Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f\in C^1(\mathbb{R})$ be a function with compact support satisfying $f(0)>0$ and $f(a)=0$ for some $a>0$. For each fixed $p\in (2,\infty)$, I want to find a continuous function $g:\mathbb{R}\rightarrow\mathbb{R}$, such that $g(0)=0$, $g$ is non-decreasing in the interval $(0,\delta)$, for some $\delta>0$, $f+g$ is non-decreasing in the interval $[0,a]$ and $$\int_0^\delta [G(s)]^{\frac{-1}{p}}ds=\infty$$

where $G(s)=\int_0^s g(r)dr$.

Thanks for your help and your patience.

share|cite|improve this question
Are you sure this is possible? For if $f'(0) < 0$, then for $f+g$ to be non-decreasing that means that $g'(0)$ has to be at least as great in magnitude. This means that $G(x)$ has at least a $x^2$ term in the Taylor expansion about 0, and in turn that says something bad about the local behavior of $G^\frac{-1}{p}$, namely, that it behaves like $x^\frac{-2}{p}$ locally, which is integrable. – Ray Yang Mar 4 '13 at 17:23
Well, I was thinking that this is not possible too, because of your example. I alread know it, but I could not prove anything. Also, there are cases where $f'(0)=0$ and $f'(s)<0$ for small $s$ and it still not workng. I think, i will have to assume the hypothesis $f'(0)>0$. Anyway, do you have any idea to prove it is not possile? – Tomás Mar 4 '13 at 17:31
@RayYang, please verify if my observation is right. $G$ must behavior (near the origin) like a polynomious of degree $q>p$? – Tomás Mar 4 '13 at 17:48
up vote 2 down vote accepted

If $f'(0)<0$ it is no possible. Since $f$ is smooth, there exists $\epsilon>0$ and an interval $[0,\alpha]$ such that $f'(x)\le-\epsilon$ if $x\in[0,\alpha]$. Without loss of generality we may assume $\alpha<a$. Since $f+g$ is non decreasing in $[0,\alpha]$ we have $f'+g'\ge0$ in $[0,\alpha]$ and $$ g'(x)\ge-f'(x)\ge\epsilon\quad 0\le x\le\alpha. $$ Then if $x\in[0,\alpha]$ $$ g(x)=g(0)+\int_0^xg'(t)\,dt\ge\epsilon\,x $$ and $$ G(s)\ge\frac{\epsilon}{2}x^2. $$ Since $p>2$, $\int_0^\delta[G(s)]^{-1/p}ds<\infty$ for any $\delta>0$.

Assume now $f'(0)>0$. There exists $\beta>0$ such that $f'(x)>0$ if $x\in[0,\beta]$. Let $g(x)=m\,x^{p-1}/(p-1)$ where $m>0$ is such that $$ m>\max\Bigl\{0,\max_{\beta\le x\le a}\Bigl(-\frac{f'(x)}{x^{p-2}}\Bigr)\Bigr\}. $$ Clearly $g$ is cotinuous and increasing. Moreover $$ f'(x)+g'(x)=f'(x)+m\,x^{p-2}>0\quad\forall x\in[0,a]. $$ Finally $G(s)$ is like $s^p$ near $0$ and $[G(s)]^{-1/p}$ is like $s^{-1}$.

The case $f'(0)=0$ seems more delicate. You can carry out the above construction if for instance $f'(x)\ge C\,x^{p-2}$ for some constant $C>0$ in some interval $[0,\beta]$.

share|cite|improve this answer
Thank you. For the case $f'(0)=0$, I think that if $f'(x)=-\sqrt{1-(x-1)^2}$ near the origin, then do not exist such $g$. – Tomás Mar 4 '13 at 18:03

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.