# A modified version of Poincare inequality

We know the general version of Poincare inequality:

$$\int_\Omega |u-u_\Omega|^2dx\leq C\int_\Omega|\nabla u|^2dx,\quad \forall u\in W^{1,2}(\Omega),$$ where $u_\Omega$ is the average of $u$ over $\Omega$. This immediately shows that $$\int_\Omega |u|^2dx\leq C\int_\Omega|\nabla u|^2dx+\frac{1}{|\Omega|}\Bigr(\int_\Omega udx\Bigr)^2,\quad \forall u\in W^{1,2}(\Omega).$$

Now, the question is how to deduce the following modified Poincare inequality: for any $m\geq2$, there holds $$\int_\Omega |u|^2dx\leq C\Bigr\{\int_\Omega|\nabla u|^2dx+\Bigr(\int_\Omega |u|^{\frac{2}{m}}dx\Bigr)^m\Bigr\},\quad \forall u\in W^{1,2}(\Omega),$$ where $C>0$ depends only on $m$ and $\Omega$. My tentative way is to use interpolation by means of the Holder inequlity to the second inequality above. But I can not get the desired result. Any help is highly appreciated.

There is a direct way to prove the modified Poincaré inequality: by noting $$\frac{2}{m}\alpha +2(1-\alpha)=1\Rightarrow \alpha=\frac{m}{2(m-1)},$$ then Hölder's inequality implies $$\Bigr(\int_\Omega |u|dx\Bigr)^2=\Bigr(\int_\Omega |u|^{\frac{2}{m}\alpha}|u|^{2(1-\alpha)}dx\Bigr)^2\leq \Bigr(\int_\Omega |u|^{\frac{2}{m}}dx\Bigr)^{\frac{m}{m-1}}\Bigr(\int_\Omega |u|^2dx\Bigr)^{\frac{m-2}{m-1}}. \tag{1}$$ Finally, the Young equality with $\epsilon$ gives $$\Bigr(\int_\Omega |u|^{\frac{2}{m}}dx\Bigr)^{\frac{m}{m-1}}\Bigr(\int_\Omega |u|^2dx\Bigr)^{\frac{m-2}{m-1}}\leq \epsilon \int_\Omega |u|^2dx +C(\epsilon)\Bigr(\int_\Omega |u|^{\frac{2}{m}}dx\Bigr)^m. \tag{2}$$ Now, combining (1), (2) and the standard Poincaré inequality, one gets the desired result.
Supposed the modified Poincaré inequality fails. Then there is a sequence of functions $u_n$ such that $$\int_\Omega |u_n|^2dx=1\tag{1}$$ and $$\int_\Omega|\nabla u_n|^2dx+\Big(\int_\Omega |u_n|^{\frac{2}{m}}dx\Big)^m\to 0\tag{2}$$ Since the mean of $u_n$ is controlled by $\int_\Omega |u_n|^2dx=1$, it is uniformly bounded; passing to a subsequence we may assume $(u_n)_\Omega \to \mu$. By the standard Poincaré inequality, $$\int_\Omega |u_n-\mu|^2\to 0 \tag{3}$$ Passing again to a subsequence, we can say that $u_n$ converges to $\mu$ a.e. The Vitali convergence theorem applies to the functions $|u_n|^{\frac{2}{m}}$, since they are uniformly integrable by $(1)$. It yields $$\int_\Omega |u_n|^{\frac{2}{m}}dx \to |\mu|^{2/m}|\Omega |$$ By $(2)$, $\mu=0$, but then $(3)$ contradicts $(1)$.
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