用户: Yao/Fourier 分析习题选解

推广形式的 Hardy-Littlewood-Sobolev 不等式:

设 $(X,\mu )$, $(Y,\nu )$ 为两个测度空间, $1<q<\infty$, $K(x,y)$ 为 $X\times Y$ 上的可测函数, 且存在 $c>0$ s.t.$$||K(x,\cdot )|{|}_{L^{q,\infty }(Y,\nu )}\le c,\text{a.e.}x\in X,$$$$||K(\cdot ,y)|{|}_{L^{q,\infty }(X,\mu )}\le c,\text{a.e.}y\in Y.$$若 $f\in L^p(Y,\nu )$ ($1\le p<+\infty )$, 则$$Tf(x)={\int }_{Y}K(x,y)f(y)d\nu (y)存在\text{a.e.}x\in X,$$且 $1<p<\infty$ 且 $1+\frac{1}{r}=\frac{1}{p}+\frac{1}{q}$, $\exists C(p,q,r,c)$ s.t.$$||Tf|{|}_{L^{r,\infty }(X,\mu )}\le C(p,q,r,c)||f|{|}_{L^p(Y,\nu )},\forall f\in L^p(Y,\nu ).$$

 
证明:

分解 $K=K_1+K_2$, $K_1=K{\chi }_{K>\beta }$, $K_2=K{\chi }_{K\le \beta }$, 其中 $\beta >0$ 待定.

首先, 估计$$||K_1(x,\cdot )|{|}_{L^1(Y,\nu )}={\int }_{\beta }^{\infty }{d}_{K_1(x,\cdot )}(\alpha )d\alpha \le {\int }_{\beta }^{\infty }{c}^q{\alpha }^{-q}d\alpha =\frac{c^q}{q-1}{\beta }^{1-q},$$同理有$$||K_1(\cdot ,y)|{|}_{L^1(X,\mu )}\le \frac{c^q}{q-1}{\beta }^{1-q},$$于是$$\begin{array}{rl}{\int }_Y{K}_1(x,y)f(y)d\nu & ={\int }_Y{K}_1(x,y{)}^{\frac{1}{p^{\prime }}}(K_1(x,y{)}^{\frac{1}{p}}f(y))d\nu 其中\frac{1}{p}+\frac{1}{p^{\prime }}=1\\ & \le ({\int }_Y{K}_1(x,y)d\nu {)}^{\frac{1}{p^{\prime }}}({\int }_Y{K}_1(x,y)f(y{)}^{p}d\nu {)}^{\frac{1}{p}}\\ & \le \left(\frac{c^q}{q-1}{\beta }^{1-q}{)}^{\frac{1}{p^{\prime }}}\right({\int }_Y{K}_1(x,y)f(y{)}^{p}d\nu {)}^{\frac{1}{p}},\end{array}$$由此得到$$\begin{array}{rl}||{\int }_Y{K}_1(x,y)f(y)d\nu |{|}_{L^p(X,\mu )}& \le \left(\frac{c^q}{q-1}{\beta }^{1-q}{)}^{\frac{1}{p^{\prime }}}\right({\int }_X{\int }_Y{K}_1(x,y)f(y{)}^{p}d\nu d\mu {)}^{\frac{1}{p}}\\ & \le \left(\frac{c^q}{q-1}{\beta }^{1-q}{)}^{\frac{1}{p^{\prime }}}\right({\int }_Y{\int }_X(K_1(x,y))d\mu f(y{)}^{p}d\nu {)}^{\frac{1}{p}}\\ & \le \frac{c^q}{q-1}{\beta }^{1-q}||f|{|}_{L^p(Y,\nu )}因\frac{1}{p}+\frac{1}{p^{\prime }}=1\stackrel{◯}{1}.\end{array}$$其次, 估计$$||K_2(x,\cdot )|{|}_{L^{p^{\prime }}(Y,\nu )}^{p^{\prime }}=p^{\prime }{\int }_0^{\beta }{\alpha }^{p^{\prime }-1}{d}_{K_2(x,\cdot )}(\alpha )d\alpha \le p^{\prime }{\int }_0^{\beta }{\alpha }^{p^{\prime }-1-q}{c}^{q}d\alpha =\frac{p^{\prime }{c}^q}{p^{\prime }-q}{\beta }^{p^{\prime }-q},$$得到$$||{\int }_Y{K}_1(x,y)f(y)d\nu |{|}_{L^{\infty }(X,\mu )}\le (\frac{p^{\prime }{c}^q}{p^{\prime }-q}{)}^{\frac{1}{p^{\prime }}{\beta }^{p^{\prime }-q}||f|{|}_{L^p}}\stackrel{◯}{2}.$$现在使用 $\stackrel{◯}{1},\stackrel{◯}{2}$ 来估计 $d_{\alpha }(Tf{)}^{\frac{1}{r}}$.

由 $Tf(x)={\int }_Y{K}_1(x,y)f(y)d\nu +{\int }_Y{K}_2(x,y)f(y)d\nu$ 得到$$d_{\alpha }(Tf)\le d_{\alpha /2}({\int }_Y{K}_1(x,y)f(y)d\nu )+d_{\alpha /2}({\int }_Y{K}_2(x,y)f(y)d\nu ).$$取 $\beta$ s.t. $(\frac{p^{\prime }{c}^q}{p^{\prime }-q}{\beta }^{p^{\prime }-q}||f|{|}_{L^p}{)}^{\frac{1}{p^{\prime }}}||f|{|}_{L^p}=\frac{\alpha }{2}$, 则由 $\stackrel{◯}{2}$ 得到 $d_{\alpha /2}({\int }_Y{K}_2(x,y)f(y)d\nu )=0$, 于是$$\begin{array}{rl}{d}_{\alpha }(Tf)& \le d_{\alpha /2}({\int }_Y{K}_1(x,y)f(y)d\nu )\\ & \le (\frac{\alpha }{2}{)}^{-p}||{\int }_Y{K}_1(x,y)f(y)d\nu |{|}_{L^p(X,\mu )}^p\\ & \le \left(\frac{\alpha }{2}{)}^{-p}\right(\frac{c^q}{q-1}{\beta }^{1-q}{)}^p||f|{|}_{L^p}^p由\stackrel{◯}{1}\\ & =C(p,p^{\prime },q,c){\alpha }^{-p}(\alpha ||f|{|}_{L^p}^{-1}{)}^{\frac{(1-q)p{p}^{\prime }}{p-q}}||f|{|}_{L^p}^p\\ & =C(p,p^{\prime },q,c)({\alpha }^{-1}||f|{|}_{L^p}{)}^r因-p+\frac{(1-q)p{p}^{\prime }}{p-q}=-r,\end{array}$$即 $||Tf|{|}_{L^{r,\infty }(X,\mu )}\le C(p,q,r,c)||f|{|}_{L^p(Y,\nu )}.$