$\gdef\leq{\leqslant}$ $\gdef\geq{\geqslant}$ $\gdef\lrb#1{\lbrace#1\rbrace}$ $\gdef\norm#1{\Vert#1\Vert}$
下设$1\leq p,q<\infty,(X,\mu),(Y,\nu)$为测度空间.
$f\in(X,\mu)'\rArr f$是$(X,\mu)$上的可测函数.
定义 设$T:L^p(X,\mu)\rarr L^q(Y,\nu)$为线性算子，若存在常数$c_{p,q}>0$使得对任意简单函数$f\in L^p(X,\mu)$有$\norm{Tf}_{L^q(Y,\nu)}\leq c_{p,q}\norm{f}_{L^p(X,\mu)}$，则称$T$是一个强$(p,q)-$型算子.
类似地设$T:L^p(X,\mu)\rarr L^{q,\infty}(Y,\nu)$为线性算子，若$\norm{Tf}_{L^{q,\infty}(Y,\nu)}\leq c_{p,q}\norm{f}_{L^p(X,\mu)}$，则称$T$是弱$(p,q)-$型算子. 强(p,q)$\rArr$弱(p,q).
实方法：Marcinkiewicz插值定理
定义 $T:(X,\mu)'\rarr(Y,\nu)'$为映射
(1) $T$为线性：$\forall f,g\in(X,\mu)',\lambda\in\R$有$T(f+g)=Tf+Tg,~T(\lambda f)=\lambda Tf$
(2) $T$为次线性：$\forall f,g\in(X,\mu)',\lambda\in\R$有$T(f+g)\leq|Tf|+|Tg|,~T(\lambda f)=|\lambda||Tf|$
(3) $T$为拟线性：$\forall f,g\in(X,\mu)',\lambda\in\R,\exist K>0~s.t.~|T(f+g)|\leq K(|Tf|+|Tg|),~|T(\lambda f)|=|\lambda||Tf|$
定理 Marcinkiewicz插值定理
假设 $1\leq p_0<p_1\leq\infty,T:(L^{p_0}+L^{p_1})(X,\mu)\rarr(Y,\nu)'$为次线性算子，并$\exist A_0,A_1>0~s.t.~T$为弱$(p_0,p_0)-$型和弱$(p_1,p_1)-$型算子，即$$
\begin{cases}
\norm{Tf}_{L^{p_0,\infty}(Y,\nu)}\leq A_0\norm{f}_{L^{p_0}(X,\mu)},~\forall f\in L^{p_0}(X)\\
\norm{Tf}_{L^{p_1,\infty}(Y,\nu)}\leq A_1\norm{f}_{L^{p_1}(X,\mu)},~\forall f\in L^{p_1}(X)
\end{cases}
$$ 则对$\forall p_0<p<p_1,~\forall f\in L^p(X)$，都有$$
\norm{Tf}_{L^p(X)}\leq A\norm{f}_{L^p(X)},A=2(\frac{p}{p-p_0}+\frac{p}{p_1-p})^\frac1p A_0^\theta A_1^{1-\theta}
$$ 且$\theta\in(0,1),\frac1p=\frac\theta{p_0}+\frac{1-\theta}{p_1}$.
注：$f\in(L^{p_0}+L^{p_1})(X)$指$f=f_1+f_2$且$f_1\in L^{p_0}(X),f_2\in L^{p_1}(X)$.
证明：第一种情况，$1\leq p_0<p_1<+\infty$. 给定$f\in L^p(X)$(目标$Tf\in L^p(Y)$)，$\alpha>0$给定，$\delta>0$待定. 分解$f=f_0^\alpha+f_1^\alpha,$其中$f_0^\alpha=\begin{cases}f(x)&|f(x)|>\delta\alpha\\0&|f(x)|<\delta\alpha\end{cases},f_1^\alpha=\begin{cases}0&|f(x)|>\delta\alpha\\f(x)&|f(x)|\leq\delta\alpha\end{cases}$，可以验证$f^\alpha_0\in L^{p_0}(X),f^\alpha_1\in L^{p_1}(X)$.
