Weak $L^q$-spaces and Lebesgue measure

Question

Let $q>1$ and $f: \mathbb R^n \rightarrow \mathbb R$ Lebesgue measurable. Let $||f||_{w, q} = \sup \lambda_n (A)^{-1/q'} \int_A |f| d \lambda_n$, where the supremum is taken over all $A \in \mathcal M_{\lambda^{*}_n} $ with $\lambda_n(A) < \infty$, and $1/q+1/q' =1$.

Why does it then hold true that $$ < f >_{w,q} \ \leq \ ||f||_{w,q} \ \leq \ \dfrac{q}{q-1} < f >_{w,q}$$

where $ < f >_{w,q} := \sup \alpha \lambda_n ($ { $ x \in \mathbb R^n: |f(x)| > \alpha $ } $)^{1/q} < \infty$.

Answer 1

The following exercises come from the book of Loukas Grafakos, Classical Fourier Analysis, and your question is a special of them.

Let $(X,\mu)$ be a measure space and let $E$ be a subset of $X$ with $\mu(E)<\infty$. Assume that $f$ is in $L^{p,\infty}(X,\mu)$ for some $0<p<\infty$. Show that for $0<q<p$ we have \begin{align*} \int_{E}|f(x)|^{q}d\mu(x)\leq\dfrac{p}{p-q}\mu(E)^{1-q/p}\|f\|_{L^{p,\infty}}^{q}. \end{align*}

Here $L^{p,\infty}(X,\mu)$ is the set of all measurable functions such that $\sup_{\alpha>0}\alpha\mu(x\in X:|f(x)|>\alpha)^{1/p}<\infty$.

Here is the proof.

First of all, we note that \begin{align*} \min(\mu(E),\alpha^{-p}\|f\|_{L^{p,\infty}}^{p})=\mu(E) \end{align*} if and only if \begin{align*} \mu(E)&\leq\alpha^{-p}\|f\|_{L^{p,\infty}}^{p}\\ \alpha&\leq\mu(E)^{-1/p}\|f\|_{L^{p,\infty}}. \end{align*} Likewise, \begin{align*} \min(\mu(E),\alpha^{-p}\|f\|_{L^{p,\infty}}^{p})=\alpha^{-p}\|f\|_{L^{p,\infty}}^{p} \end{align*} if and only if \begin{align*} \alpha\geq\mu(E)^{-1/p}\|f\|_{L^{p,\infty}}. \end{align*} Now, we write that \begin{align*} \int_{E}|f(x)|^{q}d\mu(x)&=q\int_{0}^{\infty}\alpha^{q-1}\mu(E\cap(|f|>\alpha))d\alpha\\ &=q\left(\int_{0}^{\mu(E)^{-1/p}\|f\|_{L^{p,\infty}}}+\int_{\mu(E)^{-1/p}\|f\|_{L^{p,\infty}}}^{\infty}\right)\alpha^{q-1}\mu(E\cap(|f|>\alpha))d\alpha\\ &=I_{1}+I_{2}. \end{align*} We have $\mu(E\cap(|f|>\alpha))\leq\min(\mu(E),\alpha^{-p}\|f\|_{L^{p,\infty}}^{p})$ and hence \begin{align*} I_{1}=q\int_{0}^{\mu(E)^{-1/p}\|f\|_{L^{p,\infty}}}\alpha^{q-1}\mu(E)d\alpha=\mu(E)\left(\mu(E)^{-1/p}\|f\|_{L^{p,\infty}}\right)^{q}=\mu(E)^{1-q/p}\|f\|_{L^{p,\infty}}^{q}. \end{align*} Also, \begin{align*} I_{2}&=q\int_{\mu(E)^{-1/p}\|f\|_{L^{p,\infty}}}^{\infty}\alpha^{q-1}\alpha^{-p}\|f\|_{L^{p,\infty}}^{p}d\alpha\\ &=-\dfrac{q}{q-p}\|f\|_{L^{p,\infty}}^{p}\left(\mu(E)^{-1/p}\|f\|_{L^{p,\infty}}\right)^{q-p}\\ &=-\dfrac{q}{q-p}\mu(E)^{1-q/p}\|f\|_{L^{p,\infty}}^{q}. \end{align*} As a result, \begin{align*} I_{1}+I_{2}\leq\left(1-\dfrac{q}{q-p}\right)\mu(E)^{1-q/p}\|f\|_{L^{p,\infty}}^{q}=\dfrac{p}{p-q}\mu(E)^{1-q/p}\|f\|_{L^{p,\infty}}^{q}, \end{align*} we are done.

