## Tag Archives: holder

### Compactness between holder space

Let $\Omega$ is a domain in $\mathbb{R}^n$, $u$ is function on $\Omega$, $\alpha\in[0,1]$

$\displaystyle [u]_{\alpha;\Omega}=\sup\limits_{x,y}\frac{u(x)-u(y)}{|x-y|^\alpha}$ if $\alpha>0$

$\displaystyle [u]_{\alpha;\Omega}=\sup\limits_{x}|u(x)|$ if $\alpha=0$

And for $k=0,1,2,\cdots$

$[u]_{0,\alpha;\Omega}=[u]_{\alpha;\Omega}$,

$[u]_{k,\alpha;\Omega}=\sup\limits_{|\beta|=k}\sup\limits_{\Omega}[D^\beta u]_{\alpha;\Omega}$

Then we can define a norm

$\displaystyle ||u||_{k,\alpha;\Omega}=\sum\limits_{j=0}^{k}[u]_{k,0;\Omega}+[ u]_{k,\alpha;\Omega}$

So we have generalized holder space $C^{k,\alpha}(\overline{\Omega})=\{u|||u||_{k,\alpha;\Omega}<\infty\}$. In arbitrary domain $\Omega$, we don’t have the inclusion $C^{k,\alpha}(\overline{\Omega})\subset C^{j,\beta}(\overline{\Omega})$ when $j+\beta. But for $C^{0,1}$ domain, we have

$\mathbf{Thm:}$ Suppose $j+\beta, where $j=0,1,2,\cdots,$, $k=1,2,\cdots$ and $0\leq \alpha,\beta\leq 1$. Let $\Omega$ be a $C^{0,1}$ domain and $u\in C^{k,\alpha}(\overline{\Omega})$. Then for any $\epsilon >0$ and some constant $C=C(\epsilon,j,k,\Omega)$, we have

$\displaystyle |u|_{j,\beta;\Omega}\leq C|u|_{0;\Omega}+\epsilon |u|_{k,\alpha;\Omega}$

From this interpolation result, we can prove the compactness between holder space

$\mathbf{Thm:}$ Suppose $j+\beta with $k\geq 1$. $\Omega$ as before. The  inclusion $C^{k,\alpha}(\overline{\Omega})\hookrightarrow C^{j,\beta}(\overline{\Omega})$ is compact

$\mathbf{Remark:}$GT p137

### 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

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

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.