Proof Nehari manifold of semilineal subcritical $-\Delta u = f(u)$ in $\Omega$ is not empty. Given the problem
$$
\left\{ 
\begin{array}{rll}
-\Delta u& = f(u) & \text{in }\Omega \\
 u & = 0 & \text{in } \partial\Omega
\end{array}
\right.
$$
In a bounded domain $\Omega\subset \mathbb{R}^N, N\geq 3$ with $f$ satisfying:


*

*$f\in C^1(\mathbb{R}), f(0)=0$ and  $f'(0)<\lambda_1$, $\lambda_1$ the first eigenvalue of $-\Delta$ in $H_0^1(\Omega)$

*There exists $c>0$ and $\sigma \in (1,2^{*}-1)$ such that $$|f'(s)|\leq c(|s|^{\sigma-1}+1) \quad \forall s\in \mathbb{R}$$

*There is some $\alpha\in(0,1)$ such that $$f(s)s-2F(s)\geq \alpha f(s)s \quad \forall s$$ for $F(s)=\int_0^s f$.

*For all $s$ $$f'(s)>\frac{f(s)}{s} \quad \text{and} \quad \lim_{|s|\rightarrow \infty}\frac{f(s)}{s}=\infty$$
I am to show that the Nehari manifold associated with the functional $$J(u)=\frac{1}{2}\| u\|^2-\int_{\Omega} F(u)$$ is not empty. I already proved $u\in H_0^1$ is a solution to the PDE if and only if it is a critical point of $J$. I know from a previous exercise that $J$ is $C^2$ and $$J'(u)v=\langle u,v \rangle -\int f(u)v$$ and $$J''(u)(v,w)=\langle v,w\rangle-\int f'(u)vw$$
It is hinted that I must show that given $u\not\equiv 0$, the real valued function $J_u(t):(0,\infty)\mapsto \mathbb{R}, \, J_u(t)$$
=J(tu)$ satisfies $J^\prime_u (t)=0$ for exactly one $t$ and that this is a maximum.
I used the Chain rule for generalized (Frechet) derivatives to obtain
$$\frac{d J_u}{d t}(t)=\frac{d }{d t} J\circ h \,(t)=J'(h(t))\circ h'(t)=J'(tu)u.$$ Where I defined $h(t)=tu$. I must therefore prove $$t^2\|u\|^2-\int f(tu)tu=0$$ has precisely one solution for $t$. I know that the above equation is continuous in $t$ so I thought I'd show the above goes to $0^{+}$ as $t\rightarrow 0^{+}$, and to $-\infty$ as $t\rightarrow \infty$, but I can't quite work it out.
Am I missing something? or am I just getting tunnel vision? 
Thank you in advance.
 A: If we only want to show that the Nehari manifold is not empty, we can do the following. Let $J_u(t)=J(tu)$ where $u\in H_0^1(\Omega)\cap L^\infty(\Omega)$ and $u>0$. As you alread have noted, we have that $$J'_u(t)=t^2\|u\|^2-\int f(tu)tu.$$
From 3. you have that $(1-\alpha)f(s)s\ge 2 F(s)$, which implies that $$\frac{F'(s)}{F(s)}\ge \frac{2}{(1-\alpha)s}.$$
For $s\geq1$, we conclude that $$F(s)\geq C_1s^{\frac{2}{1-\alpha}},$$
for some positive constant $C_1$. Therefore $F(s)\ge C_1s^{\mu}+C_2$ for $s>0$, $C_2\in\mathbb{R}$ a constant and $\mu>2$. By using again the inequality $(1-\alpha)f(s)\ge 2F(s)$, we have that $$-f(s)s\le -D_1s^\mu-D_2,\ s>0,$$
where $D_1>0$ and $D_2\in\mathbb{R}$ are contants. So $$J'_u(t)\le t^2\|u\|^2-D_1t^\mu\int u^\mu-D_2|\Omega|.$$
Once $\mu>2$, we must conclude that $J_u'(t)\to -\infty$ when $t\to \infty$.
On the other hand, the hypotheis $f'(0)$ implies that for small $x>0$, there is $\delta>0$ such that $$f(x)<(\lambda_1-\delta)x,$$
which implies that for small $x>0$ $$-f(x)x> -(\lambda_1-\delta)x^2.$$
Therefore $$J'_u(t)\ge t^2\|u\|^2-(\lambda_1-\delta)t^2\int u^2.$$
Once $\|u\|_2^2\le \frac{1}{\lambda_1}\|u\|$, we obtain that $$J'_u(t)\ge t^2\|u\|^2-\frac{\lambda_1-\delta}{\lambda_1}t^2\|u\|^2,$$
which implies that $J'_u(t)$ is positive near ht origin. Be cause of the continuity of $J'_u$, we conclude that there is $t>0$ such that $J'_u(t)=0$, which is to say that $$\langle J'(tu),tu\rangle =0,$$
or equivalently that $tu$ belongs to the Nehari manifold.
Edit: In the above calculations, I have assumed implicitly that $F(s)\ge 0$, which is not true, however, we can overcome this problem by using the hypothesis $f(s)/s\to \infty$ when $t\to \infty$, indeed, we can conclude that $$F(s)\ge C s^{\mu},$$
where $\mu$ is as above and $s$ is such that $F(s)=\int_0^s f(t)dt$ is positive (which will be true for bigger $s$). Therefore $F(s)\ge C s^{\mu}+D$ for all $s>0$, where $C>0$ and $D\in\mathbb{R}$ are constants.
