Some interpolation theorems

By | May 25, 2015

이 페이지는 제가 Interpolation theorem들을 하나씩 배울 때마다 업로드할 예정입니다. (This page will be updated frequently when I learned some interpolation theorems.)

조금 rough하게 말하면, interpolation theorem이란 fL^p이고 L^q일 때 그 사이 공간 L^r (p<r<q)에 들어갈 조건이나, fr-norm의 크기를 재는 정리들을 말합니다.

1. Basic Interpolation Theorems

Theorem. If f\in L^{p_{0}}\cap L^{p_{1}} for some 0<p_{0}<p_{1}\le\infty, then f\in L^{p} for all p_{0}\le p\le p_{1} and that \norm fp\le\norm f{p_{0}}^{1-\theta}\norm f{p_{1}}^{\theta}, where 0<\theta<1 is such that \frac{1}{p}=\frac{1-\theta}{p_{0}}+\frac{\theta}{p_{1}}.

Proof. We first assume p_{1}<\infty. Assume \frac{1}{p}=\frac{1-\theta}{p_{0}}+\frac{\theta}{p_{1}} for some 0<\theta<1. So 1=\frac{p\left(1-\theta\right)}{p_{0}}+\frac{p\theta}{p_{1}}.
Set a=\frac{p_{0}}{p\left(1-\theta\right)} and b=\frac{p_{1}}{p\theta}.
Then by applying Hölder’s inequality to \left|f\right|^{p\left(1-\theta\right)}\left|f\right|^{p\theta}, we have

    \begin{align*} \int_{X}\left|f\right|^{p\left(1-\theta\right)}\left|f\right|^{p\theta}d\mu & \le\left(\int_{X}\left|f\right|^{p\left(1-\theta\right)a}d\mu\right)^{\frac{1}{a}}\left(\int_{X}\left|f\right|^{p\theta b}d\mu\right)^{\frac{1}{b}}\\ & =\left(\int_{X}\left|f\right|^{p_{0}}d\mu\right)^{\frac{p\left(1-\theta\right)}{p_{0}}}\left(\int_{X}\left|f\right|^{p_{1}}d\mu\right)^{\frac{p\theta}{p_{1}}}. \end{align*}

So by taking \sqrt[p]{\cdot} to both sides, we get

    \[ \norm fp\le\norm f{p_{0}}^{1-\theta}\norm f{p_{1}}^{\theta}. \]

The equality condition holds when \alpha\left|f\right|^{pa\left(1-\theta\right)}=\beta\left|f\right|^{p\theta b} for some nonnegative \alpha,\beta a.e.

For p_{1}=\infty, note that we have \frac{1}{p}=\frac{1-\theta}{p_{0}}. So

    \begin{align*} \int_{X}\left|f\right|^{p\left(1-\theta\right)}\left|f\right|^{p\theta}d\mu & \le\left(\int_{X}\left|f\right|^{p\left(1-\theta\right)}d\mu\right)\norm f{\infty}^{p\theta}\\ & =\left(\int_{X}\left|f\right|^{p_{0}}d\mu\right)\norm f{\infty}^{p\theta}. \end{align*}

Since \frac{1}{p}=\frac{1-\theta}{p_{0}}, we have

    \[ \norm fp\le\norm f{p_{0}}^{1-\theta}\norm f{\infty}^{\theta}. \]

Theorem. Let 0<p<q\le\infty and let f\in L^{p,\infty}\left(X,\mu\right)\cap L^{q,\infty}\left(X,\mu\right), where X is \sigma-finite measure space. Then f\in L^{r}\left(X,\mu\right) for all p<r<q and

    \[ \norm fr\le\left(\frac{r}{r-p}+\frac{r}{q-r}\right)^{\frac{1}{r}}\norm f{p,\infty}^{\frac{\frac{1}{r}-\frac{1}{q}}{\frac{1}{p}-\frac{1}{q}}}\norm f{q,\infty}^{\frac{\frac{1}{p}-\frac{1}{r}}{\frac{1}{p}-\frac{1}{q}}}, \]

with the interpretation that 1/\infty=0.

