13 Riese-Thorin and Marcinkiewicz Interpolation Theorems

the three lines lemma

Let $h$ be a bounded continuous function on the strip $0⩽\mathrm{Re}z⩽1$ that is holomorphic on the interior of the strip. If $|h|\le M_0$ for $\mathrm{Re}z=0$ and $|h(z)|\le M_1$ for $\mathrm{Re}z=1$, then $|h|\le M_0^{1-t}{M}_1^t$ for Re $z=t,0<t<1$.

\begin{proof} WLOG assume that $M_0{M}_1\ne 0$. For $ϵ>0$, define

$$h_{ϵ}(z)=h(z)M_0^{z-1}{M}_1^{-z}{e}^{ϵz(z-1)}$$

and it is holomorphic inside the strip. Note that $|h_{ϵ}(iy)|⩽e^{-ϵy^2}⩽1$ and $|h_{ϵ}(1+iy)|⩽1$.

For any $x\in (0,1)$, there is

$$|h_{ϵ}(x+iy)|=|h((x+iy)M_0^{x-1}{M}_1^{-x}{e}^{ϵx(x-1)}{e}^{-ϵy^2}|⩽C{e}^{-ϵy^2}$$

because $h$ and $e^{ϵx(x-1)}$ are bounded. Since $e^{-ϵy^2}\to 0$ as $y\to \infty$, there exists $A>0$ such that $|h_{ϵ}(x+iy)|⩽1$ for any $x\in (0,1)$ and $y\in (-A,A)$ by the maximum modulus principle. It holds for any arbitrarily large $A$ and so $|h_{ϵ}(z)|⩽1$ inside the entire strip. Take $ϵ\to 0$ and then we finish the proof. \end{proof}

Definition

A measure $\mu$ on a measurable space $(X,\mathcal{A})$ is semifinite if for any measurable $A$ with $\mu (A)>0$, there exists a measurable $B\subset A$ such that $0<\mu (B)<\infty$.

Theorem

Let $(X,\mathcal{A},\mu )$ be a measure space and let $1\le p,q\le \infty$ be conjugate exponents. Suppose that $g$ is a measurable function on $X$ such that $fg\in$ $L^1$ for all $f$ in the space $\mathcal{S}$ of simple measurable functions that vanish outside a set of finite measure, and the quantity

$${\mathcal{M}}_q(g)=\sup \left\{\left|\int fg\right|,f\in \mathcal{S}\text{ and }\Vert f{\Vert }_p=1\right\}$$

is finite. Also, suppose that $S_g=\{x:g(x)\ne 0\}$ is $\sigma$-finite if $q<\infty$ and that $\mu$ is semifinite if $q=\infty$. Then $g\in L_q$ and ${\mathcal{M}}_q(g)=\Vert g{\Vert }_q$.

\begin{proof} Take a bounded measurable $f$ with finite measure support $E$ and $\Vert f{\Vert }_p=1$, then there is a sequence $\left\{f_n\right\}$ of simple measurable functions such that $\left|f_n\right|⩽|f|$. Since $\left|f_{n}g\right|⩽\Vert f{\Vert }_{\infty }{\chi }_E|g|$ and ${\chi }_E|g|\in L^1$, by the dominated convergence theorem we have

$$\begin{array}{rl}\left|\int fg\right|& =\underset{n\to \infty }{\lim }\left|\int f_{n}g\right|\\ & ⩽\underset{n\to \infty }{\lim }{\Vert f_n\Vert }_p{\mathcal{M}}_q(g)=\Vert f{\Vert }_p{\mathcal{M}}_q(g)={\mathcal{M}}_q(g)\end{array}$$

Suppose that $q<\infty$. Let $\left\{E_n\right\}$ with finite measure such that $E_n\uparrow S_g$. Let $\left\{s_n\right\}$ be a sequence of simple measurable functions such that $s_n\to g$ pointwise and $|s_n|⩽|g|$ and let $g_n=s_n{\chi }_{E_n}$. Then $g_n\to g$ pointwise, $\left|g_n\right|⩽|g|$, and $g_n$ vanishes outside $E_n$. Let

$$f_n=\frac{{\left|g_n\right|}^{q-1}\overline{\mathrm{sng}g}}{{\Vert g_n\Vert }_q^{q-1}},\text{mbox}where\mathrm{sng}(g)=\{\begin{array}{l}\frac{g}{|g|},\text{ if }g\ne 0\\ 0,\text{ if }g=0\end{array}.$$

