# 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.

## 2 Comments »

1. Hôm nay có thời gian lên mạng. Rất vui vì nhìn thấy QA vẫn cập nhật tin tức hằng ngày lên đây. Khi nào cưới thông báo nhé. Chúc vui khỏe.

Comment by Lê Minh Đức — September 1, 2010 @ 2:09

• Cảm ơn sếp, nhất định là sẽ thông báo cho mọi người rồi 😀

Comment by Ngô Quốc Anh — September 1, 2010 @ 2:11

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