Proof. First, assume q<\infty. Then by definition of weak L^{p}-norm, we have

    \[ \lambda_{f}\left(\alpha\right)\le\min\left(\frac{\norm f{p,\infty}^{p}}{\alpha^{p}},\frac{\norm f{q,\infty}^{q}}{\alpha^{q}}\right). \]

Set

    \[ B=\left(\frac{\norm f{q,\infty}^{q}}{\norm f{p,\infty}^{p}}\right)^{\frac{1}{q-p}}. \]

We estimate \norm fr as follow:

    \begin{align*} \norm fr^{r} & =r\int_{0}^{\infty}\alpha^{r-1}\lambda_{f}\left(\alpha\right)d\alpha\\ & \le r\int_{0}^{\infty}\alpha^{r-1}\min\left(\frac{\norm f{p,\infty}^{p}}{\alpha^{p}},\frac{\norm f{q,\infty}^{q}}{\alpha^{q}}\right)d\alpha\\ & \le r\int_{0}^{B}\alpha^{r-p-1}\norm f{p,\infty}^{p}d\alpha+r\int_{B}^{\infty}\alpha^{r-q-1}\norm f{q,\infty}^{q}d\alpha\\ & =\frac{r}{r-p}\norm f{p,\infty}^{p}B^{r-p}+\frac{r}{q-r}\norm f{q,\infty}^{q}B^{r-q}\\ & =\left(\frac{r}{r-p}+\frac{r}{q-r}\right)\left(\norm f{p,\infty}^{p}\right)^{\frac{q-r}{q-p}}\left(\norm f{p,\infty}^{q}\right)^{\frac{r-p}{q-p}}. \end{align*}

So taking \sqrt[r]{\cdot} and calculate some minor things, we get

    \[ \norm fr\le\left(\frac{r}{r-p}+\frac{r}{q-r}\right)^{\frac{1}{r}}\norm f{p,\infty}^{\frac{\frac{1}{r}-\frac{1}{q}}{\frac{1}{p}-\frac{1}{q}}}\norm f{q,\infty}^{\frac{\frac{1}{p}-\frac{1}{r}}{\frac{1}{p}-\frac{1}{q}}}. \]

3. The Complex Interpolation Method: The Riesz-Thorine Interpolation Theorem

Theorem (The Riesz-Thorine Interpolation Theorem) Let {\left(X,\mu\right)} and {\left(Y,\nu\right)} be two {\sigma}-finite measure spaces. Let {T} be a linear operator defined on the set of all finitely simple functions on {X} and taking values in the set of measurable functions on {Y}. Let {1\le p_{0},p_{1},q_{0},q_{1}\le\infty} and assume that

    \begin{align*} \norm{T\left(f\right)}{q_{0},Y} & \le M_{0}\norm f{p_{0},X},<br/>  \norm{T\left(f\right)}{q_{1},Y} & \le M_{1}\norm f{p_{1},X}, \end{align*}

for all finitely simple functions {f} on {X}. Then for all {0<\theta<1}, we have

\displaystyle  \norm{T\left(f\right)}{q_{\theta},Y}\le M_{\theta}\norm f{p_{\theta},X}

for all finitely simple functions {f} on {X}, where

\displaystyle  \frac{1}{p_{\theta}}=\frac{1-\theta}{p_{0}}+\frac{\theta}{p_{1}}\quad\text{and}\quad\frac{1}{q_{\theta}}=\frac{1-\theta}{q_{0}}+\frac{\theta}{q_{1}}.

Consequently, when {p_{\theta}<\infty}, by density, {T} has a unique bounded extension from {L^{p_{\theta}}\left(X,\mu\right)} to {L^{q_{\theta}}\left(Y,\nu\right)}.

Proof: We divide our proof into three steps.

Step 1. Observe that for {p_{0}=p_{1}}, it is clear by Theorem

\displaystyle  \norm{Tf}{q_{\theta}}\le\norm{Tf}{q_{0}}^{1-\theta}\norm{Tf}{q_{1}}^{\theta}\le M_{0}^{1-\theta}M_{1}^{\theta}\norm fp.

So we may assume {p_{0}\neq p_{1}} and we further assume that {p_{\theta}<\infty} for {0<\theta<1}.

