$ for some $q$? – congmingniao Aug 05 '16 at 03:36

1 Answers1

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$\frac{u''}{u'}$ can equal any continuous function $g:[0,\infty)\rightarrow(-\infty,0)$. In particular it can be increasing, decreasing or neither.

Indeed let $$ G(q)=\int_0^qg(t)\;dt, $$ $$ u(q)=\int_0^qe^{G(t)}\;dt. $$ Then $u(0)=0$, $u'=e^G>0$, $u''=ge^G<0$, $u'(0)=e^{G(0)}=1$ and $\frac{u''}{u'}=g$.

stewbasic
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