How to bound the integrand to show that $|f_u(x,u,\nabla u)|\le 1+ |u|^{p-1}+|\nabla u|^{p-1},$ implies the E-L term $f_u(x,u,\nabla u) \in L^{p'}$

Question

From the book "introduction to the calculus of variations", DACOROGNA I am having trouble giustifying how the last inequality below comes out. p and p' are conjugate exponents

Given the growth condition $|f_u(x,u,\nabla u)|\le 1+ |u|^{p-1}+|\nabla u|^{p-1},$

$$\int_\Omega|f_u(x,u,\nabla u)|^{p'}dx \le \beta^{p'}\int_\Omega(1+ |u|^{p-1}+|\nabla u|^{p-1})^{\frac{p}{p-1}}dx$$$$ \le \beta_1 (1+\|u\|^p_{W^{1,p}})< \infty$$ for some $\beta_1 >0$

I think there are using convexity of the function $x^{p'}, p'>1$ but I don't really get to anything. I would need a convex combination for that, and only two terms Can someone give details on how is done?

Answer 1

If $a,b\geq0$, then we would have $(a+b)^{p}\leq (2 \max(a,b))^{p} \leq 2^p(a^{p}+b^{p})$.