Sorry if this is too easy for MO, but I found it in a research paper, so I thought that it was worth posting here.


I was reading a paper by Rabinowitz([this one](https://apps.dtic.mil/sti/pdfs/ADA093444.pdf) to be more precise) and I came across the following theorem:



> 
> **Theorem 2.10:** Suppose $H\in C^1(\mathbb{R}^{2n}, \mathbb{R})$ and satisfies  
> 
> ($H\_1$) $H(z)\ge 0$  
> 
> ($H\_2$) $H(z)=o(|z|^2)$ at $z=0$  
> 
> ($H\_3$) There are constants $r>0$ and $\mu>2$ such that for all $|z|>r$, $$0<\mu H(z)\le (z, H\_z(z))\_{\mathbb{R}^{2n}}.$$
> Then, for any $T>0$, (HS) has a nonconstant $T$ periodic solution.
> 
> 
> 


What I don't understand is one of the remarks after this statement. It says that by integrating the inequality in ($H\_3$) we get that there are constants $a\_1, a\_2>0$ such that $$H(z)\ge a\_1 |z|^{\mu}-a\_2$$ for all $z\in \mathbb{R}^{2n}$.  

I don't understand how we actually integrate that inequality. What I am sure is that for $|z|\le r$ I just have to use the fact that $H$ is continuous on this compact set and I have a lower bound. So, the integration part is, as expected, for $|z|>r$.


I thought that maybe I should just write $z=(z\_1, z\_2, ..., z\_{2n})$ and express the scalar product using its definition, i.e. $\displaystyle (z, H\_z(z))\_{\mathbb{R}^{2n}}=\sum\_{j=1}^{2n} z\_j \frac{\partial H}{\partial z\_j}(z)$. From what I understand from the paper, $z$ is a function of time ($t$). So, I think that I should just integrate with respect to $t$ on $[r, |z|]$ and apply integration by parts, but I still don't see how this leads to the inequality in the remark.


So, I would really appreciate your help, I am sure that this is just a routine computation (I saw similar inequalities in other papers of Rabinowitz, but he always just says that they are obtained by integrating some inequality), but I haven't done this before.



How do I integrate this inequality that appears in a paper of Rabinowitz?
For any unit vector $u$ and real $t>0$, let  

\begin{equation}
 h(t):=H(tu).
\end{equation}
Then
\begin{equation}
 h'(t)=H'(tu)\cdot u=\frac{H'(tu)\cdot(tu)}t\ge\frac{\mu H(tu)}t=\frac{h(t)}t. 
\end{equation}
Recall that $H>0$ and hence $h(t)>0$ for all real $t$. So, for all real $t\ge1$
\begin{equation}
 (\ln h)'(t)\ge\frac1t 
\end{equation}
and hence
\begin{equation}
 \ln h(t)\ge \ln h(1)+\mu\ln t
\end{equation}
and
\begin{equation}
 H(tu)\ge H(u) t^\mu. \tag{1}\label{1}
\end{equation}
The function $H$ is strictly positive and continuous. So, $H(u)\ge c\_1$ for some real $c\_1>0$ and all unit vectors $u$. So, by \eqref{1}, for any vector $z$ with $|z|\ge1$,
\begin{equation}
 H(z)\ge c\_1|z|^\mu\ge c\_1|z|^\mu-c\_1. 
\end{equation}
Also, the inequality $H(z)\ge c\_1|z|^\mu-c\_1$ trivially holds when $|z|<1$, since $H>0$.


Thus, for all vectors $z$,
\begin{equation}
 H(z)\ge c\_1|z|^\mu-c\_1, 
\end{equation}
as desired.


