Marcinkiewicz space

Question

Let $\Omega\subset\mathbb{R}^N, N\geq2$ be a bounded open subset and suppose that $0<p_0<p<p_1<\infty$ and $$\dfrac1p=\dfrac{1-\theta}{p_0}+\dfrac{\theta}{p_1}\quad \forall \theta \in[0,1].$$

Question. How to prove that $$\|f\|_{M^p}\leq \|f\|^{1-\theta}_{{M}^{p_0}}\|f\|^{\theta}_{{M}^{p_1}}, \quad \forall f\in M^{p_0}(\Omega)\cap M^{p_1}(\Omega)\;?$$

Here $M^p(\Omega)$ is the Marcinkiewicz space i.e. the set of measurable functions $f$ satisfying the following inequality

$$\Phi_f(k)<c k^{-p}\quad \forall c>0,$$ where $\Phi_f(k)=\operatorname{mes}\{x\in \Omega, |f(x)|>k\}$ for all $k>0$, endowed by the norm

$$\|f\|_{M^p}=\inf\{c, \Phi_f(k)\leq c k^{-p}, \forall k>0\}.$$

Edit: I don't know how to work with this norm. Can I use integrals?

I can prove that $\|f\|_{L^p}\leq \|f\|^{1-\theta}_{L^{p_0}} \|f\|^{\theta}_{L^{p_1}}$ but i just have $L^p\subsetneq M^p$.

Answer 1

Since $f$ is Marcienkiewicz $p_0$ and $p_1$, there are constants $M_{p_0}=\|f\|_{M^{p_0}}$ and $M_{p_1}=\|f\|_{M^{p_1}}$ so that $$ \mu\{x:|f(x)|\gt\alpha\}\le\min\left\{\frac{M_{p_0}}{\alpha^{p_0}},\frac{M_{p_1}}{\alpha^{p_1}}\right\}\tag1 $$

---

$M^p\subset M^{p_0}\cap M^{p_1}$: $$ \begin{align} \|f\|_{M^p} &=\sup_\alpha\mu\{x:|f(x)|\gt\alpha\}\alpha^p\tag{2a}\\ &\le\sup_\alpha\min\left\{M_{p_0}\alpha^{p-p_0},M_{p_1}\alpha^{p-p_1}\right\}\tag{2b}\\ &=M_{p_0}^{\frac{p_1-p}{p_1-p_0}}M_{p_1}^{\frac{p-p_0}{p_1-p_0}}\tag{2c}\\ \end{align} $$ Explanation:
$\text{(2a):}$ compute the Marcinkiewicz norm of $f$
$\text{(2b):}$ apply $(1)$
$\text{(2c):}$ the sup occurs at $\alpha=\left(\frac{M_{p_1}}{M_{p_0}}\right)^{\frac1{p_1-p_0}}$

This is the bound requested in the question. However, note that if $\theta=\frac{p-p_0}{p_1-p_0}$ and $1-\theta=\frac{p_1-p}{p_1-p_0}$, then we have $$ \begin{align} (1-\theta)p_0+\theta p_1 &=\frac{p_1-p}{p_1-p_0}p_0+\frac{p-p_0}{p_1-p_0}{p_1}\tag{3a}\\ &=p\tag{3b} \end{align} $$

---

Furthermore, $L^p\subset M^{p_0}\cap M^{p_1}$: $$ \begin{align} \|f\|_p &=p\int_0^\infty\mu\{x:|f(x)|\gt\alpha\}\alpha^{p-1}\,\mathrm{d}\alpha\tag{4a}\\ &\le p\int_0^sM_{p_0}\alpha^{p-p_0-1}\,\mathrm{d}\alpha+p\int_s^\infty M_{p_1}\alpha^{p-p_1-1}\,\mathrm{d}\alpha\tag{4b}\\ &=p\frac{M_{p_0}s^{p-p_0}}{p-p_0}+p\frac{M_{p_1}s^{p-p_1}}{p_1-p}\tag{4c}\\ &=p\frac{M_{p_0}^{\frac{p_1-p}{p_1-p_0}}M_{p_1}^{\frac{p-p_0}{p_1-p_0}}}{p-p_0}+p\frac{M_{p_0}^{\frac{p_1-p}{p_1-p_0}}M_{p_1}^{\frac{p-p_0}{p_1-p_0}}}{p_1-p}\tag{4d}\\ &=\frac{p(p_1-p_0)}{(p_1-p)(p-p_0)}M_{p_0}^{\frac{p_1-p}{p_1-p_0}}M_{p_1}^{\frac{p-p_0}{p_1-p_0}}\tag{4e} \end{align} $$ Explanation:
$\text{(4a):}$ write $L^p$ norm in terms of measures
$\text{(4b):}$ apply $(1)$ while splitting the integral at $s$ so that both integrals converge
$\text{(4c):}$ evaluate the integrals
$\text{(4d):}$ $s=\left(\frac{M_{p_1}}{M_{p_0}}\right)^{\frac1{p_1-p_0}}$ minimizes the sum
$\text{(4e):}$ simplify