Note that $g\overline{\text{ sng }(g)}=|g|$. There exists some $f_n$ with ${\Vert f_n\Vert }_p=1$ satisfying $||g_n|{|}_q=\int |f_n{g}_n|$ by Riesz representation theorem. Then by Fatou’s lemma

$$\begin{array}{rl}\Vert g{\Vert }_q& ⩽\underset{n\to \infty }{\mathrm{liminf}}{\Vert g_n\Vert }_q=\underset{n\to \infty }{\mathrm{liminf}}\int \left|f_n{g}_n\right|\\ & ⩽\underset{n\to \infty }{\mathrm{liminf}}\int \left|f_{n}g\right|=\underset{n\to \infty }{\mathrm{liminf}}\int f_{n}g⩽{\mathcal{M}}_q(g).\end{array}$$

On the other hand, Holder’s inequality gives ${\mathcal{M}}_q(g)\le \Vert g{\Vert }_q$, so the proof is complete for the case $q<\infty$.

Now suppose $q=\infty$. Given $ϵ>0$, let $A=\left\{x:|g(x)|>{\mathcal{M}}_{\infty }(g)+ϵ\right\}$. If $\mu (A)$ were positive, we could choose $B\subset A$ with $0<\mu (B)<\infty$ since $\mu$ is semifinite. Setting $f=\mu (B{)}^{-1}{\chi }_B\overline{\text{ sng }g}$, we would then have $\Vert f{\Vert }_1=1$, and $\int fg=\mu (B{)}^{-1}{\int }_B|g|\ge {\mathcal{M}}_{\infty }(g)+ϵ$. But this is impossible by the argument at the beginning of the proof. Hence $\Vert g{\Vert }_{\infty }\le {\mathcal{M}}_{\infty }(g)$, and the reverse inequality is obvious. \end{proof}

Remark. Note that ${\mathcal{M}}_q(g)<\infty$ is the most crucial condition. It essentially measures how “large” $g$ is in terms of its interaction with all $f\in {\mathcal{S}}_{\text{normalized in }}{L}^p$, and finite ${\mathcal{M}}_q(g)$ implies that $g$ is “not too wild” and suggests that $g$ should belong to $L^q$. Furthermore, with some condition on the measurable space, we get ${\mathcal{M}}_q(g)=||g|{|}_q$.

Remark. 这个证明不考。

the Riesz-Thorin interpolation theorem

Suppose that $(X,\mathcal{A},\mu )$ and $(Y,\mathcal{B},\nu )$ are measure spaces and $p_0,p_1,q_0,q_1\in [1,\infty ],p_0\le p_1,q_0\le q_1$. If $q_0=q_1=\infty$, suppose also that $\nu$ is semifinite. For $0<t<1$, define $p_t$ and $q_t$ by

$$\frac{1}{p_t}=\frac{1-t}{p_0}+\frac{t}{p_1},\frac{1}{q_t}=\frac{1-t}{q_0}+\frac{t}{q_1}.$$

If $T$ is a linear map from $L^{p_0}(\mu )+L^{p_1}(\mu )$ into $L^{q_0}(\nu )+L^{q_1}(\nu )$ such that $\Vert Tf{\Vert }_{q_0}⩽M_0\Vert f{\Vert }_{p_0}$ for $f\in L^{p_0}(\mu )$ and $\Vert Tf{\Vert }_{q_1}⩽M_1\Vert f{\Vert }_{p_1}$ for $f\in L^{p_1}(\mu )$, then $\Vert Tf{\Vert }_{q_t}⩽M_0^{1-t}{M}_1^t\Vert f{\Vert }_{p_t}$ for $f\in L^{p_t}(\mu )$.