Here is the second exercise.

Let $(X,\mu)$ be a $\sigma$-finite measure space and let $0<p<\infty$. Pick $0<r<p$ and define \begin{align*} |||f|||_{L^{p,\infty}}=\sup_{0<\mu(E)<\infty}\mu(E)^{-1/r+1/p}\left(\int_{E}|f|^{r}d\mu\right)^{1/r}, \end{align*} where the supremum is taken over all measurable subsets $E$ of $X$ of finite measure. Prove that for all $f\in L^{p,\infty}(X,\mu)$ we have \begin{align*} \|f\|_{L^{p,\infty}}\leq|||f|||_{L^{p,\infty}}. \end{align*}

Here is the proof.

We write \begin{align*} X=\bigcup_{k}X_{k},~~~~\mu(X_{k})<\infty, \end{align*} where $(X_{k})$ is an increasing sequence of sets.

We have \begin{align*} \|f\|_{L^{p,\infty}}=\sup_{\alpha>0}\alpha\mu(x\in X:|f(x)|>\alpha)^{1/p}=\sup_{\alpha>0}\lim_{k\rightarrow\infty}\alpha\mu(X_{k}\cap(|f|>\alpha))^{1/p}. \end{align*} Let $\alpha>0$ and $k\in\mathbb{N}$ be given. We can assume without loss of generality that $\mu(X_{k}\cap(|f|>\alpha))>0$ and hence \begin{align*} &\alpha^{p}\mu(X_{k}\cap(|f|>\alpha))\\ &=\alpha^{p-r}\int_{X_{k}\cap(|f|>\alpha)}\alpha^{r}\chi_{(|f|>\alpha)}(x)d\mu(x)\\ &\leq\alpha^{p-r}\int_{X_{k}\cap(|f|>\alpha)}|f(x)|^{r}d\mu(x)\\ &=\alpha^{p-r}\mu(X_{k}\cap(|f|>\alpha))^{1-r/p}\mu(X_{k}\cap(|f|>\alpha))^{-1+r/p}\int_{X_{k}\cap(|f|>\alpha)}|f(x)|^{r}d\mu(x)\\ &\leq\alpha^{p-r}\mu(X_{k}\cap(|f|>\alpha))^{1-r/p}|||f|||_{L^{p,\infty}}^{r}\\ &=\left(\alpha\mu(X_{k}\cap(|f|>\alpha))^{1/p}\right)^{p-r}|||f|||_{L^{p,\infty}}^{r}\\ &\leq\|f\|_{L^{p,\infty}}^{p-r}|||f|||_{L^{p,\infty}}^{r}. \end{align*} We obtain that \begin{align*} \|f\|_{L^{p,\infty}}^{p}&\leq\|f\|_{L^{p,\infty}}^{p-r}|||f|||_{L^{p,\infty}}^{r}, \end{align*} and hence \begin{align*} \|f\|_{L^{p,\infty}}^{r}&\leq|||f|||_{L^{p,\infty}}^{r}\\ \|f\|_{L^{p,\infty}}&\leq|||f|||_{L^{p,\infty}}, \end{align*} we are done.

The $\sigma$-finitenss of the second exercise cannot be relaxed. For $X=\{1,2\}$ and $\mu(\{1\})$=1 and $\mu(\{2\})=\infty$, so $(X,\mu)$ is not $\sigma$-finite. If we let $f=1$, then
\begin{align*} |||f|||_{L^{p,\infty}}=\left(\int_{\{1\}}1d\mu\right)^{1/r}=\mu(\{1\})^{1/r}=1. \end{align*} But \begin{align*} \|f\|_{L^{p,\infty}}=\sup_{\alpha>0}\alpha\mu(1>\alpha)^{1/p}\geq\dfrac{1}{2}\mu\left(x\in X:1>\dfrac{1}{2}\right)^{1/p}=\dfrac{1}{2}\mu(X)^{1/p}=\infty. \end{align*}