Let $d\geq 2$ and denote $B_r$ the open ball centered at $0$ of radius $r>0$.
1) Prove that $$\|u\|_{L^2(B_2\backslash B_1)}\leq 2^{(d-1)/2}\|\nabla u\|_{L^2(B_2\backslash B_1)}$$ for all $u\in H_0^1(B_2\backslash B_1)$
There is thing I don't understand in the following proof :
For any $x\in B_2 \setminus B_1$ denote $\bar x=\frac{2x}{|x|},$ thus $\bar x\in\partial B_2$ and $u(\bar x)=0$ for any $u\in H.$ Denoting $h(t)=u(tx+(1-t)\bar x)$ we have that $$u(x)=h(1)=\int_0^1 h'(t)dt+h(0)=\int_0^1(x-\bar x)\cdot\nabla u(tx+(1-t)\bar x)dt,$$ thus $$|u(x)|\leq\int_0^1|x-\bar x|\cdot|\nabla u(tx+(1-t)\bar x|dt\leq \int_0^1|\nabla u(tx+(1-t)\bar x|dt$$ $$\leq \int_0^1|\nabla u(\bar x-\frac{t}{2}\bar x)|dt$$
Question 1 : Why $\int_0^1|\nabla u(tx+(1-t)\bar x|dt\leq \int_0^1|\nabla u(\bar x-\frac{t}{2}\bar x)|dt$ ?
and by the Schwartz inequality we get $$|u(x)|^2\leq\int_0^1|\nabla u(\bar x-\frac{t}{2}\bar x)|^2dt.$$ We have now
\begin{align*} \int_{B_2\setminus B_1}u^2(x)dx\leq \int_{B_2\setminus B_1}\int_0^1|\nabla u((1-t/2)\bar x)|^2dtdx. \end{align*} Denoting $x=r\omega,$ where $r\in[1,2]$ and $\omega\in\mathbb S^{d-1}$ we have by polar coordinate integration $$\int_{B_2\setminus B_1}|\nabla u((1-t/2)\bar x)|^2dx=\int_1^2\int_{\partial B_1}r^{d-1}|\nabla u((2-t)\omega)|^2d\omega dr,$$
Question 2 : I don't understand this last equality, neither why $dx=r^{d-1}d\omega dr$.
thus by the Fubini theorem \begin{align*} \int_{B_2\setminus B_1}u^2(x)dx&\underset{(*)}{\leq} \int_1^2\int_1^2\int_{\partial B_1}r^{d-1}|\nabla u(s\omega)|^2d\omega drds \\ &\leq \int_1^2\int_1^2\int_{\partial B_1}r^{d-1}s^{d-1}|\nabla u(s\omega)|^2d\omega drds\\ &\leq 2^{d-1}\int_1^2\int_{\partial B_1}s^{d-1}|\nabla u(s\omega)|^2d\omega ds\\ &=2^{d-1}\int_{B_2\setminus B_1}|\nabla u(x)|^2dx. \end{align*}
Question 3 : Where come from the first inequality, i.e. $(*)$ ?