\begin{proof} See here. \end{proof}

Remark. 这个证明不考。

Recall that $f(x)=1/x$ is not integrable over $\mathbb{R}$, and to include such functions we slightly weaken the $L^p$ norms. Let $(X,\mathcal{A},\mu )$ be a measure space and $f:X\to \mathbb{C}$ be a measurable function. We consider the distribution function ${\mu }_f(t)=\mu (\{x\in X:|f(x)|>t\})$ with $t⩾0$. Recall that there is

$$\Vert f{\Vert }_p^p=p{\int }_0^{\infty }{t}^{p-1}{\mu }_f(t)dt,1⩽p<\infty$$

by layer cake representation. Notice that

$$\begin{array}{rl}\Vert f{\Vert }_p^p& ={\int }_X|f{|}^{p}d\mu ⩾{\int }_{\{x\in X:|f(x)|>t\}}|f{|}^{p}d\mu \\ & ⩾{\int }_{\{x\in X:|f(x)|>t\}}{t}^{p}d\mu =t^p{\mu }_f(t)\end{array}$$

and so $\Vert f{\Vert }_p⩾t{\left({\mu }_f(t)\right)}^{1/p}$, for all $t>0$. This leads the following definition.

Definition

Let $(X,\mathcal{A},\mu )$ be a measure space and $f:X\to \mathbb{C}$ be a measurable function. Define the distribution function as ${\mu }_f(t)=\mu (\{x\in X:|f(x)|>t\})$, and define

$$\Vert f{\Vert }_{p,w}=\{\begin{array}{ll}{\sup }_{t>0}t{\left({\mu }_f(t)\right)}^{1/p}& 1⩽p<\infty ,\\ \Vert f{\Vert }_{\infty }& p=\infty .\end{array}$$

which is called the weak $p$-norm of $f$. Furthermore, let us denote

$$L^{p,w}(X,\mu )=\left\{f:X\to \mathbb{C}:\Vert f{\Vert }_{p,w}<\infty \right\}.$$

As in the $L^p$-space, we also regard functions in $L^{p,w}(X,\mu )$ which are equal a.e. as one function.

By construction we have $\Vert f{\Vert }_{p,w}\le \Vert f{\Vert }_p$ when $1⩽p<\infty$. We have seen that $L^p(X,\mu )\subseteq L^{p,w}(X,\mu )$. Note that $f(x)=1/x\notin L^1(\mathbb{R})$ with $t{\mu }_f(t)=2$, we have $\Vert f{\Vert }_{1,w}=2$ and it deduces that this inclusion is in general strict, that is $L^p(X,\mu )\subset L^{p,w}(X,\mu )$.

Definition

Let $T$ be a map from some vector space $\mathcal{D}$ of measurable functions on $(X,\mathcal{A},\mu )$ to the space of all measurable functions on $(Y,\mathcal{B},\nu )$. $T$ is called subadditive if $|T(f+g)|⩽|T(f)|+|T(g)|$ and $|T(cf)|=c|T(f)|$ for all $f,g\in \mathcal{D}$ and $c>0$.

A subadditive operator $T$ is strong type $(p,q)$ with $1⩽p,q⩽\infty$, if $L^p(\mu )\subset \mathcal{D},T$ maps $L^p(\mu )$ into $L^q(\nu )$, and there exists $C>0$ such that

$$\Vert T(f){\Vert }_q⩽C_{p,q}\Vert f{\Vert }_p,\forall f\in L^p(\mu ).$$

A subadditive operator $T$ is weak type $(p,q)$ with $1⩽p,q<\infty$, if $L^p(\mu )\subset \mathcal{D},T$ maps $L^p(\mu )$ into $L^{q,w}(\nu )$, and there exists $C_{p,q,w}>0$ such that

$$\Vert T(f){\Vert }_{q,w}⩽C_{p,q,w}\Vert f{\Vert }_p,\forall f\in L^p(\mu ).$$

Also, we shall say that $T$ is weak type $(p,\infty )$ iff $T$ is strong type $(p,\infty )$.

