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

  1. 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}

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    The last ask is because |DV|_2 < \infty because $v\in H^1$.2012-06-27

1 Answers 1

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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.

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    With respct to answer 1), I agree that $W^{1,\infty} $ has a lipschtz representativ. However, the solution is, apriori, only $H^1=W^{1,2}$. So, Is the Fanghua's argument correct?2018-06-19