# How can I find such function?

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

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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

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 $[0,\beta]$.
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