Why does $ \frac{1 + |u|^p}{1 + |u|} \in L^{N/2}_{loc}(\Omega) $?

Question

Let $u \in H^1_{loc}(\Omega)$ where $\Omega \subset \mathbb R^N$ is an open set and $p < 2^* -1$. Why does it hold that $$ \frac{1 + |u|^p}{1 + |u|} \in L^{N/2}_{loc}(\Omega) \quad ? $$

Of course, if $\omega \subset \Omega$ is compact then $$ \int_\omega \left(\frac{1 + |u|^p}{1 + |u|}\right)^{N/2} \leq \int_\omega (1 + |u|^p)^{N/2} \leq \int_\omega (1 + |u|^p)^N \leq 2^N \int_\omega (1 + |u|^{pN}). $$ But $pN$ may be greater then $2^*$, and in this case we don't know if $u \in L^{pN}(\omega)$.

Any hints will be the most appreciated.

Context: I am trying to understand Struwe's presentation of the Brezis-Kato regularity argument.

Thanks in advance and kind regards.

Answer 1

Your bound is bad when $|u|$ is big. Try using: $$ \frac{1 + |u|^p}{1 + |u|} \leq {1 + |u|^{p-1}} $$ which can be proved multiplying both LHS and RHS by ${1 + |u|}$.