$\gdef\leq{\leqslant}$ $\gdef\lb{\lbrace}$ $\gdef\rb{\rbrace}$ $\gdef\lrb#1{\lbrace#1\rbrace}$ $\gdef\geq{\geqslant}$ $\gdef\norm#1{\Vert#1\Vert}$ $\gdef\wnorm#1{{\Vert#1\Vert}_{L^{p,\infty}}}$ $\gdef\wnormq#1{{\Vert#1\Vert}_{L^{q,\infty}}}$
定义3.1 对$1\leq p<\infty$，弱-$L^p(X,\mu)$或$L^{p,\infty}(X,\mu)$定义为一个集合：对空间上所有$\mu$-可测函数$f$有$$\inf\lbrace c>0:\alpha d_f^{\frac1p}(\alpha)\leq c\rbrace=\sup_{\alpha>0}\lbrace\alpha d_f^{\frac1p}(\alpha)\rbrace<\infty~(\forall\alpha>0)$$
$L^{\infty,\infty}=L^\infty$，记$\inf\lbrace c>0:\alpha d_f^{\frac1p}(\alpha)\leq c\rbrace=\norm{f}_{L^{p,\infty}(X,\mu)}$
考察$\wnorm{\cdot}:$所有$\mu$-可测函数$\larr[,+\infty)$
$1\degree \forall f\in L^{p,\infty},\wnorm{f}\geq0,\wnorm{f}=0\lrArr f=0$
证明：$f=0~i.e.~f(x)=0~a.e.~x\in X.~\wnorm{f}=\sup_{\alpha>0}\lbrace\alpha d_f^{\frac1p}(\alpha)\rbrace,d_f(\alpha)=\mu(\lbrace x \in X:|f(x)|>\alpha\rbrace)=0$. 从而$f=0\rArr\wnorm{f}=0$
另一方面，$\wnorm{f}=0\rArr\sup_{\alpha>0}\lbrace\alpha d_f^\frac1p(\alpha)\rbrace=0\rArr\alpha d_f^\frac1p(\alpha)=0,\forall\alpha>0\rArr d_f(\alpha)=0\rArr\forall\alpha>0,\mu(\lbrace x\in X:|f(x)|>\alpha\rbrace)=0$
可以证明：$\lb x\in X:|f(x)|\neq0\rb=\bigcup_{n=1}^\infty\lb x\in X:|f(x)|>\frac1n\rb=\bigcup_{n=1}^\infty E_f(\frac1n)$
从而$\mu(\lb x\in X:|f(x)|\neq0\rb)\leq\sum_{n=1}^\infty\mu(E_f{\frac1n})=\sum_{n=1}^\infty d_f(\frac1n)=0\rArr\mu(\lb x\in X:|f(x)|\neq0\rb)=0\rArr f=0~a.e.x\in X\rArr f=0$.
$2\degree~\forall k\in\R,f\in L^{p,\infty}(X,\mu)\rArr kf\in L^{p,\infty}$且$\wnorm{kf}=|k|\wnorm{f}$
证明：$k=0$显然，$\wnorm{kf}=\sup_{\alpha>0}\lb\alpha d_{kf}^\frac1p(\alpha)\rb$
对$\forall\alpha>0,\alpha d_{kf}^\frac1p(\alpha)=\alpha d_f^\frac1p(\frac\alpha{|k|})\overset{\beta=\frac\alpha{|k|}}{=}|k|\beta d_f^\frac1p(\beta)$
两边对$\alpha$取sup,$\sup_{\alpha>0}\lb\alpha d_{kf}^\frac1p(\alpha)\rb=|k|\sup_{\beta>0}\lb\beta d_f^\frac1p(\beta)\rb\rArr\wnorm{kf}=|k|\wnorm{f}$
$3\degree~1\leq p<\infty,\wnorm{f+g}\leq2(\wnorm{f}+\wnorm g)$
证明：$\wnorm{f+g}=\sup_{\alpha>0}\lb\alpha d_{f+g}^\frac1p(\alpha)\rb$，事实上$d_{f+g}(\alpha)\leq d_f(\frac\alpha2)+d_g(\frac\alpha2)$.