事实上，$$
\begin{aligned}
\norm{f^\alpha_0}^{p_0}_{L^{p_0}(X)}&=\int_X|f^\alpha_0(x)|^{p_0}d\mu\\&=\int_{|f(x)|>\delta\alpha}|f(x)|^{p_0}d\mu\\
&=\int_{|f(x)|>\delta\alpha}|f(x)|^p\cdot|f(x)|^{p_0-p}d\mu\\
&\leq(\delta\alpha)^{p_0-p}\int_{|f(x)|>\delta\alpha}|f(x)|^pd\mu\\
&\leq(\delta\alpha)^{p_0-p}\norm{f}^p_{L^p}<\infty
\end{aligned}
$$ 类似有$$
\begin{aligned}
\norm{f^\alpha_1}^{p_1}_{L^{p_1}}&=\int_X|f^\alpha_1(x)|^{p_1}d\mu\\&=\int_{|f(x)|\leq\delta\alpha}|f(x)|^{p_1}d\mu\\
&=\int_{|f(x)|\leq\delta\alpha}|f(x)|^p|f(x)|^{p_1-p}d\mu\\
&\leq(\delta\alpha)^{p_1-p}\int_{|f(x)|\leq\delta\alpha}|f(x)|^pd\mu
\end{aligned}
$$ 因为$T$为次线性算子，$|Tf(x)|\leq|Tf^\alpha_0(x)|+|Tf^\alpha_1(x)|$，所以$$
\begin{aligned}
E_{Tf}(\alpha)&=\lrb{x\in Y:|Tf(x)|>\alpha}\\&\subset\lrb{x\in Y:|Tf^\alpha_0(x)|>\frac\alpha2}\cup\lrb{x\in Y:|Tf^\alpha_1|>\frac\alpha2}\\
&=E_{Tf^\alpha_0}(\frac\alpha2)\cup E_{Tf^\alpha_1}(\frac\alpha2)
\end{aligned}
$$ 从而$d_{Tf}(\alpha)\leq d_{Tf^\alpha_0}(\frac\alpha2)+d_{Tf^\alpha_1}(\frac\alpha2)$
所以$$
\begin{aligned}
\norm{Tf}^p_{L^p}&=p\int^{+\infty}_0\alpha^{p-1}d_{Tf}(\alpha)d\alpha\\
&\leq p\int^{+\infty}_0\alpha^{p-1}d_{Tf^\alpha_0}(\frac\alpha2)d\alpha+p\int^{+\infty}_0\alpha^{p-1}d_{Tf^\alpha_1}(\frac\alpha2)d\alpha
\end{aligned}
$$ 考虑$d_{Tf^\alpha_0}(\frac\alpha2)$，由$f^\alpha_0\in L^{p_0}(X,\mu)$，$T$为弱$(p_0,p_0)-$型算子得$\norm{Tf^\alpha_0}_{L^{p_0,\infty}}\leq A_0\norm{f^\alpha_0}_{L^{p_0}}$
$\rArr\norm{Tf_0^\alpha}_{L^{p_0,\infty}}=\displaystyle\sup_{\frac\alpha2>0}(\frac\alpha2)d^{\frac{1}{p_0}}_{Tf^\alpha_0}(\frac\alpha2)\leq A_0\norm{f^\alpha_0}_{L^{p_0}}$
$\rArr \frac\alpha2d^{\frac{1}{p_0}}_{Tf^\alpha_0}(\frac\alpha2)\leq A_0\norm{f^\alpha_0}_{L^{p_0}}\rArr(\frac\alpha2)^{p_0}d_{Tf^\alpha_0}(\frac\alpha2)\leq A_0^{p_0}\norm{f_0^\alpha}^{p_0}_{L^{p_0}}$
$$\begin{aligned}\rArr d_{Tf^\alpha_0}(\frac\alpha2)&\leq(\frac\alpha2)^{-p_0}A_0^{p_0}\norm{f_0}^{p_0}_{L^{p_0}}\\
&=(2A_0)^{p_0}\alpha^{-p_0}\norm{f^\alpha_0}^{p_0}_{L^{p_0}}\\
&\leq(2A_0)^{p_0}\alpha^{-p_0}\int_X|f^\alpha_0(x)|^{p_0}d\mu\\
&=(2A_0)^{p_0}\alpha^{-p_0}\int_{|f(x)|>\delta\alpha}|f(x)|^{p_0}d\mu
\end{aligned}
