Let $\Omega$ be a bounded domain in $\mathbb{R}^N$, $N\geq 2$. Let $u\in W^{1,p}(\Omega)$ which is the class of all measurable functions $u:\Omega\to\mathbb{R}$ such that $$ \|u\|_{W^{1,p}(\Omega)}:=\|u\|_{L^p(\Omega)}+\|\nabla u\|_{L^p(\Omega)}<\infty. $$ Let $1<q<p^*=\frac{N-p}{N-p}$ if $1<p<N$ and $1<q<\infty$ if $p\geq N$. Assume that $$ \int_{\Omega}|u|^{q-2}u\,dx=0. $$ Then can we obtain $\int_{\Omega}u\,dx=0$? If so, can someone please help me with this?
If we assume $$ \int_{\Omega}|u|^{q-2}u\,dx=0, $$ then I can get $\int_{\Omega}(u_+^{q-1}-u_-^{q-1})\,dx=0$, where $u_{+}=\max\{u,0\}$ and $u_-=\max\{-u,0\}$. After this I get no clue. Kinldy help. Thanks.
Not an explicit counterexample, but a way to construct one.
Consider $$ u(x)= \begin{cases} 1 & \text{ if } & 0\leq x<1,\\ -2^{-1/q} & \text{ if } & 1\leq x\leq 3. \end{cases} $$ Notice that $\int_0^3 |u|^{q-1}u\, dx =0$, but $u$ doesn't have zero average. To build a function in $W^{1,p}$ you can linearly interpolate between the two values as $$ u(x)= \begin{cases} 1 & \text{ if } & 0\leq x<a,\\ \ell(x) & \text{ if } & a<x<b,\\ -2^{-1/q} & \text{ if } & b\leq x\leq 3, \end{cases} $$ for some $0<a<1<b<3$ and $\ell$ is the linear function with value $1$ at $x=a$ and $-2^{-1/q}$ at $x=b$. This function is always in $W^{1,\infty}$, and by choosing $a,b$ appropriately you can check that $\int_0^3|u|^{q-1} u\, dx =0$, but $u$ doesn't have zero average.