$\rArr\alpha^p d_{f+g}(\alpha)\leq2^p[(\frac\alpha2)^p d_f(\frac\alpha2)+(\frac\alpha2)^p d_g(\frac\alpha2)]\overset{\beta=\frac\alpha2}{=}2^p(\beta^p d_f(\beta)+\beta^p d_g(\beta))$
$\rArr\alpha d_{f+g}^\frac1p(\alpha)\leq2(\beta^p d_f(\beta)+\beta^p d_g(\beta))^\frac1p$
两边对$\alpha$取上确界，得$\wnorm{f+g}\leq2(\wnorm{f}^p+\wnorm{g}^p)^\frac1p\leq2(\wnorm{f}+\wnorm{g})$
称$\wnorm{\cdot}$为拟范数，$(L^{p,\infty},\wnorm{\cdot})$是一个完备的拟赋范线性空间.
注：$L^p(X,\mu)\subsetneqq L^{p,\infty}(X,\mu)$
证明：首先证$L^p(X,\mu)\subseteqq L^{p,\infty}(X,\mu)$
$\forall\alpha>0,f\in L^p(X,\mu),d_f(\alpha)\leq\alpha^{-p}\int_{|f(x)|>\alpha}|f(x)|^pd\mu$
$\alpha^p d_f(\alpha)\leq\int_{|f(x)|>\alpha}|f(x)|^pd\mu\leq\int_X|f(x)|^pd\mu\rArr\alpha d_f^\frac1p(\alpha)\leq\norm{f}_{L^p}$，两边对$\alpha$取sup得$\wnorm{f}\leq\norm{f}<+\infty\rArr f\in L^{p,\infty}(X,\mu)$
然后证$\exist g\in L^{p,\infty}$且$g\notin L^p$.
考虑$X=\R^n,\mu=dx,h(x)=|x|^{-\frac{n}{p}},x\in\R^n,1\leq p<\infty$，有$|h(x)|^p\notin L(\R^n), \int_{\R^n}|x|^{-n}dx$发散$\rArr h(x)\notin L^p(\R^n)$
$$
\begin{aligned}
\forall\alpha>0,d_h(\alpha)&=\mu(\lb x\in\R^n:|h(x)|>\alpha\rb)\\
&=\mu(\lb x\in\R^n:|x|<\alpha^{-\frac{p}{n}}\rb)\\
&=\int_{|x|<\alpha^{-\frac{p}{n}}}1dx=\nu_n\cdot[\alpha^{-\frac{p}{n}}]^n
\end{aligned}
$$ $\rArr\alpha d_h^\frac1p(\alpha)=\nu_n^\frac1p$，取sup得$\wnorm{h}=\nu^\frac1p<\infty$从而$h\in L^{p,\infty}$
几乎处处收敛：$f_n\overset{a.e.}{\rarr}f~(f_n(x)\rarr f(x)(n\rarr\infty)~a.e.~x\in X)$
依$L^p$收敛：$f_n\overset{L^p}{\rarr}f\lrArr\norm{f_n(x)-f(x)}_{L^p(X,\mu)}\overset{n\rarr\infty}{\rarr}0$
依测度收敛：$f_n\overset{\mu}{\rarr}f(n\rarr\infty)\lrArr\lrb{f_n}_{n\in\N},f$可测且$\forall\varepsilon>0,\lim\limits_{n\rarr\infty}d_{f_n-f}(\varepsilon)=0~i.e.~\forall\varepsilon>0,\lim\limits_{n\rarr\infty}\mu(\lrb{x\in X:|f_n(x)-f(x)|>\varepsilon})=0$
$i.e.~\forall\varepsilon>0,\exist n_0\in\N,s.t.~\forall n>n_0,\mu(\lrb{x\in X:|f_n-f|>\varepsilon})<\varepsilon$
命题4.2 (1) $f_n\overset{L^p}{\rarr}f\rArr f_n\overset{L^{p,\infty}}{\rarr}f\lrArr\wnorm{f_n-f}\rarr0(n\rarr\infty)$
(2) $f_n\overset{L^{p,\infty}}{\rarr}f\rArr f_n\overset{\mu}{\rarr}f$
证明：(1) $p=+\infty$显然，设$1\leq p<+\infty$
$$
\begin{aligned}
d_{f_n-f}(\alpha)&=\int_{E_{f_n-f}(\alpha)}1d\mu\quad(E_{f_n-f}(\alpha)\lrArr|f_n-f|\geq\alpha\lrArr(\frac{|f_n-f|}{\alpha})^p\geq1)\\
&\leq\alpha^{-p}\int_{E_{f_n-f}(\alpha)}|f_n-f|^pd\mu\\
&\leq\alpha^{-p}\int_X|f_n-f|^pd\mu
\end{aligned}
$$ $\forall\alpha>0,\alpha d_{f_n-f}^\frac1p(\alpha)\leq(\int_X|f_n-f|^pd\mu)^\frac1p=\norm{f_n-f}_{L^p}$
两边对$\alpha$取sup得$0\leq\wnorm{f_n-f}\leq\norm{f_n-f}_{L^p}\rarr0(n\rarr\infty)$
由两边夹，$\wnorm{f_n-f}\rarr0(n\rarr\infty)$.