$$ 即$d_{Tf^\alpha_0}(\frac\alpha2)\leq(2A_0)^{p_0}\alpha^{-p_0}\int_{|f(x)|>\delta\alpha}|f(x)|^{p_0}d\mu$
另一方面，$\norm{Tf^\alpha_1}_{L^{p_1,\infty}}\leq A_1\norm{f^\alpha_1}_{L^{p_1}}\rArr\displaystyle\sup_{\beta>0}\beta d^{\frac{1}{p_1}}_{Tf_1^\alpha}(\beta)\leq A_1\norm{f^\alpha_1}_{L^{p_1}}$
$\rArr \frac\alpha2d^\frac1{p_1}_{Tf^\alpha_1}(\frac\alpha2)\leq A_1\norm{f^\alpha_1}_{L^{p_1}}$
$$ \begin{aligned} \rArr d_{Tf^\alpha_1}(\frac\alpha2)&\leq(2A_1)^{p_1}\alpha^{-p_1}\int_X|f^\alpha_1(x)|^{p_1}d\mu\\ &\leq(2A_1)^{p_1}\alpha^{-p_1}\int_{|f(x)|\leq\delta\alpha}|f(x)|^{p_1}d\mu \end{aligned} $$ 从而$$ \begin{aligned} \norm{Tf}^p_{L^p}&\leq p\int^{+\infty}_0\alpha^{p-1}(2A_0)^{p_0}\alpha^{-p_0}\int_{|f(x)|>\delta\alpha}|f(x)|^{p_0}d\mu d\alpha\\ &+p\int^{+\infty}_0\alpha^{p-1}(2A_1)^{p_1}\alpha^{-p_1}\int_{|f(x)|\leq\delta\alpha}|f(x)|^{p_1}d\mu d\alpha\\ &=I_1+I_2 \end{aligned} $$
$$ \begin{aligned} I_1&=p\int^{+\infty}_0\alpha^{p-1}(2A_0)^{p_0}\alpha^{-p_0}\int_{|f(x)|>\delta\alpha}|f(x)|^{p_0}d\mu d\alpha\\ &=(2A_0)^{p_0}p\int^{+\infty}_0\alpha^{p-p_0-1}\int_{|f(x)|>\delta\alpha}|f(x)|^{p_0}d\mu d\alpha\\ &=(2A_0)^{p_0}p\int_X|f(x)|^{p_0}\int_0^{\frac{|f(x)|}{\delta}}\alpha^{p-p_0-1}d\alpha d\mu\\ &=(2A_0)^{p_0}\frac{p}{p-p_0}\delta^{p_0-p}\int_X|f(x)|^pd\mu\\ &=(2A_0)^{p_0}\frac{p}{p-p_0}\norm{f}^p_{L^p(X)}\delta^{p_0-p} \end{aligned} $$
$$ \begin{aligned} I_2&=p\int^{+\infty}_0\alpha^{p-1}(2A_1)^{p_1}\alpha^{-p_1}\int_{|f(x)|\leq\delta\alpha}|f(x)|^{p_1}d\mu d\alpha\\ &=p(2A_1)^{p_1}\int^{+\infty}_0\alpha^{p-p_1-1}\int_{|f(x)|\leq\delta\alpha}|f(x)|^{p_1}d\mu d\alpha\\ &=p(2A_1)^{p_1}\int_X|f(x)|^{p_1}\int^{+\infty}_\frac{|f(x)|}{\delta}\alpha^{p-p_1-1}d\alpha d\mu\\ &=\frac{p(2A_1)^{p_1}}{p_1-p}\int_X|f(x)|^{p_1}(\frac{|f(x)|}{\delta})^{p-p_1}d\mu\\ &=\frac{p}{p_1-p}(2A_1)^{p_1}\delta^{p_1-p}\int_X|f(x)|^pd\mu=\frac{p}{p_1-p}(2A_1)^{p_1}\delta^{p_1-p}\norm{f}^p_{L^p} \end{aligned} $$ $\rArr\norm{Tf}^p_{L^p}\leq[\frac{p}{p_1-p}(2A_1)^{p_1}\delta^{p_1-p}+\frac{p}{p-p_0}(2A_0)^{p_0}\delta^{p_0-p}]\norm{f}^p_{L^p}$
取$\delta~s.t.~(2A_1)^{p_1}\delta^{p_1-p}=(2A_0)^{p_0}\delta^{p_0-p}$
