TITLE: Poincare-type inequality on ball.
QUESTION [5 upvotes]: Let $B_r$ be an open ball in $\Bbb R^d$ with radius $r$. Let $u\in W^{1,p}(B_r)$ and define
$
\bar u=\frac 1{|B_r|}\int u\ dx,
$
so that by Poincare inequality we have 
$$
||u-\bar u||_{L^p(B_r)} \le C||Du ||_{L^p(B_r)}\quad;\quad C=C(r,p).
$$ Actually, $C(r,p)=C(p)\cdot r$ by dilating $B_r$ to $B_1$ and change of variables.
Now, by Sobolev embedding we have $W^{1,p}(\Omega) \hookrightarrow L^q(\Omega)$ for $q\in [1,\frac{dp}{d-p}]$, i.e.
$$
||u||_{L^q} \le C'||u||_{W^{1,p}}\quad;\quad C'=C'(\Omega,p,q,d).
$$
Combining these we can easily get
$$
||u-\bar u||_{L^q(B_r)} \le C'' ||Du||_{L^p(B_r)}
$$
for some constant $C''$ depending on $r,p,q,d$.

However, I've seen a version that says
  $$
\left(\frac 1{|B_r|}\int |u-\bar u|^q \right)^{1/q} \le Cr \left(\frac 1{|B_r|}\int |Du|^p \right)^{1/p}
$$
  where $C=C(d,p,q)$ does not depend on $r$.

Is that an easy consequence of the above or does it require another approach entirely? I tried scaling argument but the term $||u||_p$ and $||Du||_p$ in $||u||_{W^{1,p}}$ scale differently. I seem to somewhat stuck, could anyone please help?

REPLY [2 votes]: The inequality you want is only true for $q=\frac{dp}{d-p}$. 
Assume first that $g\in W^{1,p}(B_1)$. Then you can extend $g$ to a function $G\in W^{1,p}(\mathbb{R}^d)$ with 
$$\Vert G\Vert_{W^{1,p}(\mathbb{R}^d)}\le c \Vert g\Vert_{W^{1,p}(B_1)}.$$
Next apply Sobolev embedding to find
$$\Vert g\Vert_{L^{p^*}(B_1)}\le\Vert G\Vert_{L^{p^*}(\mathbb{R}^d)}\le c \Vert \nabla G\Vert_{L^{p}(\mathbb{R}^d)}\le C(\Vert \nabla g\Vert_{L^{p}(B_1)}+\Vert g\Vert_{L^{p}(B_1)}).$$
Now you apply this inequality to $u-\bar u$ to get
$$\Vert u-\bar u\Vert_{L^{p^*}(B_1)}\le C(\Vert \nabla u\Vert_{L^{p}(B_1)}+\Vert u-\bar u\Vert_{L^{p}(B_1)})
\le C(\Vert \nabla u\Vert_{L^{p}(B_1)},$$
where in the last inequality you apply Poincare's inequality (the one you wrote at the beginning).
Now you do a rescaling and use the fact that $p^*=\frac{dp}{d-p}$.