Do we have a name for this space?

Question

Let $\Omega$ be a bounded domain in $\mathbb{R}^{n}$. Consider the class

$$ \mathcal{F}=\{f\in L^{1}(\Omega):\exists C>0 \text{ s.t. } \int_{U}|f|\leq C\sqrt{|U|},\text{ for any }U\subset \Omega.\} $$ Here $U$ runs over all open subsets of $\Omega$.

It is clear that $L^{2}(\Omega)\subset \mathcal{F}$.

My question: do we have a name for this class $\mathcal{F}$? Did anyone study it before?

Thanks.

Answer 1

$\newcommand{\Om}{\Omega} \newcommand{\F}{\mathcal{F}}$Prompted by a comment by Nate Eldredge, I found this in Wikipedia :

For any $0<r<p$ the expression $$\||f|\|_{L^{p,\infty}}=\sup_{0<\mu(E)<\infty}\mu(E)^{-1/r+1/p}\Big(\int_E |f|^r\,d\mu\Big)^{1/r}$$ is comparable to the $L^{p,w}$-norm.

Taking here $r=1$ and $p=2$, we conclude that

$\mathcal F$ is the weak $L^2$ space $L^{2,w}$.

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Let us also provide a direct proof of the latter statement. Recall that \begin{equation*} L^{2,w}=\Big\{f\in L^0(\Om)\colon\;\exists C\in(0,\infty)\ \forall t\in(0,\infty)\ P_f(t)\le C/t^2\Big\}, \end{equation*} where $$P_f(t):=\big|[|f|>t]\big|$$ and $[|f|>t]:=\{x\in\Om\colon|f(x)|>t\}$. On the other hand, as was shown in my other answer on this web page, \begin{equation*} \F=\Big\{f\in L^1(\Om)\colon\;\exists C\in(0,\infty)\ \forall t\in(0,\infty)\,\int_{[|f|>t]}|f|\le C\sqrt{P_f(t)}\Big\}. \tag{0} \end{equation*}

Take any $f\in L^0(\Om)$ and any $t\in(0,\infty)$. Then \begin{align*} \int_{[|f|>t]}|f|&=\int_\Om 1_{[|f|>t]}\,|f| \\ &=\int_\Om 1_{[|f|>t]}\int_0^\infty ds\,1_{[|f|>s]} \\ &=\int_0^\infty ds\,\int_\Om 1_{[|f|>t]}1_{[|f|>s]} \\ &=\int_0^t ds\,\int_\Om 1_{[|f|>t]} + \int_t^\infty ds\,\int_\Om 1_{[|f|>s]} \\ &=tP_f(t) + \int_t^\infty ds\,P_f(s). \tag{1} \end{align*}

If now $f\in\F$ then, by (1),
\begin{equation*} tP_f(t)\le \int_{[|f|>t]}|f|\le C\sqrt{P_f(t)}, \end{equation*} so that $P_f(t)\le C^2/t^2$ and hence $f\in L^{2,w}$. Thus, \begin{equation*} \F\subseteq L^{2,w}. \tag{2} \end{equation*}

To prove that $\F\supseteq L^{2,w}$, take any $f\in L^{2,w}$, so that \begin{equation*} P(t):=P_f(t)\le C/t^2 \tag{3} \end{equation*} for some real $C>0$ and all real $t>0$. The function $P$ is nonincreasing and right-continuous on $(0,\infty)$, with $P(\infty-)=0$. Consider the generalized inverse $P^{-1}$ of $P$ given by the formula \begin{equation*} t_u:=P^{-1}(u):=\inf\{s\ge0\colon P(s)\le u\}=\min\{s\ge0\colon P(s)\le u\} \end{equation*} for $u\in(0,P(0))$. Then for all real $s\ge0$ and all $u\in(0,P(0))$ we have \begin{equation*} s<t_u\iff u<P(s). \tag{4} \end{equation*} So, for any real $t>0$ \begin{align*} \int_t^\infty ds\,P_f(s)&=\int_t^\infty ds\,\int_0^{P(s)}du \\ &=\int_0^{P(t)}du \int_t^{t_u} ds\,\\ &=\int_0^{P(t)}du\, (t_u-t)\,\\ &\le\int_0^{P(t)}du\,t_u. \end{align*} By (3), for any real $u>0$ we have $P(\sqrt{C/u})\le u$, whence, by (4), $t_u\le\sqrt{C/u}$. Now the latter multi-line display yields \begin{equation*} \int_t^\infty ds\,P_f(s)\le2\sqrt C\,\sqrt{P_f(t)}. \end{equation*} Also, (3) is obviously equivalent to $tP_f(t)\le\sqrt C\,\sqrt{P_f(t)}$. So, by (1), \begin{align*} \int_{[|f|>t]}|f|\le3\sqrt C\,\sqrt{P_f(t)}, \end{align*} so that, by (0), $f\in\F$. Thus, indeed $\F\supseteq L^{2,w}$. Now (2) yields \begin{equation*} \F=L^{2,w}, \end{equation*} as claimed.

Answer 2

$\newcommand{\F}{\mathcal F}\newcommand{\Om}{\Omega}$Take any $f\in L^2(\Om)$. Then for any measurable subset $U$ of $\Om$, by Hölder's inequality we have $$\int_U|f|\le\|f\|_{L^2(\Om)}|U|^{1/2}, $$ so that $f\in\F$. Thus, $\F\supseteq L^2(\Om)$, as you noted.

The purpose of this partial answer is to show that $\F\not\subseteq L^2(\Om)$. Indeed, let $\Om=(0,1)$ and $f(x)=1/\sqrt x$ for $x\in\Om$. Take any measurable subset $U$ of $\Om$ with $u:=|U|\ne0$, so that $u\in(0,1]$. Let $U_t:=\{x\in\Om\colon f(x)>t\}=(0,u)$, where $t:=1/u^2>0$. Then $$\int_U|f|-\int_{U_t}|f|=\int_U f-\int_{U_t}f =\int_{U\setminus U_t} f-\int_{U_t\setminus U}f \le\int_{U\setminus U_t}t-\int_{U_t\setminus U}t=0. $$ So, $$\int_U|f|\le\int_{U_t}|f|=\int_0^u\frac{dx}{\sqrt x}=2\sqrt u=2\sqrt{|U|}, $$ so that $f\in\F$. However, $f\notin L^2(\Om)$.

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It also follows from the above reasoning that (generally, for any $\Om$) $$\F=\Big\{f\in L^1(\Om)\colon\;\exists C\in(0,\infty)\ \forall t\in(0,\infty)\,\int_{[|f|>t]}|f|\le C\big|[|f|>t]\big|^{1/2}\Big\}, $$ where $[|f|>t]:=\{x\in\Om\colon|f(x)|>t\}$. That is, in the definition of $\F$, instead of arbitrary open or, equivalently, arbitrary measurable subsets $U$ of $\Om$, one may consider the subsets of $\Om$ of the special form $[|f|>t]$.