How do I integrate this inequality that appears in a paper of Rabinowitz?

Question

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 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.

Answer 1

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.