(2) $p=+\infty$显然，设$1\leq p<+\infty$
因为$f_n\overset{L^{p,\infty}}{\rarr}f$，所以$\wnorm{f_n-f}\rarr0~i.e.~\sup_{\alpha>0}\alpha d_{f_n-f}^\frac1p(\alpha)\rarr0$.
$\forall\varepsilon>0$给定，$\varepsilon d_{f_n-f}^\frac1p(\varepsilon)\rarr0(n\rarr\infty)\rArr\exist n_0\in\N~s.t.~\forall n>n_0,\varepsilon d_{f_n-f}^\frac1p(\varepsilon)<\varepsilon^{1+\frac1p}\rArr d_{f_n-f}(\varepsilon)<\varepsilon~i.e.~f_n\overset{\mu}{\rarr}f~(n\rarr\infty)$
$H\ddot{o}lder$不等式：$$
\int_X|fg|d\mu\leq\norm{f}_{L^p}\norm{g}_{L^{p'}},\forall f\in L^p,g\in L^{p'}~(1\leq p,p'\leq\infty,\frac1p+\frac1{p'}=1)
$$ 引理 $f\in L^p,g\in L^{p'},\norm{f}_{L^p}=1,\norm{g}_{L^{p'}}=1$，则$\int_X|fg|d\mu\leq1$.
若引理成立，事实上对$\forall f\in L^p,g\in L^{p'}$令$\tilde{f}=\frac1{\norm{f}_{L^p}}f,\tilde{g}=\frac1{\norm{g}_{L^{p'}}}g$
只考虑$\norm{f}_{L^p}>0,\norm{g}_{L^{p'}}>0$，从而$\norm{\tilde{f}}_{L^p}=1,\norm{\tilde{g}}_{L^{p'}}=1\rArr\int_X|\tilde{f}\tilde{g}|d\mu\leq1\rArr\int_X|fg|d\mu\leq\norm{f}_{L^p}\norm{g}_{L^{p'}}$
引理证明：$p=1,p'=\infty$和$p=\infty,p'=1$易证. 设$1<p,p'<+\infty,p'=\frac{p}{p-1}$
(Young不等式：$1<p<\infty,a>0,b>0,ab\leq\frac1pa^p+\frac1{p'}b^{p'}$)
注意$\norm{f}_{L^p}=1,\norm{g}_{L^{p'}}=1$
$$
\begin{aligned}
\int_X|f||g|d\mu&\leq\int_X(\frac1p|f|^p+\frac1{p'}|g|^{p'})d\mu\\
&\leq\frac1p\int_X|f|^pd\mu+\frac1{p'}\int_X|g|^{p'}d\mu\\
&=\frac1p\norm{f}^p_{L^p}+\frac1{p'}\norm{g}^{p'}_{L^{p'}}=\frac1p+\frac1{p'}=1
\end{aligned}
$$
$H\ddot{o}lder$不等式：设$1\leq p<q\leq+\infty,f\in L^p(X,\mu)\bigcap L^q(X,\mu)$，对$\forall r,p<r<q$有$f\in L^r(X,\mu)$且$\norm{f}_{L^r}\leq\norm{f}^\theta_{L^p}\norm{f}^{1-\theta}_{L^q},\theta\in(0,1),\frac1r=\frac\theta{p}+\frac{1-\theta}q$
证明：设$1<p<q<+\infty$，$\norm{f}^r_{L^r}=\int_X|f|^rd\mu=\int_X|f|^{r\theta}|f|^{r(1-\theta)}d\mu\leq\norm{|f(x)|^{r\theta}}_{L^\frac{p}{r\theta}}\norm{|f(x)|^{r(1-\theta)}}_{L^{(\frac{p}{r\theta})'}}$
$\norm{|f(x)|^{r\theta}}_{L^\frac{p}{r\theta}}=(\int|f(x)|^{r\theta\cdot\frac{p}{r\theta}}d\mu)^\frac{r\theta}{p}=(\int_X|f|^pd\mu)^{\frac1p\cdot r\theta}=\norm{f}^{r\theta}_{L^p}$
$$
\begin{aligned}
\norm{|f(x)|^{r(1-\theta)}}_{L^{(\frac{p}{r\theta})'}}&=\norm{|f(x)|^{r(1-\theta)}}_{L^{\frac{p}{p-r\theta}}}\\
&=(\int|f|^{r(1-\theta)\frac{p}{p-r\theta}}d\mu)^\frac{p-r\theta}{p}\\
&=(\int_X|f|^qd\mu)^{\frac1qr(1-\theta)}\\
&=\norm{f}^{r(1-\theta)}_{L^q}
\end{aligned}
$$ $\rArr\norm{f}^r_{L^r}\leq\norm{f}^{r\theta}_{L^p}\norm{f}^{r(1-\theta)}_{L^q}\rArr\norm{f}_{L^r}\leq\norm{f}^\theta_{L^p}\norm{f}^{1-\theta}_{L^q}$
$p=1,1<r<q\leq+\infty$和$q=\infty$略.
