$u\in W^{1,p}(\Omega)$ with $u\in L^q(\Omega)$. Does this implies that $v\in L^q(\Omega)$ for $v$ close to $u$?

Question

Let $\Omega\subset\mathbb{R}^n$ be a bounded domain and suppose that $p<q$ with, $p,q\in [1,\infty]$. Assume that for some $u\in W^{1,p}(\Omega)$, we have that $u\in L^q(\Omega)$. Is it possible to find $r>0$ such that for all $v\in W^{1,p}(\Omega)$ with $\|u-v\|_{1,p}<r$, we have that $v\in L^q(\Omega)?$

Answer 1

Set $u=0$ and pick $w \in W^{1,p}(\Omega)$ such that $w \notin L^q(\Omega)$ so $\varepsilon w$ is not in $L^q(\Omega)$ either. Then $v=u+\varepsilon w$ is a counterexample for any $u$.