还是先来了解概念上的东西.

先回忆一下卷积的概念:$f*g=\int_{E}{f(x-y)g(y)dy}$,剩下的内容中只引入一个新概念.

设$K(x)$是定义在$R^n$上的函数,$\epsilon>0$,令

\[K_{\epsilon}(x)=\epsilon^{-n}K(\frac{x}{\epsilon})=\epsilon^{-n}K(\frac{x_1}{\epsilon},\frac{x_2}{\epsilon},\cdots,\frac{x_n}{\epsilon}),\]

称$K_{\epsilon}(x)$为$K(x)$的展缩函数.

下面研究的是$L^p(E)$空间的一些性质.

1. 关于$L^p$的范数$\Vert\cdot\Vert$.

若$f \in L^p(E)$,($1 \le p < \infty$),则存在$g \in L^{p’}(E)$,且$\Vert{g}\Vert_{p’}=1$,使得

\[\Vert{f}\Vert_p=\int_{E}{f(x)g(x)dx},\]

注意这里$p’$与$p$是共轭指数的关系:$\frac{1}{p} + \frac{1}{p’} = 1$.

根据Holder不等式:

\[|\int_{E}{f(x)g(x)dx}| \le \Vert{f}\Vert_p\Vert{g}\Vert_{p’}\]

证明是使用了构造法,直接构造出了$g(x)$.

当$p=1$,取$g(x)=\text{sign}{f(x)}$,当$1<p<\infty$,设$\Vert{f}\Vert_p \neq 0$,$g(x)=(\frac{|f(x)|}{\Vert{f}\Vert_p})^{p-1}\cdot\text{sign}{f(x)}$.

对于$p=\infty$,有下列结论:

\[\Vert{f}\Vert_{\infty}=\sup_{\Vert{g}\Vert_1=1}{\{|\int_{E}{f(x)g(x)dx}|\}},\]

书中的证明充分利用$\Vert{f}\Vert_{\infty}$的定义,注意$g(x)$的构造

\[g(x)=\frac{1}{a}\chi_{A}(x)\text{sign}{f(x)}.\]

另外,当$p=\infty$时,$f \in L^p(E)$,可能不存在$g \in L^1(E)$,$\Vert{g}\Vert_1=1$,使得$\Vert{f}\Vert_{\infty}=\int_{E}{f(x)g(x)dx}$.书中给出了一个例子.$f(x)=x$,$x \in [0,1]$.

2. Holder不等式的逆命题

设$g(x)$是$E$上的可测函数,若存在$M>0$,使得对一切在$E$上可积的简单函数$\varphi(x)$,都有

\[|\int_{E}{g(x)\varphi(x)dx}| \le M\Vert{\varphi}\Vert_p,\]

则$g \in L^{p’}(E)$,$p’$是$p$的共轭指标,且$\Vert{g}\Vert_{p’} \le M$.

$g \in L^{p’}(E)$,即要证明$|g(x)|^{p’}$可积.

令$\{\varphi_k(x)\}$逼近$|g(x)|^{p’}$,$\varphi_k(x)$单调上升.

\[\begin{gather*}
\psi_k(x)=[\varphi_k(x)]^{1/p}\text{sign}{g(x)},\\
\Vert{\psi_k}\Vert_p=[\int_{E}{\varphi_k(x)dx}]^{1/p} \\
0 \le \varphi_k(x) = [\varphi_k(x)]^{1/p}[\varphi_k(x)]^{1/p’}\le \psi_k(x)g(x) \\
\int_{E}{\varphi_k(x)dx} \le \int_{E}{\psi_k(x)g(x)dx} \le M\Vert{\psi_k}\Vert_p
\end{gather*}\]

由此得到:$\int_{E}{\varphi_k(x)dx} \le M^{p’}$,令$k \rightarrow \infty$,$\int_{E}{|g(x)|^{p’}dx} \le M^{p’}$,可积.

对于$p=1$,书中采取了反证法,那么$g*L^{\infty}(E)$意味着什么呢?

存在$E$中的可测集列$\{A_k\}$,$\infty > m(A_k) > 0$,使得$g(x) \ge k$,$x \in A_k$, $k=1,2,\cdots$.$\varphi_k(x)=\chi_{A_k}(x)$,则

\[\frac{\int{\varphi_k(x)g(x)dx}}{\Vert{\varphi_k}\Vert_1} \ge \frac{km(A_k)}{m(A_k)}=k,\]

矛盾.