$L^{p,\infty}(X,\mu)$上的插值不等式
$1\leq p<q\leq\infty,f\in L^{p,\infty}(X,\mu)\bigcap L^{q,\infty}(X,\mu)$
$\forall p<r<q,f\in L^r(X,\mu)$且$\norm{f}_{L^r}\leq(\frac{r}{r-p}+\frac{r}{q-r})^\frac1r\norm{f}^\theta_{L^{p,\infty}}\norm{f}^{1-\theta}_{L^{q,\infty}},\theta\in(0,1),\frac1r=\frac\theta{p}+\frac{1-\theta}q$
证明：设$1\leq p<r<q<+\infty$
$f\in L^{p,\infty}\bigcap L^{q,\infty}\rArr\sup_{\alpha>0}\alpha d_f^\frac1p(\alpha)\leq min\lrb{\wnorm{f},\wnormq{f}}\rArr\forall\alpha>0,d_f(\alpha)\leq min\lrb{\frac{\wnorm{f}^p}{\alpha^p},\frac{\wnormq{f}^q}{\alpha^q}}$
从而$$
\begin{aligned}
\norm{f}^r_{L^r}&=\int_X|f|^rd\mu=r\int^{+\infty}_0\alpha^{r-1}d_f(\alpha)d\alpha\\
&=r(\int^M_0+\int_M^{+\infty})\alpha^{r-1}d_f(\alpha)d\alpha\\
&\leq r\int_0^M\alpha^{r-1}\frac{\wnorm{f}^p}{\alpha^p}d\alpha+r\int_M^{+\infty}\alpha^{r-1}\frac{\wnormq{f}^q}{\alpha^q}d\alpha\\
&=\wnorm{f}^pr\int_0^M\alpha^{r-p-1}d\alpha+\wnormq{f}^qr\int_M^{+\infty}\alpha^{r-p-1}d\alpha\\
&=\wnorm{f}^p\frac{r}{r-p}M^{r-p}+\wnormq{f}^q\frac{r}{q-r}M^{r-q}
\end{aligned}
$$ 取$M~s.t.~\wnorm{f}^pM^{r-p}=\wnormq{f}^qM^{r-q}\rArr M=(\frac{\wnormq{f}^q}{\wnorm{f}^p})^\frac1{q-p}$
$$
\begin{aligned}
\norm{f}^r_{L^r}&\leq(\frac{r}{r-p}+\frac{r}{q-r})\wnorm{f}^pM^{r-p}\\
&=(\frac{r}{r-p}+\frac{r}{q-r})\wnorm{f}^p(\frac{\wnormq{f}^q}{\wnorm{f}^p})^\frac1{q-p}\\
&=(\frac{r}{r-p}+\frac{r}{q-r})\wnorm{f}^{p(1-\frac{r-p}{q-p})}\wnormq{f}^{q\frac{r-p}{q-p}}
\end{aligned}
$$ $\rArr\norm{f}^r_{L^r}\leq(\frac{r}{r-p}+\frac{r}{q-r})^\frac1r\wnorm{f}^{\frac{p}{r}(1-\frac{r-p}{q-p})}\wnormq{f}^{\frac{q}{r}\frac{r-p}{q-p}}$
可以验证$\frac{p}{r}(1-\frac{r-p}{q-p})+\frac{q}{r}\frac{r-p}{q-p}=1$