$\rArr \norm{Tf}_{L^p}\leq(\frac{p}{p-p_0}+\frac{p}{p_1-p})^\frac1p\norm{f}_{L^p}(2A_1)^\frac{p_1}{p}[(2A_0)^{p_0}(2A_1)^{-p_1}]^{\frac{p_1-p}{p_1-p_0}\cdot\frac1p}$
其中$2^{\frac{p_1}{p}+(p_0-p_1)\frac1p\frac{p_1-p}{p_1-p_0}}=1,A_1^{\frac{p_1}{p}+(-p_1)\frac1p\frac{p_1-p}{p_1-p_0}}=A_1^{1-\theta},A_0^{p_0\frac1p\frac{p_1-p}{p_1-p_0}}=A_0^\theta$，且有$\frac1p=\frac{\theta}{p_0}+\frac{1-\theta}{p_1}$
第二种情况，$p_1=+\infty,\gamma>0$待定，$T:L^{p_0}\rarr L^{p_0,\infty}$
$f=f^\alpha_0+f^\alpha_1,f_0^\alpha=\begin{cases}f(x)&|f(x)|>\gamma\alpha\\0&|f(x)|\leq\gamma\alpha\end{cases},f_1^\alpha=\begin{cases}0&|f(x)|>\gamma\alpha\\f(x)&|f(x)|\leq\gamma\alpha\end{cases}$
$\rArr\norm{Tf^\alpha_1}_{L^\infty}\leq A_1\norm{f^\alpha_1}_{L^\infty}\leq A_1\gamma\alpha,$取$\gamma=\frac{1}{2A_1}$，则$\norm{Tf^\alpha_1}_{L^\infty}\leq\frac\alpha2\rArr\mu\lrb{x:|Tf^\alpha_1(x)|>\frac\alpha2}=0$
$\rArr d_{Tf^\alpha_1}(\frac\alpha2)=0\rArr d_{Tf}(\alpha)\leq d_{Tf^\alpha_0}(\frac\alpha2)+d_{Tf^\alpha_1}(\frac\alpha2)=d_{Tf^\alpha_0}(\frac\alpha2)$
其中$d_{Tf^\alpha_0}(\frac\alpha2)\leq(2A_0)^{p_0}\alpha^{-p_0}\norm{f^\alpha_0}^{p_0}_{L^{p_0}}=(2A_0)^{p_0}\alpha^{-p_0}\int_{|f|>\gamma\alpha}|f(x)|^{p_0}d\mu$
从而$$ \begin{aligned} \norm{Tf}^p_{L^p}&=p\int^{+\infty}_0\alpha^{p-1}d_{Tf}(\alpha)d\alpha\\ &\leq p\int^{+\infty}_0\alpha^{p-1}d_{Tf^\alpha_0}(\frac\alpha2)d\alpha\\ &\leq p\int^{+\infty}_0\alpha^{p-p_0-1}(2A_0)^{p_0}\int_{|f|>\frac{\alpha}{2A_1}}|f(x)|^{p_0}d\mu d\alpha\\ &=p(2A_0)^{p_0}\int_X|f(x)|^{p_0}\int_0^{2A_1|f(x)|}\alpha^{p-p_0-1}d\alpha d\mu\\ &=\frac{p(2A_1)^{p-p_0}(2A_0)^{p_0}}{p-p_0}\int_X|f(x)|^pd\mu \end{aligned} $$ $\rArr\norm{Tf}_{L^p}\leq A\norm{f}_{L^p(X)},A=2(\frac{p}{p-p_0})^\frac1pA_1^{1-\frac{p_0}{p}}A_0^\frac{p_0}{p}$
复方法
Riesz-Thorin定理 $1\leq p_0,q_0,p_1,q_1\leq\infty,p_0<p_1,q_0<q_1,T:(X,\mu)'\rarr(Y,\nu)'$线性映射且满足对任意简单函数$f\in X,\norm{Tf}_{L^{q_0}(Y,\nu)}\leq M_0\norm{f}_{L^{p_0}(X,\mu)},\norm{Tf}_{L^{q_1}(Y,\nu)}\leq M_1\norm{f}_{L^{p_1}(X,\mu)}$，即$T$为强(p_0,q_0)和(p_1,q_1)型算子，则$\forall f\in L^p(X,\mu),\norm{Tf}_{L^q(Y,\nu)}\leq M_0^\theta M_1^{1-\theta}\norm{f}_{L^p(X,\mu)},\theta\in(0,1)$且$\frac1p=\frac\theta{p_0}+\frac{1-\theta}{p_1},\frac1q=\frac{\theta}{q_0}+\frac{1-\theta}{q_1}$.