3. 关于卷积$f*g$,我们已经证明:

\[\Vert{f*g}\Vert_1 \le \Vert{f}\Vert_1\Vert{g}\Vert_1,\]

即书中的定理4.31.

\[\int_{R^n}{|(f*g)(x)|dx} \le \int_{R^n}{|f(x)|dx}\int_{R^n}{|g(x)|dx}.\]

对于$L^p(R^n)$,我们有类似的Young不等式.

设$f \in L^1(R^n)$,$g \in L^p(R^n)$,($1 < p < \infty$),则$\Vert{f \* g}\Vert_p \le \Vert{f}\Vert_1\Vert{g}\Vert_p$.

证明过程先用Holder不等式,后用Fubini定理.

\[\begin{aligned}
|(f*g)(x)| &\le \int_{R^n}{|f(x-y)||g(y)|dy}\\
&=\int_{R^n}{|f(x-y)|^{1/p}|g(y)||f(x-y)|^{1/p’}dy}\\
&\le [\int_{R^n}{|f(x-y)||g(y)|^pdy}]^{1/p}[\int_{R^n}{|f(x-y)|dy}]^{1/p’}\\
\int_{R^n}{|(f*g)(x)|^pdx} &\le \int_{R^n}{(\int_{R^n}{|f(x-y)||g(y)|^pdy})(\int_{R^n}{|f(x-y)|dy})^{p/p’}dx} \\
&= \Vert{f}\Vert_1^{p/p’}\int_{R^n}{[\int_{R^n}{|f(x-y)||g(y)|^pdy}]dx} \\
&= \Vert{f}\Vert_1^{p/p’}\int_{R^n}{|g(y)|^p[\int_{R^n}{|f(x-y)|dx}]dy} \\
&= \Vert{f}\Vert_1^{p/p’+1}\int_{R^n}{|g(y)|^pdy}=\Vert{f}\Vert_1^p\Vert{g}\Vert_p^p
\end{aligned}\]

这里主要应注意$\int_{R^n}{|f(x-y)|dy}=\Vert{f}\Vert_1$,即

\[\int_{R^n}{|f(x-y)|dy} =\int_{R^n}{||f(y)dy}.\]

变量替换.

这说明$f*g$是属于$L^p(R^n)$的.

4. 广义Minkowski不等式:设$f(x,y)$是$R^n \times R^n$上的可测函数,若对几乎处处的$y \in R^n$,$f(x,y)$属于$L^p(R^n)$($1 \le p < \infty$),且有$\int_{R^n}{[\int_{R^n}{|f(x,y)|^pdx}]^{1/p}dy}=M<\infty$,则

\[[\int_{R^n}{|\int_{R^n}{f(x,y)dy}|^pdx}]^{1/p} \le \int_{R^n}{[\int_{R^n}{|f(x,y)|^pdx}]^{1/p}dy}.\]

证明过程中使用了前面的结论,对于$p=1$它是显然的.

$F(x)=\int_{R^n}{f(x,y)dy}$,则左边即为$\Vert{F}\Vert_p$,右边为$M$.

对任意的简单可积函数$\varphi(x)$有

\[\begin{aligned}
|\int_{R^n}{F(x)\varphi(x)dx}| &\le \int_{R^n}{|F(x)\varphi(x)|dx}\\
&\le \int_{R^n}{[\int_{R^n}{|f(x,y))||\varphi(x)|dx}]dy}\\
&\le\int_{R^n}{[\int_{R^n}{|f(x,y)|^pdx}]^{1/p}[\int_{R^n}{|\varphi(x)|^{p’}}]^{1/p’}dy} \\
&=M\Vert{\varphi}\Vert_{p’}
\end{aligned}\]

由此$\Vert{F}\Vert_p \le M$.

5. 对前面提到的概念:展缩函数进行简单的讨论:

$K(x)=\chi_{B(0,1)}(x)$,则

\[K_{\epsilon}(x)=\begin{cases}
\epsilon^{-n}, &|x|<\epsilon\\
0,&|x|\ge \epsilon.
\end{cases}\]

设$K \in L^1(R^n)$,则有

(i) $\int_{R^n}{K_{\epsilon}(x)dx}=\int_{R^n}{K(x)dx}$.