Let {\Sigma_{X}} denote the space of all finite simple function on {X}. Then clearly, {\Sigma_{X}\subset L^{p}\left(X\right)} and {\overline{\Sigma_{X}}=L^{p}\left(X\right)}. So is {\Sigma_{Y}}. We will show that for any {f\in\Sigma_{X}}, we have

\displaystyle  \norm{Tf}{q_{\theta}}\le M_{0}^{1-\theta}M_{1}^{\theta}\norm f{p_{\theta}}.

By considering the dual space of {L^{q_{\theta}}\left(Y\right)}, we have

\displaystyle  \norm{Tf}{q_{\theta}}=\sup\left\{ \left|\int\left(Tf\right)gd\nu\right|:g\in\Sigma_{Y},\,\norm g{q_{\theta}^{\prime}}=1\right\}

where {q_{\theta}^{\prime}} is a conjugate exponent of {q_{\theta}}.

We further assume that {f\neq0} with {\norm f{p_{\theta}}=1}. We will show the following claim in Step 2:

  • If {f\in\Sigma_{X}} and {\norm f{p_{\theta}}=1}, then {\left|\int\left(Tf\right)gd\nu\right|\le M_{0}^{1-\theta}M_{1}^{\theta}} for all {g\in\Sigma_{Y}} such that {\norm g{q_{\theta}^{\prime}}=1}.

Step 2. Let {f\in\Sigma_{X}} and {g\in\Sigma_{Y}} with {\norm f{p_{\theta}}=1} and {\norm g{q_{\theta}^{\prime}}=1}. Let {f=\sum_{j=1}^{m}c_{j}\chi_{E_{j}}} and {g=\sum_{k=1}^{n}d_{k}\chi_{F_{k}}} be the canonical representation. Write {c_{j}=\left|c_{j}\right|e^{i\theta_{j}}}, {d_{k}=\left|d_{k}\right|e^{i\psi_{k}}}. Also, let

\displaystyle  \alpha\left(z\right)=\left(1-z\right)p_{0}^{-1}+zp_{1}^{-1},\quad\beta\left(z\right)=\left(1-z\right)q_{0}^{-1}+zq_{1}^{-1};

thus {\alpha\left(\theta\right)=p_{\theta}^{-1}} and {\beta\left(\theta\right)=q_{\theta}^{-1}} for {0<\theta<1}.

Fix {0<\theta<1}. Since {p_{\theta}<\infty}, {\alpha\left(\theta\right)>0}. So we may define

\displaystyle  f_{z}=\sum_{1}^{m}\left|c_{j}\right|^{\frac{\alpha\left(z\right)}{\alpha\left(\theta\right)}}e^{i\theta_{j}}\chi_{E_{j}}.

If {\beta\left(\theta\right)<1}, we define

\displaystyle  g_{z}=\sum_{1}^{n}\left|d_{k}\right|^{\frac{1-\beta\left(z\right)}{1-\beta\left(\theta\right)}}e^{i\psi_{k}}\chi_{F_{k}},

while if {\beta\left(\theta\right)=1}, we define {g_{z}=g} for all {z}.

We first show our claim when {\beta\left(\theta\right)<1}. Finally, we set

\displaystyle  \phi\left(z\right)=\int\left(Tf_{z}\right)g_{z}d\nu.

Thus,

\displaystyle  \phi\left(z\right)=\sum_{j,k}A_{j,k}\left|c_{j}\right|^{\frac{\alpha\left(z\right)}{\alpha\left(\theta\right)}}\left|d_{k}\right|^{\frac{1-\beta\left(z\right)}{1-\beta\left(\theta\right)}}

where

\displaystyle  A_{j,k}=e^{i\left(\theta_{j}+\psi_{k}\right)}\int\left(T\chi_{E_{j}}\right)\chi_{F_{k}}d\nu,

so that {\phi} is an entire holomorphic function of {z} that is bounded in the strip {0\le\mathrm{Re}\left(z\right)\le1}. Since {\phi\left(\theta\right)=\int\left(Tf\right)gd\nu}, by Three lines lemma, it suffices to show {\left|\phi\left(z\right)\right|\le M_{0}} for {\mathrm{Re}\left(z\right)=0} and {\left|\phi\left(z\right)\right|\le M_{1}} for {\mathrm{Re}\left(z\right)=1} to show the claim.

