Is there any bilinear Poincaré/Sobolev inequality?

Question

Is the following, I call it bilinear Poincaré inequality, true?

Let $\Omega$ be an open bounded set in $\mathbf R^n\DeclareMathOperator{\dL}{d\!}$. There exists $C > 0$ such that for any $u, v \in H^2(\Omega)$, \begin{equation} \int_{\Omega} (u - \bar{u})^2(v - \bar{v})^2 \dL x \le C\int_{\Omega} \lvert Du\rvert^2\lvert Dv\rvert^2 \dL x \end{equation} where $\bar{u}$, $\bar{v}$ are the mean values of $u$ and $v$ over $\Omega$ respectively.

If the above is false, is there any other form of "bilinear Poincare inequality" or "bilinear Sobolev inequality" and even for $p$ other than 2?

Answer 1

Note: My answer was posted before the question was edited to a different question. My counterexample still works for edit 6 of the question.

Let $\Omega=(0,2)$. Let $$\DeclareMathOperator{\dL}{d\!}u(x)=\begin{cases}0&\text{ if }x<1\\ x-1&\text{ if }x>1\end{cases}$$ and $$ v(x)=\begin{cases}x&\text{ if }x<1\\ 1&\text{ if }x>1\end{cases} $$ Then $$ \int_0^2 u(x)\dL x=\frac{1}{2} $$ and $$ \int_0^2 v(x)\dL x=\frac{3}{2} $$ so $\bar u=\frac{1}{4}$ and $\bar v=\frac{3}{4}$. However $$ \begin{split} \int_0^2(u(x)-\bar u)(v(x)-\bar v)\dL x & =\int_0^1 -\frac{1}{4}\left(x-\frac{3}{4}\right)\dL x+\int_1^2 \left(x-\frac{5}{4}\right)\frac{1}{4} \dL x \\ &=\frac{1}{16}+\frac{1}{16}=\frac{1}{8}. \end{split} $$ But $$ \int_0^2 |u'(x)||v'(x)|\dL x=0. $$

Answer 2

From a higher level point of view, the Poincare inequality is about comparing seminorms of functions. For example, it is reasonable to expect something like:

$$\DeclareMathOperator{\dL}{d\!}\int_\Omega ( u - \overline u )^p \dL x \leq C \int_\Omega | D u |^p \dL x$$

Your suggested inequality looks like trying to compare two scalar products, except it has been modified on the right-hand side to be positive.