(ii) 对于固定的$\delta>0$,有

\[\lim_{\epsilon \rightarrow 0}{\int_{|x|>\delta}{|K_{\epsilon}(x)|dx}}=0.\]

(i)简单的积分换元:$y = \frac{x}{\epsilon}$,则

\[J=\begin{pmatrix}
\frac{1}{\epsilon} && \\
&\ddots&\\
&&\frac{1}{\epsilon}
\end{pmatrix}=\epsilon^{-n}.\]

\[\int_{R^n}{K(x)dx}=\int_{R^n}{K(\frac{x}{\epsilon})|J|dx}=\int_{R^n}{\epsilon^{-n}K(\frac{x}{\epsilon})dx}=\int_{R^n}{K_{\epsilon}(x)dx}.\]

(ii) $K_{\epsilon}(x)=\epsilon^{-n}K(\frac{x}{\epsilon})$,当$|x|>\delta$时,$|\frac{x}{\epsilon}|>\frac{\delta}{\epsilon}$,$x=\epsilon{}t$,则$J=\epsilon^n$.

\[\begin{aligned}
int_{|x|>\delta}{|K_{\epsilon}(x)|dx} &= \int_{|x|>\delta}{\epsilon^{-n}|K(\frac{x}{\epsilon})|dx} \\
&=\int_{|t|>\frac{\delta}{\epsilon}}{\epsilon^{-n}|K(t)|\epsilon^ndt}\\
&=\int_{|t|>\frac{\delta}{\epsilon}}{|K(t)|dt}
\end{aligned}\]

当$\epsilon \rightarrow 0$时,$\frac{\delta}{\epsilon} \rightarrow \infty$,故有

\[\int_{|t|>\frac{\delta}{\epsilon}}{|K(t)|dt}=0.\]

6. 设$K \in L^1(R^n)$,且$\Vert{K}\Vert_1=1$,若$f \in L^p(R^n)$,$1 \le p < \infty$,则有

\[\lim_{\epsilon \rightarrow 0}{\Vert{K_{\epsilon}*f-f}\Vert_p}=0.\]

证明中使用了广义Minkowski不等式和Lebesgue控制收敛定理.

\[\begin{aligned}
(K_{\epsilon}*f)(x)-f(x) &= \int_{R^n}{[f(x-y)-f(x)]K_{}(y)dy}\\
&=\int_{R^n}{[f(x-\epsilon{}y)-f(x)]K(y)dy}\quad ?
\end{aligned}\]

这只有在$K(x)\ge0$时成立.

\[\Vert{K(x)}\Vert_1=\int_{R^n}{|K(x)|dx}=1=\int_{R^n}{K(x)dx}=\int_{R^n}{K_{\epsilon}(y)dy},\]

\[\begin{gather*}
F_{\epsilon}(y)=[\int_{R^n}{|f(x-\epsilon{}y)-f(x)|^pdx}]^{1/p},\\
0 \le F_{\epsilon}(y)|K(y)|\le 2\Vert{f}\Vert_p|K(y)|,
\end{gather*}\]

当$\epsilon \rightarrow 0$时,$F_{\epsilon}(y) \rightarrow 0$.

\[\rho(x)=\begin{cases}
c\exp(-\frac{1}{1-|x|^2}) &|x|<1\\
0&|x| \ge 1,
\end{cases}\]

这里$c$使得$\Vert{\rho}\Vert_1=1$.当$f(x)$是具有紧支集$F$且属于$L^p(R^n)$时,$(\rho_{\epsilon}*f)(x)$也具有紧支集.

\[(\rho_{\epsilon}*f)(x)=\int_{|x|\le 1}{f(x-\epsilon{}t)\rho(t)dt},\]

$(\rho_{\epsilon}*f)(x)$的紧支集是$F$的$\epsilon$-邻域.即$F$与$B(x,\epsilon)$不相交时,必有$f(x-\epsilon{}t)\rho(t)=0$.只要支集是有界的,必然就是紧支集.

7. 具有紧支集且无限次可微的函数类$C_c^{\infty}(R^n)$在$L^p(R^n)$中稠密.

证明是构造性的.

