In Evans PDE book there is the following theorem:
(Poincaré's inequality for a ball). Assume $1 \leq p \leq \infty .$ Then there exists a constant $C,$ depending only on $n$ and $p,$ such that $$ \left\|u-(u)_{x, r}\right\|_{L^{p}(B(x, r))} \leq C r\|D u\|_{L^{p}(B(x, r))} $$ for each ball $B(x, r) \subset \mathbb{R}^{n}$ and each function $u \in W^{1, p}\left(B^{0}(x, r)\right)$
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The proposed proof is the following: The case $U=B^{0}(0,1)$ follows from the previous Poincaré inequality. In general, if $u \in W^{1, p}\left(B^{0}(x, r)\right)$ write $$ v(y):=u(x+r y) \quad(y \in B^0(0,1)) $$ Then $v \in W^{1, p}\left(B^{0}(0,1)\right),$ and we have $$ \left\|v-(v)_{0,1}\right\|_{L^{p}(B(0,1))} \leq C\|D v\|_{L^{p}(B(0,1))} $$ Changing variables, we recover our result.
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My question is if I am doing the changing variables correctly: $$\left\|v-(v)_{0,1}\right\|_{L^{p}(B(0,1))} \leq C\|D v\|_{L^{p}(B(0,1))}$$ hence by definition we have $$\left\|u(x+ ry)-(u)_{0,1}\right\|_{L^{p}(B(0,1))} \leq C\|D u(x+ry)\|_{L^{p}(B(0,1))}$$ now I change variable $z=x+ ry$ and since $y \in B(0,1)$ then $z \in B(x,r)$ and in the formula by the change of variable in $\mathbb{R}^n$ we get the $det(J(z))= r^n$ and so we get on the right hand side:
$$r^n \left\|u(z)-(u)_{x,r}\right\|_{L^{p}(B(x,r))} $$ but I cannot work the LHS $$ C\|D u(x+ry)\|_{L^{p}(B(0,1))} = C \left(\int_{B(x,r)}(D u(x+ry))^p\right)^{\frac{1}{p}}$$
How do I change variable here? Can I apply some chain rule even if we are taking the weak derivative? Hwere do I put the determinant of the jacobian? I am a bit stuck. Thank you so much for your time!
The change of variable is not correct. Just write $z =x +ry$, where $x,r$ are fixed. Hence, the Jacobian becomes $r^{-n}$ and the term $r^{-n}$ cancels out on the right and left side. The term In the derivative. Do not substitute, but use the chain rule so that $Du(z)= r Du(x+ry)$ and you get the dependence on the radius $r$ on the left hand side.