Last step in $\nabla_z f(x,y,z), \partial_y f(x,y,z)\le c(1+|y|^{p-1}+|z|^{p-1})$ imply $ f(x,y,z)\le c(1+|y|^p+|z|^p)$

Question

Let $\Omega \subset \mathbb{R}^n$ open and bounded with Lipschitz boundary $\partial \Omega$ and the functional

$$I(u)=\int_\Omega f(x,u(x),\nabla u(x))dx$$ with $f: \bar\Omega\times\mathbb{R}\times \mathbb{R}^n \to\mathbb{R}$ of class $C^1$

$f$ satisfies the following grow conditions:

$\partial_y f(x,y,z)\le c(1+|y|^{p-1}+|z|^{p-1}) \tag{4.5}$

$\nabla_z f(x,y,z)\le c(1+|y|^{p-1}+|z|^{p-1}), \forall (x,y,z) \in \bar\Omega\times\mathbb{R}\times \mathbb{R}^n \tag{4.5},$

for some $1<p<\infty$ and a constant $c >0$

Consider $u \in W^{1,p}(\Omega)$ in the class of functions $\Phi=\{u \in W^{1,p}(\Omega): u-g \in W^{1,p}_0(\Omega) \}$ with $g \in W^{1,p}(\Omega)$ given

I want to prove the following theorem the conditions (4.5) imply that

$f(x,y,z)\le c(1+|y|^p+|z|^p), \forall (x,y,z) \in \bar\Omega\times\mathbb{R}\times \mathbb{R}^n \tag{4.4},$ for some $1<p<\infty$ and a constant $c >0$ .

I did as instructed and bounded the integral using the triangle inequality, (4.5) and Young's inequality and I got this: $$|f(x,y,z)|\le |f(x,0,0)|+c(1+|y|^p+|z|^p)$$

I just don't know how to get rid of $|f(x,0,0)|$ term. Any idea ?

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Edit. The problem is just following analogous steps to what was done in the begining of the chapter. Here it is:

Answer 1

I think the part people sometimes ignore is the given conditions. You are given the domain of $x$ variable as $\Omega$ which is open and bounded. $f$ itself is also continuously differentiable, meaning that $f$ is bounded, say by $C$ ,on $\overline{\Omega}.$ Therefore, $$|f|\leq C + c(1+|y|^p + |z|^p)\leq c'(1+|y|^p + |z|^p) $$ with $c' = c + C.$