Cut-off function and interior holder estimate

Suppose we know the following theorem

$\bf{Thm 1:}$ Let $v\in C^2_0(\mathbb{R}^n)$ , $g\in C^\alpha_0(\mathbb{R}^n)$ satisfy the Possion equation $\Delta v=g$ in $\mathbb{R}^n$ . Then $v\in C^{2,\alpha}_0(\mathbb{R}^n)$ and if $\text{supp }g\subset B=B_{R}(x_0)$ , we have

$\displaystyle |D^2v|'_{0,\alpha;B}\leq C|g|'_{0,\alpha;B}$ , $C=C(n,\alpha)$
$|v|'_{1;B}\leq CR^2|g|_{0;B}$ , $C=C(n)$

Use the above theorem and cut-off function

$\eta(|x|)\in C^2(\mathbb{R}^n)$ ,

$\eta(|x|)=1$ when $\displaystyle |x|<\frac{3}{2}$ and $\eta(|x|)=0$ when $|x|>2$

to prove the interior holder estimate of possion equation

$\bf{Thm 2:}$ Let $\Omega$ be a domain in $\mathbb{R}^n$ , $\Delta u=f$ in $\Omega$ . If $u\in C^2(\Omega)$ and $f\in C^\alpha(\Omega)$ , then $u\in C^{2,\alpha}(\Omega)$ and for any two concentric balls $B_{R}\subset B_{2R}(x_0)\subset \Omega$ , we have

$\displaystyle |u|'_{2,\alpha;B_{R}}\leq C(|u|_{0;B_{2R}}+R^2|f|'_{0,\alpha;B_{2R}})\quad (1)$

$\bf{Proof:}$ WLOG assume $x_0=0$ , let $v(x)=u(x)\eta(|x|/R)$ , $\eta$ is the cut off fuction.

then $\displaystyle v_i(x)=u_i(x)\eta\left(\frac{|x|}{R}\right)+u(x)\eta'\left(\frac{|x|}{R}\right)\frac{x_i}{|x|}\frac 1R$ .

$\displaystyle v_{ii}(x)=u_{ii}(x)\eta\left(\frac{|x|}{R}\right)+2u_i(x)\eta'\left(\frac{|x|}{R}\right)\frac{x_i}{|x|}\frac 1R+u(x)\eta''\left(\frac{|x|}{R}\right)\frac{x^2_i}{|x|^2}\frac {1}{R^2}+u(x)\eta'\left(\frac{|x|}{R}\right)\frac 1R\frac{|x|^2-x^2_i}{|x|^3}$

$\displaystyle \Delta v(x)=\Delta u\cdot\eta\left(\frac{|x|}{R}\right)+2\nabla u\cdot\nabla |x|\eta'\left(\frac{|x|}{R}\right)\frac 1R+u(x)\eta''\left(\frac{|x|}{R}\right)\frac{1}{R^2}+u(x)\eta'\left(\frac{|x|}{R}\right)\frac{1}{R}\frac{n-1}{|x|}:=g$

Since $v$ has compact support, we have representation $\displaystyle v=\int_{B_{2R}}\Gamma(x-y)g(y)dy$

Let $\displaystyle w_1=\int_{B_{2R}}\Gamma(x-y)\Delta u\cdot\eta\left(\frac{|y|}{R}\right)dy$

$\displaystyle w_2=\int_{B_{2R}}\Gamma(x-y)\nabla u\cdot\nabla |y|\eta'\left(\frac{|y|}{R}\right)\frac 1Rdy$

$\displaystyle w_3=\int_{B_{2R}}\Gamma(x-y)u(y)\left[\eta''\left(\frac{|y|}{R}\right)\frac{1}{R^2}+\eta'\left(\frac{|y|}{R}\right)\frac{1}{R}\frac{n-1}{|y|}\right]dy$

Then $v=w_1+2w_2+w_3$ from the representation. We will bound them respectively.

For $w_1$ , note that $\displaystyle |w_1|_{0;B_{R}}\leq |u|_{0;B_{2R}}$ , and theorem 4.5 implies $\displaystyle R|Dw_1|_{0;B_{R}}+R^2|D^2w_1|_{0,\alpha;B_{R}}\leq CR^2|f\eta|'_{0,\alpha;B_{2R}}\leq CR^2|f|'_{0,\alpha;B_{2R}}$

So $w_1$ satisfies (1)

For $w_3$ , since $\eta'(x)=0$ when $\displaystyle |x|<\frac 32$ , we get that in fact

$\displaystyle w_3=\int_{B_{2R}\backslash B_{\frac 32 }R}\Gamma(x-y)u(y)\left[\eta''\left(\frac{|y|}{R}\right)\frac{1}{R^2}+\eta'\left(\frac{|y|}{R}\right)\frac{1}{R}\frac{n-1}{|y|}\right]dy$

Since we have $\displaystyle |x-y|\geq \frac 12 R$ when $x\in B_R$ and $\displaystyle y\in B_{2R}\backslash B_{\frac 32 }R$ , this means

$\displaystyle |w_3|\leq C\frac{|u|_{0;B_{2R}}}{R^2}\int_{B_{2R}\backslash B_{\frac 32 }R}\Gamma(x-y)dy\leq C|u|_{0;B_{2R}}$

$\displaystyle |Dw_3|'_{1,\alpha;B_{2R}}\leq C\frac{|u|_{0;B_{2R}}}{R^2}\int_{B_{2R}\backslash B_{\frac 32 }R}|D\Gamma(x-y)|'_{1,\alpha;B_{2R}}dy\leq C|u|_{0;B_{2R}}$

So $w_3$ satisfies (1)

For $w_2$ , by the same reason,

$\displaystyle w_2=\int_{B_{2R}\backslash B_{\frac 32 R}}\Gamma(x-y)\nabla u\cdot\nabla |y|\eta'\left(\frac{|y|}{R}\right)\frac 1Rdy$

Apply the Green identity on the domin $B_{2R}\backslash B_{\frac 32 R}$

$\displaystyle w_2+\int_{B_{2R}\backslash B_{\frac 32 R}}\nabla \Gamma(x-y)\cdot \nabla |y|u(y)\eta'\left(\frac{|y|}{R}\right)\frac 1Rdy+\int_{B_{2R}\backslash B_{\frac 32 R}}\Gamma(x-y)u(y)\eta''\left(\frac{|y|}{R}\right)\frac {1}{R^2}dy$

$\displaystyle +\int_{B_{2R}\backslash B_{\frac 32 R}}\Gamma(x-y)u(y)\eta'\left(\frac{|y|}{R}\right)\frac{n-1}{|y|}\frac 1Rdy=\left(\int_{\partial B_{2R}}-\int_{\partial B_{\frac 32R}}\right)\Gamma(x-y)u(y)\eta'\left(\frac{|y|}{R}\right)\frac 1Rds_y=0$

So we can estimate $w_2$ as $w_3$ , which means $w_2$ also satisfies (1).

$\bf{Remark:}$ Gilbarg Trudinger’s book. chapter 4, exercise 4.3.

lim Practice= Perfect