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 $G(s)=\int_0^s g(r)dr$. 
Thanks for your help and your patience. 
 A: 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 $[0,\beta]$.
