Prove if there exists $h\in L^1(\mathbb R)$ such that $|u(x)| +|u(x)|^{p-1}\le h(x)$

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Let $s\in (0, 1/2)$ and $p\in (2, 2^*_s)$, where $2^*_s= 2/(1-2s)$. Let $q\in [1, 2^*_s)$ and let $u$ be a function such that there exists $l\in L^q(\mathbb R)$ satisfying $|u(x)|\le l(x)$ a.e. in $\mathbb R$.

The question is: does $h\in L^1(\mathbb R)$ exist such that $$|u(x)| +|u(x)|^{p-1}\le h(x)?$$

It seems true to me for $|u|$ but I am not sure if it is the same for the power $|u|^{p-1}$. Anyone could help with that?

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Not true. Take $u \in L^{q}$ such that $u \notin L^{1}$. Take $l=u$. If your conlcusion holds then $\int |u|\leq \int h <\infty$, which is a contradiction.