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Let $\Omega$ be a bounded convex (non-empty) open subset of $\mathbb{R}^n$ ($\Omega$ can be as smooth as you like). Let also $p, q > 1$ be conjugate exponents.

Is it true that there exists a constant $C > 0$ such that the following holds:

Assume given a probability measure $\omega(x) dx$ with $\omega \in L^p(\Omega)$. Then, for any function $f$ in $W^{1, q}(\Omega)$, we have

$$ \left\| f - f_\omega\right\|_{L^q(\Omega)} \leq C \|df\|_{L^q(\Omega)}, $$

where $f_\omega$ is the average of $f$ with respect to $\omega(x) dx$: $$ f_\omega = \int_\Omega f(y) \omega(y) dy $$

A simple contradiction argument similar to the one used to prove the standard Poincaré inequality can be used to prove that the constant $C$ exists for any given $\omega$. My interest is in showing that $C$ can be chosen independently of $\omega$.

Any insight would be invaluable.

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    $\begingroup$ Are $p,q$ conjugate? Do you want $C$ to be independent of the weight? $\endgroup$
    – Giorgio Metafune
    Commented Sep 19, 2023 at 8:42
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    $\begingroup$ @GiorgioMetafune: Thank you very much for pointing these oversights! I corrected my question accordingly. $\endgroup$
    – Romain Gicquaud
    Commented Sep 19, 2023 at 9:12
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    $\begingroup$ Is the weight only entering the average $f_\omega$ or does it also enter the norm $L^q(\Omega)$? Typically weighted Poincaré inequalities are stated in $L^q(\omega)$. $\endgroup$
    – André Schlichting
    Commented Sep 19, 2023 at 9:50
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    $\begingroup$ @AndréSchlichting: No, there is no weight in the definition of $L^q$. I want to compare $f$ with its average value (w.r.t. $\omega$) in the standard $L^q$-norm. $\endgroup$
    – Romain Gicquaud
    Commented Sep 19, 2023 at 9:59
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    $\begingroup$ As $\Omega$ is convex and bounded and as we are in flat space, a good deal of the problem is about the regularity (in fact, the geometric structure) of its boundary, I think. Also, I think that $p$ can't be arbitrary, as I'm seeing the weight $\omega$ as a family of riemannian metrics on the boundary. A consequence of this later view entails the study of the weight at some "characteristic" inner point of $\Omega$. This make the link with capacity theory, and then, with quasiconformal geometry, and then... $\endgroup$
    – Eric Arnéo Vespira Kengne
    Commented Sep 19, 2023 at 10:26

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Such an inequality cannot exist. Take $\Omega=B_1(0)\subset \mathbb{R}^n$ and assume find a constant $C>0$ independent of $\omega$, then taking a sequence $(w_k)_k \subset L^p(B_1)$ weakly converging to a delta at $0$ you would prove that for every $f\in C^0(B_1)$ there holds $$ \|f-f(0) \|_{L^q(B_1)} \le C \|\nabla f \|_{L^q(B_1)} \,.$$ But this is not true whenever points have zero capacity (e.g. $p=q=2$ and $n\ge 2$), since it would be equivalent to a Poincaré inequality without any condition.

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    $\begingroup$ You can make your argument easier by using Minkowski's inequality to conclude that there exists a constant $C > 0$ such that $|f(0)| \leq \|f\|_{W^{1, q}}$. As the point $0$ has nothing special, you get that $W^{1, q}$ embeds into $L^\infty$. This rules out the possibility of choosing $q < n$. For $q > n$, the proof is fairly easy I guess using the embedding into $C^\gamma$. $\endgroup$
    – Romain Gicquaud
    Commented Sep 19, 2023 at 11:31

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