First, note that

    \[ \alpha\left(is\right)=\frac{1}{p_{0}}+is\left(\frac{1}{p_{1}}-\frac{1}{p_{0}}\right),\quad1-\beta\left(is\right)=\left(1-\frac{1}{q_{0}}\right)-is\left(\frac{1}{q_{1}}-\frac{1}{q_{0}}\right) \]

for s\in\mathbb{R}. This shows

    \begin{align*} \left|f_{is}\right| & =\left|f\right|^{\mathrm{Re}\left[\alpha\left(is\right)/\alpha\left(\theta\right)\right]}=\left|f\right|^{p_{\theta}/p_{0}},\\ \left|g_{is}\right| & =\left|g\right|^{\mathrm{Re}\left[\left(1-\beta\left(is\right)\right)/\left(1-\beta\left(\theta\right)\right)\right]}=\left|g\right|^{q_{\theta}^{\prime}/q_{\theta}}. \end{align*}

Therefore, by Hölder’s inequality, we have

    \begin{align*} \left|\phi\left(is\right)\right| & \le\norm{Tf_{is}}{q_{0}}\norm{g_{is}}{q_{0}^{\prime}}\le M_{0}\norm{f_{is}}{p_{0}}\norm{g_{is}}{q_{0}^{\prime}}\\  & =M_{0}\norm f{p_{\theta}}\norm g{q_{\theta}^{\prime}}=M_{0}. \end{align*}

Similarly, we get \left|\phi\left(1+is\right)\right|\le M_{1}.

The case \beta\left(\theta\right)=1 is similar.

So we show the claim.

Step 3. From Step 2, we have shown that {\norm{Tf}{q_{\theta}}\le M_{0}^{1-\theta}M_{1}^{\theta}\norm f{p_{\theta}}}for {f\in\Sigma_{X}}. Then {T\mid_{\Sigma_{X}}} has a unique extension to {L^{p_{\theta}}\left(X\right)} satisfying the same estimate there. It remains to show that this extension is {T} itself.

Given {f\in L^{p_{\theta}}\left(X\right)}, choose a sequence {\left\{ f_{n}\right\} } in {\Sigma_{X}} such that {\left|f_{n}\right|\le\left|f\right|} and {f_{n}\rightarrow f} pointwise. Also, let {E=\left\{ x:\left|f\left(x\right)\right|>1\right\} }, {g=f\chi_{E}}, {g_{n}=f_{n}\chi_{E}}, {h=f-g}, and {h_{n}=f_{n}-g_{n}}. Then if {p_{0}<p_{1}}, we have {g\in L^{p_{0}}\left(X\right)}, {h\in L^{p_{1}}\left(X\right)} and by DCT, {\norm{f_{n}-f}{p_{\theta}}\rightarrow0}, {\norm{g_{n}-g}{p_{0}}\rightarrow0}, and {\norm{h_{n}-h}{p_{1}}\rightarrow0}. Hence, {\norm{Tg_{n}-Tg}{q_{0}}\rightarrow0} and {\norm{Th_{n}-Th}{q_{1}}\rightarrow0}. So by passing to a suitable subsequence, we may assume that {Tg_{n}\rightarrow Tg} a.e. and {Th_{n}\rightarrow Th} a.e. But then {Tf_{n}\rightarrow Tf} a.e. So by Fatou’s lemma, we get

\displaystyle  \norm{Tf}{q_{\theta}}\le\liminf\norm{Tf_{n}}{q_{\theta}}\le\liminf M_{0}^{1-\theta}M_{1}^{\theta}\norm{f_{n}}{p_{\theta}}=M_{0}^{1-\theta}M_{1}^{\theta}\norm f{p_{\theta}},

and we are done. \Box

4. The Real Interpolation Method: The Marcinkiewicz Interpolation Theorem

Definition 1 Let {\left(X,\mu\right)} and {\left(Y,\nu\right)} be two measure spaces. Suppose we are given an operator {T:L^{p}\left(X,\mu\right)\rightarrow L^{q}\left(Y,\nu\right)}. Operators that map {L^{p}} to {L^{q}} are called of strong type {\left(p,q\right)} and operators that map {L^{p}} to {L^{q,\infty}} are called weak type {\left(p,q\right)}.