Marcinkiewicz

Let $(X,\mathcal{A},\mu )$ and $(Y,\mathcal{B},\nu )$ be measure spaces and $1⩽p_0<p_1⩽\infty$. Let $T$ be a subadditive operator defined on $L^p(\mu ),p\in \left[p_0,p_1\right]$. If $T$ is of weak type $\left(p_0,p_0\right)$ and $\left(p_1,p_1\right)$ then it is also of strong type $(p,p)$ for every $p_0<p<p_1$.

\begin{proof} Firstly, assume that $p_1<\infty$. Fix $f\in L^p(\mu )$. By ^a9aid4, we can take some number $r>0$ and decompose $f=f_0+f_1$ where $f_0=f{\chi }_{\{x:|f(x)|>r\}}$ and $f_1=f{\chi }_{\{x:|f(x)|\le r\}}$. Then $f_0\in L^{p_0}\cap L^p$ and $f_1\in L^p\cap L^{p_1}$. We have

$$\begin{array}{rl}\Vert T(f){\Vert }_p^p& =p{\int }_0^{\infty }{r}^{p-1}{\nu }_{T(f)}(r)dr\\ & =p{\int }_0^{\infty }{r}^{p-1}\nu (\{y\in Y:|T(f)(y)|>r\})dr\\ & =p{2}^p{\int }_0^{\infty }{r}^{p-1}{\nu }_{T(f)}(2r)dr.\end{array}$$

Since $T$ is subadditive, there is $|T(f)|=\left|T\left(f_0+f_1\right)\right|⩽\left|T\left(f_0\right)\right|+\left|T\left(f_1\right)\right|$ and so ${\nu }_{T(f)}(2r)⩽{\nu }_{T\left(f_0\right)}(r)+{\nu }_{T\left(f_1\right)}(r)$. Since $T$ is of weak type $\left(p_0,p_0\right)$ and of weak type $(p_1,p_1)$, we have

$${\nu }_{T\left(f_0\right)}(r)⩽{\left(\frac{C_0}{r}\right)}^{p_0}\cdot {\Vert f_0\Vert }_{p_0}^{p_0}\text{mbox}and{\nu }_{T\left(f_1\right)}(r)⩽{\left(\frac{C_1}{r}\right)}^{p_1}\cdot {\Vert f_1\Vert }_{p_1}^{p_1},\forall r>0.$$

Thus we have

$$\begin{array}{rl}\Vert T(f){\Vert }_p^p& ⩽p{2}^p\left({\int }_0^{\infty }{r}^{p-1}{\nu }_{T\left(f_0\right)}(r)dr+{\int }_0^{\infty }{r}^{p-1}{\nu }_{T\left(f_1\right)}(r)dr\right)\\ & ⩽p{2}^p\left({\int }_0^{\infty }{r}^{p-1}{\left(\frac{C_0}{r}\right)}^{p_0}\cdot {\Vert f_0\Vert }_{p_0}^{p_0}+{\int }_0^{\infty }{r}^{p-1}{\left(\frac{C_1}{r}\right)}^{p_1}\cdot {\Vert f_1\Vert }_{p_1}^{p_1}\right)dr\end{array}$$

Observe that ${\Vert f_0\Vert }_{p_0}^{p_0}={\int }_{\{x\in X:|f(x)|>r\}}|f{|}^{p_0}d\mu$ and ${\Vert f_1\Vert }_{p_1}^{p_1}={\int }_{\{x\in X:|f(x)|\le r\}}|f{|}^{p_1}d\mu$, then $\Vert T(f){\Vert }_p^p\le p{2}^p\left(C_0^{p_0}{I}_0+C_1^{p_1}{I}_1\right)$, where

$$\begin{array}{rl}{I}_0& ={\int }_0^{\infty }\left[{\int }_{\{x\in X:|f(x)|>r\}}{r}^{p-p_0-1}|f{|}^{p_0}d\mu \right]dr\\ & ={\int }_0^{\infty }\left[{\int }_X{r}^{p-p_0-1}|f{|}^{p_0}{\chi }_{\left\{(x,r)\in X\times {\mathbb{R}}^{+}:|f(x)|>r\right\}}d\mu \right]dr\\ & ={\int }_X\left[{\int }_0^{\infty }{r}^{p-p_0-1}|f{|}^{p_0}{\chi }_{\left\{(x,r)\in X\times {\mathbb{R}}^{+}:|f(x)|>r\right\}}dr\right]d\mu \\ & ={\int }_X\left[{\int }_0^{|f|}{r}^{p-p_0-1}|f{|}^{p_0}dr\right]d\mu =\frac{1}{p-p_0}\Vert {f\Vert }_p^p,\end{array}$$

