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Let $\Omega$ be a bounded Lipschitz domain and let $$\mathcal{S}=\{u\in W^{1,2}(\Omega)\colon\int_\Omega u^3dx=0\}$$ Then there exists some constant $C(\Omega)>0$ such that $$ \int_\Omega|u|^2dx\leq C\int_\Omega \|\nabla u\|^2dx. $$ This already really looks likes Poincaré inequality, besides the fact that $u\in\mathcal{S}$ does not need to have non-zero trace. Or is this non-zero trace property inherited by the fact that the $\int_\Omega u^3dx$ vanishes?

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  • $\begingroup$ What is the dimension of $\Omega$? Is $\Omega$ connected? $\endgroup$ – gerw Feb 13 at 7:54
  • $\begingroup$ $\Omega\subset\mathbb{R}^3$ is a bounded Lipschitz domain. $\endgroup$ – Lucas Smits Feb 13 at 7:56
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The inequality is indeed a Poincare inequality, but not the classical one for functions that vanish on the boundary. When $\Omega$ is a bounded Lipschitz domain, Poincare's inequality holds for any subspace $$ S:=\{u\in W^{1,2}(\Omega)\ |\ G(u)=0 \}$$ where $G:W^{1,2}(\Omega)\rightarrow\mathbb{R}$ is weakly continuous and has the property $$ u=const\ \wedge\ G(u) =0\quad\Rightarrow\quad u=0.$$ Here is proof by contradiction: Assume Poincare's inequality does not hold for all $u\in S$. Then there is a sequence $(u_n)\subset S$ such that $||u_n||_2 \geq n ||\nabla u_n||_2.$ Put $v_n:=\frac{u_n}{||u_n||_2}$. Then $(v_n)$ is a bounded sequence in $W^{1,2}(\Omega)$, and you can therefore find a weakly convergent subsequence (still denoted by $(v_n)$) with $v_n \rightharpoonup v$ for some $v\in W^{1,2}(\Omega)$. Since $G$ is weakly continuous, we even have $G(v)=0$. By the compact embedding $W^{1,2}(\Omega)\hookrightarrow L^2(\Omega)$, we further deduce $||v||_2=1$. Since $||\nabla v_n||_2\rightarrow 0$, we also have $\nabla v$=0. But this means that $v=const$, whence the property of $G$ implies v=0. This contradicts $||v||_2=1$.

In your case, you can put $G(u):=\int_\Omega u^3 dx$. The weak continuity of G then follows from the compact embedding $W^{1,2}(\Omega)\hookrightarrow L^3(\Omega)$, which holds when $\Omega\subset\mathbb{R}^3$.

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