Definition Let {T} be an operator defined on a vector space of complex-valued measurable functions on a measure space {\left(X,\mu\right)} and taking values in the set of all complex-valued finite almost everywhere measurable functions on a measure space {\left(Y,\nu\right)}. Then {T} is called linear if for all {f,g} in the domain of {T} and all {\lambda\in\mathbb{C}} we have

\displaystyle  T\left(f+g\right)=T\left(f\right)+T\left(g\right)\quad\text{and}\quad T\left(\lambda f\right)=\lambda T\left(f\right).

{T} is called sublinear if for all {f,g} in the domain of {T} and all {\lambda\in\mathbb{C}} we have

\displaystyle  \left|T\left(f+g\right)\right|\le\left|T\left(f\right)\right|+\left|T\left(g\right)\right|\quad\text{and}\quad\left|T\left(\lambda f\right)\right|=\left|\lambda\right|\left|T\left(f\right)\right|.

{T} is called quasi-linear if for all {f,g} in the domain of {T} and all {\lambda\in\mathbb{C}} we have

\displaystyle  \left|T\left(f+g\right)\right|\le K\left(\left|T\left(f\right)\right|+\left|T\left(g\right)\right|\right)\quad\text{and}\quad\left|T\left(\lambda f\right)\right|=\left|\lambda\right|\left|T\left(f\right)\right|

for some constant {K>0}.

We are now ready to state the Marcinkiewicz Interpolation Theorem.

Theorem (Marcinkiewicz Interpolation Theorem) Let {\left(X,\mu\right)} be a {\sigma}-finite measure space, let {\left(Y,\nu\right)} be another measure space, and let {0<p_{0}<p_{1}\le\infty}. Let {T} be a sublinear operator defined on {L^{p_{0}}\left(X\right)+L^{p_{1}}\left(X\right)} and taking values in the space of measurable functions on {Y}. Assume that there exist {A_{0},A_{1}<\infty} such that

    \begin{align*} \norm{T\left(f\right)}{L^{p_{0},\infty}\left(Y\right)} & \le A_{0}\norm f{L^{p_{0}}\left(X\right)}\quad\text{for all }f\in L^{p_{0}}\left(X\right),<br/>  \norm{T\left(f\right)}{L^{p_{1},\infty}\left(Y\right)} & \le A_{1}\norm f{L^{p_{1}}\left(X\right)}\quad\text{for all }f\in L^{p_{1}}\left(X\right). \end{align*}

Then for all {p_{0}<p<p_{1}} and for all {f\in L^{p}\left(X\right)}, we have the estimate

\displaystyle  \norm{T\left(f\right)}{L^{p}\left(Y\right)}\le A\norm f{L^{p}\left(X\right)},

where

\displaystyle  A=2\left(\frac{p}{p-p_{0}}+\frac{p}{p_{1}-p}\right)^{\frac{1}{p}}A_{0}^{\frac{\frac{1}{p}-\frac{1}{p_{0}}}{\frac{1}{p_{0}}-\frac{1}{p_{1}}}}A_{1}^{\frac{\frac{1}{p_{0}}-\frac{1}{p}}{\frac{1}{p_{0}}-\frac{1}{p_{1}}}}.

Proof: We divide our proof into two cases: when {p_{1}<\infty} and {p_{1}=\infty}.

Case 1. {p_{1}<\infty}

Let {f\in L^{p}\left(X\right)} and let {\alpha>0}. Split {f=f_{0}^{\alpha}+f_{1}^{\alpha}} to be {f_{0}^{\alpha}\in L^{p_{0}}\left(X\right)}, {f_{1}^{\alpha}\in L^{p_{1}}\left(X\right)} by defining

\displaystyle  f_{0}^{\alpha}\left(x\right)=\begin{cases} f\left(x\right) & \text{for }\left|f\left(x\right)\right|>\delta\alpha,\\ 0 & \text{for }\left|f\left(x\right)\right|\le\delta\alpha, \end{cases}

and

\displaystyle  f_{1}^{\alpha}\left(x\right)=\begin{cases} f\left(x\right) & \text{for }\left|f\left(x\right)\right|\le\delta\alpha,\\ 0 & \text{for }\left|f\left(x\right)\right|>\delta\alpha. \end{cases}

The constant {\delta>0} will be determined later.

