doubts in the book of fanghua lin I'd like to ask questions about some steps in the following article, book.


*

*In the lemma 2.1
$$
\int_{B_{\rho}} | w - w_{\rho}|^{2} \le \int_{B_{\rho}} | w - w(0)|^{2}
$$
and 
\begin{equation}
\int_{B_\rho} | w - w(0)|^{2} \le \rho^{n+2} |Dw|_{L^{\infty}(B_{1/2}}^{2}
\end{equation}
As $w \in H^1(B_r), w$ could be redefined in $ 0$, could not?


Incidentally,  is there a kind of inequality of the mean value for $w \in H^{1}(B_r)$?
2 In the theorem 2.4 we have
\begin{equation}
\lambda \int_{B_r(x_0)} |Dv|^{2} \le \int_{B_r(x_0)} (| a_{ij}(x_0) -a_{ij}(x))D_i u D_j v| + \int_{B_r(x_0)} |fv|
\end{equation}
Then, why
\begin{equation}
\int_{B_r(x_0)} |Dv|^{2} \le C\left \{ \tau^{2}(r) \int_{B_r(x_0)} |Du|^{2} + \Bigl( \int_{B_r(x_0)} |f|^{2n/(n+2)} \Bigr) \right \}?
\end{equation}
I know that
\begin{eqnarray}
\int_{B_r(x_0)} |Dv|^{2} \le \int_{B_r(x_0)} (| a_{ij}(x_0) -a_{ij}(x))D_i u D_j v| 
& \le &  \int_{B_r(x_0)} \tau(r) |Du| |Dv|
\end{eqnarray}
Can I use Young's inequality whit $\varepsilon$ here?
And I know that
\begin{eqnarray}
\int_{B_r(x_0)} |fv| \le |v|_{2^{*}} |f|_{2^*/((2^* -1)} = |v|_{2^*} |f|_{2n/(n+2)}
\end{eqnarray}
But
\begin{equation}
|f|_{2n/(n+2)} = \Bigl( \int_{B_r(x_0} |f|^{2n/(n+2)}\Bigl)^{n+2/2n} \neq \Bigl( \int_{B_r(x_0} |f|^{2n/(n+2)}\Bigl)^{n+2/n}
\end{equation}
Finally. Why do we have to use Sobolev Inequality of the form
\begin{equation}
|v|_{2^*} \le C(n)|Dv|_{2}?
\end{equation}
 A: 1) The estimates involving $w(0)$ are based on the claim which includes, in particular, $\|Dw\|_{L^{\infty}(B_{1/2})}<\infty$. A function in $W^{1,\infty}$ has a Lipschitz representative, with which it is identified without hesitation. 
2) The author wants the right hand side to have a multiple of $\left(\int |Dv|^2\right)^{1/2}$, so that he can absorb it on the left by division: he will divide both sides by this square root and then square both sides. Sneaky! So it goes: $\tau(r)\int |Du||Dv|\le \tau(r) \|Du\|_2 \|Dv\|_2$ and $$\int |fv|\le \|f\|_{2n/(n+2)}\|v\|_{2n/(n-2)}\le  C\|f\|_{2n/(n+2)}\|Dv\|_{2}$$ Notice that the Hölder exponents were chosen precisely so that the subsequent Sobolev inequality gives $\|Dv\|_2$. 
Now cancel $\|Dv\|_2$ on both sides (watch out: we must know that it's finite! which it is.)
$$
\left(\int |Dv|^2\right)^{1/2} \le \left\{\tau(r) \|Du\|_2+ C\|f\|_{2n/(n+2)}\right\}
$$
Then square both sides. Since we are PDE people, $(a+b)^2$ is the same as $a^2+b^2$ for us (they agree up to the factor of $2$). The result is 
$$
\int |Dv|^2  \le C\left\{\tau^2(r) \|Du\|_2^2+ \|f\|_{2n/(n+2)}^2\right\}
$$
as claimed. 
