# Ngô Quốc Anh

## August 29, 2010

### Achieving regularity results via bootstrap argument, 4

Filed under: Giải Tích 6 (MA5205), PDEs — Tags: — Ngô Quốc Anh @ 2:36

Let us consider the following equation

$\displaystyle u(x) = \int_{{\mathbb{R}^n}} {\frac{{u{{(y)}^{\frac{{n + \alpha }}{{n - \alpha }}}}}}{{{{\left| {x - y} \right|}^{n - \alpha }}}}dy} , \quad x \in {\mathbb{R}^n}$

for $n\geqslant 1$ and $0<\alpha. In this entry, by using boothstrap argument, we show that

Theorem. If positive function $u \in L_{loc}^\frac{2n}{n-\alpha}(\mathbb R^n)$ solves the equation, then $u \in C^\infty(\mathbb R^n)$.

In the process of proving the result, we need the following result

Proposition. Let $V \in L^\frac{n}{\alpha}(B_3)$ be a non-negative function and set

$\displaystyle \delta(V)=\|V\|_{L^\frac{n}{\alpha}(B_3)}$.

For $\nu >r>\frac{n}{n-\alpha}$, there exist positive constants $\overline \delta<1$ and $C \geqslant 1$ depending only on $n, \alpha, r$ and $\nu$ such that for any $0 \leqslant V \in L^\frac{n}{\alpha}(B_3)$ with $\delta(V) \leqslant \overline \delta$, $h \in L^\nu(B_2)$ and $0 \leqslant u \in L^r(B_3)$ satisfying

$\displaystyle u(x) \leqslant \int_{{B_3}} {\frac{{V(y)u(y)}}{{{{\left| {x - y} \right|}^{n - \alpha }}}}dy} + h(x), \quad x \in {B_2}$

we have

$\displaystyle {\left\| u \right\|_{{L^\nu }({B_{1/2}})}} \leqslant C\left( {{{\left\| u \right\|}_{{L^r}({B_3})}} + {{\left\| h \right\|}_{{L^\nu }({B_2})}}} \right)$.

As a consequence, we have

Corollary. For $n \geqslant 1$, $0<\alpha, $\nu >r>\frac{n}{n-\alpha}$ and $R_2>R_1>0$, let $0 \leqslant V \in L^\frac{n}{\alpha}(B_{R_2})$, $h\in L^\nu(B_{R_1})$ and $0 \leqslant u \in L^r(B_{R_2})$ satisfy

$\displaystyle u(x) \leqslant \int_{B_{R_2}} {\frac{{V(y)u(y)}}{{{{\left| {x - y} \right|}^{n - \alpha }}}}dy} + h(x), \quad x \in B_{R_1}$.

Then for some $\varepsilon>0$, $u \in L^\nu(B_\varepsilon)$.

Proof. For some $|\overline x| <1$ we have, by the PDE,

$\displaystyle\int_{|y| > 2} {\frac{{u{{(y)}^{\frac{{n + \alpha }}{{n - \alpha }}}}}}{{{{\left| y \right|}^{n - \alpha }}}}dy} \leqslant C\int_{|y| > 2} {\frac{{u{{(y)}^{\frac{{n + \alpha }}{{n - \alpha }}}}}}{{{{\left| {\overline x - y} \right|}^{n - \alpha }}}}dy} \leqslant \int_{{\mathbb{R}^n}} {\frac{{u{{(y)}^{\frac{{n + \alpha }}{{n - \alpha }}}}}}{{{{\left| {\overline x - y} \right|}^{n - \alpha }}}}dy} = u(\overline x ) < \infty$.

For any $R>0$, we write

$\displaystyle u(x) = \underbrace {\int_{|y| \leqslant 2R} {\frac{{u{{(y)}^{\frac{{n + \alpha }}{{n - \alpha }}}}}}{{{{\left| {x - y} \right|}^{n - \alpha }}}}dy} }_{{I_R}(x)} + \underbrace {\int_{|y| > 2R} {\frac{{u{{(y)}^{\frac{{n + \alpha }}{{n - \alpha }}}}}}{{{{\left| {x - y} \right|}^{n - \alpha }}}}dy} }_{I{I_R}(x)}$.

Take

$\displaystyle V(x) = u{(x)^{\frac{{2\alpha }}{{n - \alpha }}}}$

and

$\displaystyle h(x) = \int_{|y| > 2R} {\frac{{u{{(y)}^{\frac{{n + \alpha }}{{n - \alpha }}}}}}{{{{\left| {x - y} \right|}^{n - \alpha }}}}dy}$.

Since

$\displaystyle u \in L_{loc}^\frac{2n}{n-\alpha}(\mathbb R^n)$

by the Holder inequality

$\displaystyle V \in L_{loc}^\frac{n}{\alpha}(\mathbb R^n)$.

Obviously $h \in L^\infty(B_R)$. For any $\nu > \frac{n}{n-\alpha}$, we have by Corollary $u \in L^\nu(B_{\varepsilon(\nu)})$ for some $\varepsilon(\nu)>0$. Since any point can be taken as the origin, we have proved that $u \in L_{loc}^\nu(\mathbb R^n)$ for all $\frac{n}{n-\alpha}<\nu<\infty$.

By the Holder inequality $I_R \in L^\infty(B_R)$. Note that we can differentiate $II_R$ under the integral sign for $|x| so

$II_R \in C^\infty(\mathbb R^n)$.

Therefore, the remaining task is to prove

$I_R \in C^\infty(\mathbb R^n)$.

Clearly, $u \in L^\infty(B_R)$. Since $R$ is arbitrary, $u \in L^\infty_{loc}(\mathbb R^n)$. Back to the decomposition $u=I_R+II_R$ by using the regularity theory for integral equations we see that $u$ is at least Holder continuous in $B_R$. Since $R>0$ is arbitrary, $u$ is at least Holder continuous in $\mathbb R^n$.

It now follows from this entry that $u^\frac{n+\alpha}{n-\alpha}$ is $\frac{2\alpha}{n-\alpha}$-Holder continuous in $B_{2R}$. Note that since $u$ is locally Holder continuous in $\mathbb R^n$, $u$ is bounded from above. Since we have no strictly positive lower bound, we find that $u$ cannot be $\beta$-Holder continuous for arbitrary $\beta$. Thanks to a corollary in this entry we find that $u^\frac{n+\alpha}{n-\alpha}$ is $\frac{2\alpha}{n-\alpha}$-Holder continuous in $B_{2R}$. Now a regularity result tells us that

$I_R \in C^{2,\frac{2\alpha}{n-\alpha}}_{loc}(\mathbb R^n)$

which implies by the decomposition

$u \in C^{2,\frac{2\alpha}{n-\alpha}}_{loc}(\mathbb R^n)$.

In other words, the regularity of $u$ further improves. We now differentiate $I_R$ to further improve the regularity of $u$. We have

$\displaystyle {D_i}{I_R}(x) = \int_{|y| \leqslant 2R} {\frac{1}{{{{\left| {x - y} \right|}^{n - \alpha }}}}\left[ {\frac{{{x_i} - {y_i}}}{{{{\left| {x - y} \right|}^2}}}u{{(y)}^{\frac{{n + \alpha }}{{n - \alpha }}}}} \right]dy}$.

Eventually we have $u \in C^\infty(\mathbb R^n)$ via a bootstrap argument.