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Let $\Omega$ be a ball of radius $r$. It is well known that given a function $f \in W^{1,p} (\Omega)$, it holds the Poincaré-Wirtinger inequality

$$ \left(\int_\Omega |f - f_\Omega|^p dx \right)^{1/p} \lesssim r \left(\int_\Omega |\nabla f|^p dx \right)^{1/p} ,$$

and actually on the right one could go up to $p^*$, where $f_\Omega$ is the mean of $f$ in $\Omega$. Let now $\phi$ be a non negative test function supported in $\Omega$ with $\int_\Omega \phi =1$. Interpreting $ \phi$ as a probability density function, I could define

$$ f_{\phi, \Omega} = \int_\Omega f(x) \phi(x) dx. $$

In this case, the mean $f_\Omega$ would correspond to the choice $\phi= \mathbb{1}_{\Omega}/|\Omega|$. My question is if there is a more general Poincaré type estimate in this case, so something like:

$$ \left(\int_\Omega |f - f_{\phi,\Omega}|^p dx \right)^{1/p} \leq C(\phi) \left(\int_\Omega |\nabla f|^p dx \right)^{1/p} .$$

EDIT: There is a very detailed analysis of this problem in Ziemer's "Weakly differentiable functions". However, I actually got carried away and asked a more general question than what I really need, for which I hope there is a more precise, quantitative answer, maybe somewhere else. I wonder, in particular, what the constant would be if $\Omega= B_R $ and $\phi= \mathbb{1}_{B_r} /|B_r|$ with $r<R$. I.e. when I integrate the difference of $f$ and the mean of $f$ on a ball of radius $r$ over a ball of different (for instance, bigger) radius. The precise constant would be nice, but knowing the scaling with respect to $R$ and $r$ would be sufficient. One should get something between the regular Poincaré-Wirtinger inequality when $r=R$ and when $r\to 0$ a bound for $\left(\int_{B_R} |f - f(0)|^p \right)^{1/p} $, which should be treatable only when $p>d$, i.e. $f$ is Holder.

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    $\begingroup$ Yes, this is e.g. included in Theorem 8.11 of Lieb-Loss Analysis. $\endgroup$
    – unwissen
    Commented Jul 7 at 21:59
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    $\begingroup$ It also follows from Theorem 4.2.1 in Ziemer's Weakly differentiable functions which has a whole (and very nice) chapter dedicated to Poincare-type inequalities. (Maybe it is the same theorem as in Lieb-Loss, which I do not have access to right now, if so, then disregard this comment.) $\endgroup$
    – Hannes
    Commented Jul 8 at 11:42
  • $\begingroup$ @Hannes unwissen thanks. Especially Ziemer's book goes in great detail, in a very general setting. Actually, I edited the question since I got carried away asking a question way more general than what I really need. Hope that in the simplified case presented there is a more explicit answer. It might be possible to recover it by going through Ziemer's book, but I am not 100% sure $\endgroup$
    – tommy1996q
    Commented Jul 9 at 10:24

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