NNote that p_{0}<p<p_{1} and we estimate f_{0}^{\alpha}, f_{1}^{\alpha}
as follows:

    \begin{align*} \norm{f_{0}^{\alpha}}{p_{0}}^{p_{0}} & =\int_{X}\left|f_{0}^{\alpha}\right|^{p_{0}}d\mu\\  & \le\int_{\left|f\right|>\delta\alpha}\left|f\right|^{p_{0}-p}\left|f\right|^{p}d\mu\\  & \le\left(\delta\alpha\right)^{p_{0}-p}\norm fp^{p}. \end{align*}

Similarly,

    \begin{align*} \norm{f_{1}^{\alpha}}{p_{1}}^{p_{1}} & =\int_{X}\left|f_{1}^{\alpha}\right|^{p_{1}}d\mu\\  & =\int_{\left|f\right|>\delta\alpha}\left|f\left(x\right)\right|^{p_{1}-p}\left|f\left(x\right)\right|^{p}d\mu\\  & \le\left(\delta\alpha\right)^{p_{1}-p}\norm fp^{p}. \end{align*}

Since T is sublinear by subadditivity we get

    \[ \left|T\left(f\right)\right|\le\left|T\left(f_{0}^{\alpha}\right)\right|+\left|T\left(f_{1}^{\alpha}\right)\right|. \]

So from this,

    \begin{align*} \lambda_{T\left(f\right)}\left(\alpha\right) & \le\lambda_{T\left(f_{0}^{\alpha}\right)}\left(\frac{\alpha}{2}\right)+\lambda_{T\left(f_{1}^{\alpha}\right)}\left(\frac{\alpha}{2}\right). \end{align*}

So by definition of weak L^{p} norm, we get

    \begin{align*} \lambda_{T\left(f_{0}^{\alpha}\right)}\left(\frac{\alpha}{2}\right) & \le\frac{\norm{T\left(f_{0}^{\alpha}\right)}{L^{p_{0},\infty}\left(Y\right)}^{p_{0}}}{\left(\alpha/2\right)^{p_{0}}}\le\frac{\left(2A_{0}\right)^{p_{0}}}{\alpha^{p}}\norm{f_{0}^{\alpha}}{L^{p}\left(X\right)}^{p_{0}}\\  & =\frac{\left(2A_{0}\right)^{p_{0}}}{\alpha^{p_{0}}}\int_{\left|f\right|>\delta\alpha}\left|f\left(x\right)\right|^{p_{0}}d\mu\left(x\right). \end{align*}

Similarly, we get

    \[ \lambda_{T\left(f_{1}^{\alpha}\right)}\left(\frac{\alpha}{2}\right)\le\frac{\left(2A_{1}\right)^{p_{1}}}{\alpha^{p_{1}}}\int_{\left|f\right|\le\delta\alpha}\left|f\left(x\right)\right|^{p_{1}}d\mu\left(x\right). \]

Finally, using Fubini’s theorem since \left(X,\mu\right) is \sigma-finite,
we have