and

$$\begin{array}{rl}{I}_1& ={\int }_0^{\infty }\left[{\int }_{\{x\in X:|f(x)|\le r\}}{r}^{p-p_1-1}|f{|}^{p_1}d\mu \right]dr\\ & ={\int }_X\left[{\int }_{|f|}^{\infty }{r}^{p-p_1-1}|f{|}^{p_1}dr\right]d\mu =\frac{1}{p_1-p}\Vert f{\Vert }_p^p\end{array}$$

Hence, $\Vert T(f){\Vert }_p⩽C\Vert f{\Vert }_p$ and so $T$ is of strong type $(p,p)$.

Now assume that $p_1=\infty$. In this case, we have $\Vert T(g){\Vert }_{\infty }⩽C_1\Vert g{\Vert }_{\infty }$ for all $g\in L^{\infty }(\mu )$, and $\Vert T(h){\Vert }_{p_0,w}⩽C_0\Vert h{\Vert }_{p_0}$ for all $h\in L^{p_0}(\mu )$. The case $C_1=0$ can be settled easily, and then we assume that $C_1>0$. Let $f\in L^p(\mu )$ and set

$$f_0=f{\chi }_{\left\{x\in X:|f(x)|>\frac{r}{C_1}\right\}}\in L^{p_0}\cap L^p\text{mbox}and{f}_1=f{\chi }_{\left\{x\in X:|f(x)|\le \frac{r}{C_1}\right\}}\in L^p\cap L^{\infty }.$$

Since ${\Vert f_1\Vert }_{\infty }⩽r/C_1$, we have ${\Vert T\left(f_1\right)\Vert }_{\infty }⩽r$, which implies that ${\nu }_{T\left(f_1\right)}(r)=\nu \left(\left\{y\in Y:\left|T\left(f_1\right)(y)\right|>r\right\}\right)=0$. Hence

$$\begin{array}{rl}{\nu }_{T(f)}(2r)& ⩽{\nu }_{T\left(f_0\right)}(r)+{\nu }_{T\left(f_1\right)}(r)={\nu }_{T\left(f_0\right)}(r)\\ & ⩽\frac{C_0^{p_0}}{r^{p_0}}{\Vert f_0\Vert }_{p_0}^{p_0}=\frac{C_0^{p_0}}{r^{p_0}}{\int }_{\left\{x\in X:|f(x)|>\frac{r}{C_1}\right\}}|f{|}^{p_0}d\mu .\end{array}$$

It then follows from the same calculations as in the case 1 that

$$\begin{array}{rl}\Vert T(f){\Vert }_p^p& =p{2}^p{\int }_0^{\infty }{r}^{p-1}{\nu }_{T(f)}(2r)dr\\ & ⩽p{2}^p{C}_0^{p_0}{\int }_0^{\infty }\left[{\int }_{\left\{x\in X:|f(x)|>\frac{r}{C_1}\right\}}{r}^{p-p_0-1}|f{|}^{p_0}d\mu \right]dr\\ & =p{2}^p{C}_0^{p_0}{\int }_X|f(x){|}^{p_0}\left[{\int }_0^{C_1|f(x)|}{r}^{p-p_0-1}dr\right]d\mu \\ & =\frac{p{2}^p{C}_0^{p_0}{C}_1^{p-p_0}}{p-p_0}{\int }_X|f{|}^{p}d\mu .\end{array}$$

Therefore, $\Vert T(f){\Vert }_p⩽2{\left(\frac{p}{p-p_0}\right)}^{1/p}{C}_0^{p_0/p}{C}_1^{1-p_0/p}\Vert f{\Vert }_p$ and so $T$ is of strong type $(p,p)$. \end{proof}