In what follows, I transcribe an example that illustrates the application of the Nehari Manifold method to solve a nonlinear partial differential equation contained in the text (in Portuguese) that I am reading.
My question is: Why can we conclude that $\Phi(u) \geq 0$ in the proof of Lemma 1.2 below?
Let $\Omega \subset \Bbb{R}^N$ be an smooth, bounded, connected open set. Consider the following problem:
$$ (*)\begin{cases} - \Delta u = f(u) \quad \text{ in } \Omega \\ u = 0 \quad \text{ on } \partial \Omega \end{cases} $$ Assume that the nonlinearity $f$ satisfies the following:
($f_1$) $f$ is of class $C^1$ and there exist $C, r \in \Bbb{R}$ such that $2 < r < 2^*$ and \begin{equation} |f(t)| \leq C(1 + |t|^{r-1}); \qquad (1) \end{equation}
($f_2$) $\displaystyle \lim_{|t| \to 0} \frac{|f(t)|}{|t|} = 0$;
($f_3$) $\displaystyle \lim_{t \to + \infty} \frac{F(t)}{t^2} = + \infty$ where $F(t) = \int_0^tf(s) ds$;
($f_4$) The map $\displaystyle t \mapsto \frac{f(t)}{t}$ is increasing for $t > 0$.
By ($f_1$) and ($f_2$), given $\varepsilon > 0$ there exists $C_\varepsilon > 0$ such that $$ |f(t)| \leq \varepsilon |t| + C_\varepsilon |t|^{r-1}. $$
Theorem: If ($f_1$) - ($f_4$) hold, then $(*)$ has a Ground State solution.
A weak solution to $(*)$ is $u \in H_0^1(\Omega)$ such that $$ \int_\Omega \nabla u \nabla \phi dx - \int_\Omega f(u)\phi dx \quad \forall \phi \in H_0^1(\Omega), $$ that is, if it is a critical point of the functional $$ \Phi(u) = \frac12 ||u||^2 - \int_\Omega F(u) dx $$ where $||\cdot||$ is the usual norm in $H_0^1(\Omega)$.
The Nehari manifold is the set $$ \mathcal{N} = \{u \in H_0^1(\Omega) \setminus \{0\} \ : \ \Phi'(u)u = 0 \} $$
The Nehari manifold is not empty:
Lemma 1.1: If ($f_1$) - ($f_4$) hold, then for $u \in H_0^1(\Omega) \setminus \{0\}$ there exists a unique $t_0 = t_0(u) > 0$ such that $t_0u \in \mathcal{N}$ and $\Phi(t_0u) = \max_{t \geq 0}\Phi(tu)$.
I get the proof of this lemma so I will omit it here. I can add it upon requests.
Lemma 1.2: There exists a constant $C >0$ such that $$ 0 < C < ||u|| \quad \forall u \in \mathcal{N}. $$
Also, for all $u \in \mathcal{N}$ we have that $\Phi(u) \geq 0$.Proof: By $(1)$ and the Sobolev immersions, for $u \in \mathcal{N}$ we have: $$ ||u||^2 = \int_\Omega f(u)u \ dx \leq \varepsilon C_1 ||u||^2 + C_\varepsilon C_2 ||u||^r $$ and hence $$ 0 < \left(\frac{1 - \varepsilon C_1}{C_\varepsilon C_2} \right)^{1/(r-2)} \leq ||u||. $$ Now, from ($f_4$) we have that $$ f'(t)t - f(t) > 0 $$ for $t > 0$, which is equivalent to the function $$ \frac12 f(t)t - F(t) $$ being increasing for $t > 0$. From this, we conclude that $$ \Phi(u) = \Phi(u) - \frac12 \Phi'(u)u = \int_\Omega \left(\frac12 f(u)u - F(u)\right) dx \geq 0 $$