    \begin{align*} \norm{T\left(f\right)}{L^{p}\left(X\right)}^{p} & =p\int_{0}^{\infty}\alpha^{p-1}\lambda_{T\left(f\right)}\left(\alpha\right)d\alpha\\  & \le p\left(2A_{0}\right)^{p_{0}}\int_{0}^{\infty}\alpha^{p-1-p_{0}}\int_{\left|f\right|>\delta\alpha}\left|f\left(x\right)\right|^{p_{0}}d\mu\left(x\right)d\alpha\\  & \relphantom{=}+p\left(2A_{1}\right)^{p_{1}}\int_{0}^{\infty}\alpha^{p-1-p_{1}}\int_{\left|f\right|\le\delta\alpha}\left|f\left(x\right)\right|^{p_{1}}d\mu\left(x\right)d\alpha\\  & =p\left(2A_{0}\right)^{p_{0}}\int_{X}\left|f\left(x\right)\right|^{p_{0}}\int_{0}^{\left|f\left(x\right)\right|/\delta}\alpha^{p-1-p_{0}}d\alpha d\mu\left(x\right)\\  & \relphantom{=}+p\left(2A_{1}\right)^{p_{1}}\int_{X}\left|f\left(x\right)\right|^{p_{1}}\int_{\left|f\left(x\right)\right|/\delta}^{\infty}\alpha^{p-1-p_{1}}d\alpha d\mu\left(x\right)\\  & =\left(\frac{\left(2A_{0}\right)^{p_{0}}}{p-p_{0}}\frac{p}{\delta^{p-p_{0}}}+\frac{\left(2A_{1}\right)^{p_{1}}}{p_{1}-p}\frac{p}{\delta^{p-p_{1}}}\right)\norm fp^{p}. \end{align*}

Choose \delta>0 so that

    \[ \left(2A_{0}\right)^{p}\frac{1}{\delta^{p-p_{0}}}=\left(2A_{1}\right)^{p_{1}}\delta^{p_{1}-p}. \]

Then we get

    \[ \norm{T\left(f\right)}{L^{p}\left(Y\right)}\le A\norm f{L^{p}\left(X\right)}, \]

where

    \[ A=2\left(\frac{p}{p-p_{0}}+\frac{p}{p_{1}-p}\right)^{\frac{1}{p}}A_{0}^{\frac{\frac{1}{p}-\frac{1}{p_{0}}}{\frac{1}{p_{0}}-\frac{1}{p_{1}}}}A_{1}^{\frac{\frac{1}{p_{0}}-\frac{1}{p}}{\frac{1}{p_{0}}-\frac{1}{p_{1}}}}. \]

Finally, using Fubini’s theorem since {\left(X,\mu\right)} is {\sigma}-finite, we have

    \begin{align*} \norm{T\left(f\right)}{L^{p}\left(X\right)}^{p} & =p\int_{0}^{\infty}\alpha^{p-1}\lambda_{T\left(f\right)}\left(\alpha\right)d\alpha\\  & \le p\left(2A_{0}\right)^{p_{0}}\int_{0}^{\infty}\alpha^{p-1-p_{0}}\int_{\left|f\right|>\delta\alpha}\left|f\left(x\right)\right|^{p_{0}}d\mu\left(x\right)d\alpha\\  & \relphantom{=}+p\left(2A_{1}\right)^{p_{1}}\int_{0}^{\infty}\alpha^{p-1-p_{1}}\int_{\left|f\right|\le\delta\alpha}\left|f\left(x\right)\right|^{p_{1}}d\mu\left(x\right)d\alpha\\  & =p\left(2A_{0}\right)^{p_{0}}\int_{X}\left|f\left(x\right)\right|^{p_{0}}\int_{0}^{\left|f\left(x\right)\right|/\delta}\alpha^{p-1-p_{0}}d\alpha d\mu\left(x\right)\\  & \relphantom{=}+p\left(2A_{1}\right)^{p_{1}}\int_{X}\left|f\left(x\right)\right|^{p_{1}}\int_{\left|f\left(x\right)\right|/\delta}^{\infty}\alpha^{p-1-p_{1}}d\alpha d\mu\left(x\right)\\  & =\left(\frac{\left(2A_{0}\right)^{p_{0}}}{p-p_{0}}\frac{p}{\delta^{p-p_{0}}}+\frac{\left(2A_{1}\right)^{p_{1}}}{p_{1}-p}\frac{p}{\delta^{p-p_{1}}}\right)\norm fp^{p}. \end{align*}

Choose {\delta>0} so that

\displaystyle  \left(2A_{0}\right)^{p}\frac{1}{\delta^{p-p_{0}}}=\left(2A_{1}\right)^{p_{1}}\delta^{p_{1}-p}.