设$f(x) \in L^p(R^n)$,令

\[\left.
f_N(x)=\begin{cases}
f(x), &|x| \le N, \\
0,&|x|>N,
\end{cases}
\right.
\quad (\rho_{\epsilon}*f_N)(x) \in C_c^{\infty}(R^n)
\]

则从$f \in L^p(R^n)$可得$\Vert{f-f_N}\Vert_p<\epsilon$.从前一个定理的结论可得

\[\Vert{\rho_{\epsilon}*f_N-f_N}\Vert_p<\epsilon.\]

8. Hardy-Littlewood极大函数

$f \in L^p(R^n)$,

\[(Mf)(x)=\sup_{r>0}{\frac{1}{|B(x,r)|}\int_{B(x,r)}{|f(y)|dy}},\]

该函数对于$L^1(R^n)$已经有结论了,估计式5.22,下面对于$f \in L^p(R^n)$的情形讨论.

设$f \in L^p(R^n)$($1 < p \le \infty$),则$(Mf) \in L^p(R^n)$,且有$\Vert{Mf}\Vert_p \le A_p\Vert{f}\Vert_p$.

(1)$p=\infty$时,

\[\begin{gather*}
\Vert{f}\Vert_p=\sup_{m(E)=0}{\{f(x):x\in R^n \backslash E\}} \\
|f(y)|\le M \quad \frac{1}{|B(x,r)|}\int_{B(x,r)}{|f(y)|dy} \le \frac{M}{|B(x,r)|}\int_{B(x,r)}{dy}\\
|Mf| \le \Vert{f}\Vert_{\infty} \Rightarrow \Vert{Mf}\Vert_{\infty} \le \Vert{f}\Vert_{\infty},A_{\infty}=1.
\end{gather*}\]

(2)$1<p<\infty$时,书中存在印刷错误.

$E_{\lambda} = \{x : |f(x)|>\frac{\lambda}{2}\}$,令

\[
f^{\lambda}(x)=\begin{cases}
f(x),&x \in E_{\lambda} \\
0,&x \notin E_{\lambda}
\end{cases}\]

\[\begin{gather*}
|f(x)| \le |f^{\lambda}(x)| + \frac{\lambda}{2} \\
(Mf)(x) \le (Mf^{\lambda})(x) + \frac{\lambda}{2} \\
\{x: (Mf)(x) > \lambda\} \subset \{x : (Mf^{\lambda})(x) > \frac{\lambda}{2}\}
\end{gather*}\]

由$f^{\lambda} \in L^1(R)$可得

\[m(\{x:(Mf)(x) > \lambda\}) \le \frac{2A}{\lambda}\int_{R^n}{|f^{\lambda}(x)|dx}=\frac{2A}{\lambda}\int_{E_{\lambda}}{|f(x)|dx},\]

$m(\{x:(Mf)(x)>\lambda\})$为$(Mf)(x)$的分布函数,记为$(Mf)_*(\lambda)$,则有(4.32)

\[\begin{aligned}
\int_{R^n}{|(Mf)(x)|^pdx} &= p\int_{0}^{\infty}{\lambda^{p-1}(Mf)_*(\lambda)d\lambda} \\
&\le p\int_{0}^{\infty}{\lambda^{p-1}[\frac{2A}{\lambda}\int_{E_{\lambda}}{|f(x)|dx}]d\lambda}\\
&=2A\int_{0}^{\infty}{\lambda^{p-2}[\int_{R^n}{|f(x)|\chi_{E_{\lambda}}(x)dx}]d\lambda} \\
&=2A\int_{R^n}{|f(x)|[\int_{0}^{\infty}{\lambda^{p-2}\chi_{E_{\lambda}}(x)dlambda}]dx}
\end{aligned}\]

而

\[\int_{0}^{\infty}{\lambda^{p-2}\chi_{E_{\lambda}}(x)d\lambda} = \int_{0}^{2|f(x)|}{\lambda^{p-2}d\lambda}=\frac{2^{p-1}}{p-1}|f(x)|^{p-1},\]

因此

\[\int_{R^n}{|(Mf)(x)|^pdx} \le \frac{2^pA_p}{p-1}\int_{R^n}{|f(x)|dx}, \quad A_p=2(\frac{A}{p-1})^{1/p}.\]