# doubts in the book of fanghua lin

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 $$\int_{B_\rho} | w - w(0)|^{2} \le \rho^{n+2} |Dw|_{L^{\infty}(B_{1/2}}^{2}$$ 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 $$\lambda \int_{B_r(x_0)} |Dv|^{2} \le \int_{B_r(x_0)} (| a_{ij}(x_0) -a_{ij}(x))D_i u D_j v| + \int_{B_r(x_0)} |fv|$$ Then, why $$\int_{B_r(x_0)} |Dv|^{2} \le C\left \{ \tau^{2}(r) \int_{B_r(x_0)} |Du|^{2} + \Bigl( \int_{B_r(x_0)} |f|^{2n/(n+2)} \Bigr) \right \}?$$ 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 $$|f|_{2n/(n+2)} = \Bigl( \int_{B_r(x_0} |f|^{2n/(n+2)}\Bigl)^{n+2/2n} \neq \Bigl( \int_{B_r(x_0} |f|^{2n/(n+2)}\Bigl)^{n+2/n}$$ Finally. Why do we have to use Sobolev Inequality of the form $$|v|_{2^*} \le C(n)|Dv|_{2}?$$

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In my opinion, on page 10 the big bracket $\{ \ldots \}$ should be $\{\ldots \}^{1/2}$. There is no reason why Cauchy-Schwartz should give $\tau(r)^2$. – Siminore Jun 27 '12 at 16:56
Thank you. I thought again and I think he used Hölder inequality after. – user29999 Jun 27 '12 at 17:03
The last ask is because $|DV|_2 < \infty$ because $v\in H^1$. – user29999 Jun 27 '12 at 17:06

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.
Ok. But, Let $f=1$ in [0,1] and $f=0$ em (1,2]. Has $f$ a Lipschitz representative in [1,2]? I think that $f\in W^{1,\infty}(B_{1/2}) \Leftrightarrow f \in Lip$ in $B_{1/2}$. Even so, why $$\int_{B_{\rho}} | w - w_{\rho}|^{2} \le \int_{B_{\rho}} | w - w(0)|^{2}?$$ If $w \in C^1$ I could use mean value theorem. Can I use a sort of mean vaue theorem here to justify? $$\int_{B_\rho} | w - w(0)|^{2} \le \rho^{n+2} |Dw|_{L^{\infty}(B_{1/2})}^{2}.$$ – user29999 Aug 2 '12 at 11:53
@user29999 (a) I fixed a typo in the second line, where it had $L^{\infty}$ instead of $W^{1,\infty}$. (b) The minimum of $\int_X |w-c|^2$ over all constants $c$ is attained when $c$ is the average of $w$ over $X$. (Just expand the square). (c) The $L^{\infty}$ norm of $|Dw|$ is the Lipschitz constant of $w$ on the ball; let's denote it by $L$. The inequality $|w(x)-w(0)|\le L|w|$ can be squared and integrated in polar coordinates, justifying the last line. See Lipschitz Analysis by Heinonen for more on Lipschitz functions. – user31373 Aug 2 '12 at 16:12
Very good I have only one question. \begin{eqnarray} F(c)=\int_X |w-c|^{2} &=& \int_X |w|^2 -2 \int_X w \cdot c +c^2 |X|.\\ \end{eqnarray} The minimum of $F$ is $\dfrac{4|X|\int_X |w|^2 - 4\Bigl(\int_Xw\Bigr)^2}{4|X|}$, isn't it? By the way, I started a new bounty. Maybe you can help me. The reference is in the comments. – user29999 Aug 2 '12 at 16:56
@user29999 $F'(c)=-2\int_Xw+2c|X|$, hence $F'(c)=0$ when $c=\frac{1}{|X|}\int_X w$. – user31373 Aug 2 '12 at 17:31