Then we get

\displaystyle  \norm{T\left(f\right)}{L^{p}\left(Y\right)}\le A\norm f{L^{p}\left(X\right)},

where

\displaystyle  A=2\left(\frac{p}{p-p_{0}}+\frac{p}{p_{1}-p}\right)^{\frac{1}{p}}A_{0}^{\frac{\frac{1}{p}-\frac{1}{p_{0}}}{\frac{1}{p_{0}}-\frac{1}{p_{1}}}}A_{1}^{\frac{\frac{1}{p_{0}}-\frac{1}{p}}{\frac{1}{p_{0}}-\frac{1}{p_{1}}}}.

Case 2. {p_{1}=\infty}

Write f=f_{0}^{\alpha}+f_{1}^{\alpha}, where

    \begin{align*} f_{0}^{\alpha}\left(x\right) & =\begin{cases} f\left(x\right) & \text{for }\left|f\left(x\right)\right|>\gamma\alpha,\\ 0 & \text{for }\left|f\left(x\right)\right|\le\gamma\alpha, \end{cases}\\ f_{1}^{\alpha}\left(x\right) & =\begin{cases} f\left(x\right) & \text{for }\left|f\left(x\right)\right|\le\gamma\alpha,\\ 0 & \text{for }\left|f\left(x\right)\right|>\gamma\alpha. \end{cases} \end{align*}

We have

\displaystyle  \norm{T\left(f_{1}^{\alpha}\right)}{L^{\infty}\left(Y\right)}\le A_{1}\norm{f_{1}^{\alpha}}{L^{\infty}\left(X\right)}\le\gamma\alpha=\frac{\alpha}{2}

where provided we choose {\gamma=\left(2A_{1}\right)^{-1}}. So {\left\{ x\in X:\left|T\left(f_{1}^{\alpha}\right)\left(x\right)\right|>\frac{\alpha}{2}\right\} } has measure zero. Therefore,

\displaystyle  \lambda_{T\left(f\right)}\left(\alpha\right)\le\lambda_{T\left(f_{0}^{\alpha}\right)}\left(\frac{\alpha}{2}\right).

Since {T} maps {L^{p_{0}}} to {L^{p_{0},\infty}} with norm at most {A_{0}}, it follows that

    \begin{align*} \lambda_{T\left(f_{0}^{\alpha}\right)}\left(\frac{\alpha}{2}\right) & \le\frac{\left(2A_{0}\right)^{p_{0}}\norm{f_{0}^{\alpha}}{p_{0}}^{p_{0}}}{\alpha^{p_{0}}}\\  & =\frac{\left(2A_{0}\right)^{p_{0}}}{\alpha^{p_{0}}}\int_{\left|f\right|>\gamma\alpha}\left|f\left(x\right)\right|^{p_{0}}d\mu\left(x\right). \end{align*}

Therefore, we obtain

    \begin{align*} \norm{T\left(f\right)}p^{p} & =p\int_{0}^{\infty}\alpha^{p-1}\lambda_{T\left(f\right)}\left(\alpha\right)d\alpha\\  & \le p\int_{0}^{\infty}\alpha^{p-1}\frac{\left(2A_{0}\right)^{p_{0}}}{\alpha^{p_{0}}}\int_{\left|f\right|>\alpha/\left(2A_{1}\right)}\left|f\left(x\right)\right|^{p_{0}}d\mu\left(x\right)d\alpha\\  & =p\left(2A_{0}\right)^{p_{0}}\int_{X}\left|f\left(x\right)\right|^{p_{0}}\int_{0}^{2A_{1}\left|f\left(x\right)\right|}\alpha^{p-p_{0}-1}d\alpha d\mu\left(x\right)\\  & =\frac{p\left(2A_{1}\right)^{p-p_{0}}\left(2A_{0}\right)^{p_{0}}}{p-p_{0}}\int_{X}\left|f\left(x\right)\right|^{p}d\mu\left(x\right). \end{align*}

This proves the theorem with constant

\displaystyle  A=2\left(\frac{p}{p-p_{0}}\right)^{\frac{1}{p}}A_{1}^{1-\frac{p_{0}}{p}}A_{0}^{\frac{p_{0}}{p}}.\qedhere

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