Can we find $f: [0,1] \to [0, \infty) \notin L^{\frac{q}{q-1}}[0,1]$ such that $\int_A f \le |A|^{1/q}$ for some $q \ge 2$ and all $A \subset [0,1]$?

Question

Let $f: [0,1] \to [0, \infty)$ be a measurable function satisfying $$\int_A f \le |A|^{1/q}$$ for some $q \ge 2$ and all measurable subsets $A \subset [0,1]$. Show that $f\in L^p[0,1]$ for all $1 < p < \frac{q}{q-1}$. Is $f$ necessarily in $L^{\frac{q}{q-1}}[0,1]$?

I solved the first part. Apply the given inequality to the sets $A_t := \{f > t\}$ for $t\in \Bbb R$, to get $$t |A_t| \le \int_{A_t} f \le |A_t|^{1/q},$$ i.e., $$|A_t| \le t^{\frac{q}{1-q}}.$$ Using $$\int_0^1 f^p = p\int_0^\infty t^{p-1} |A_t| \, dt$$ we get $$\int_0^1 f^p \le p + \frac{p}{\frac{q}{q-1} - p} < \infty$$ for $1 < p < \frac{q}{q-1}$.

Could someone help me produce a measurable function $f: [0,1] \to [0, \infty) \notin L^{\frac{q}{q-1}}[0,1]$ satisfying $$\int_A f \le |A|^{1/q}$$ for some $q \ge 2$ and all measurable subsets $A \subset [0,1]$?

Answer 1

Let $q>1$. Consider $$f(x)=\begin{cases}\frac1q x^{\frac1q-1}, &x>0,\\ 0, & x=0.\end{cases}$$ Clearly we have $f: [0,1] \to [0, \infty) \notin L^{\frac{q}{q-1}}[0,1]$. Now we show that $$\int_A f \le |A|^{1/q}\tag{$*$}$$ for all measurable $A\subset [0,1]$.

This example is motivated by $f(x)^{\frac q{q-1}}=Cx^{-1}$ lies on the boardline of $L^1$ and $(*)$ holds for $A=[0,1]$ (to determine the constant $C$).

• For any interval $[a,b]\subset \mathbb R$, since $f$ is decreasing, we have $$\int_{[a,b]}f\leq \int_0^{b-a}f=(b-a)^{1/q},$$ hence $(*)$ holds for all sub-interval of $[0,1]$.

• Let $\{I_j=[a_j, b_j]\}_{1\leq j\leq n}$ be a finite collection of disjiont sub-intervals of $[0,1]$, and let $E=\cup_{j=1}^n I_j$, we show that $(*)$ holds for $E$. We take $n=2$ as an example, the proof for general $n\geq 2$ is analogous. If $n=2$, we can assume WLOG that $a_1<b_1<a_2<b_2$, i.e., $I_1$ lies on the left of $I_2$. Since $f$ is decreasing, we have $\int_{I_2}f\leq \int_{[b_1, b_1+b_2-a_2]}f$, hence $$\int_{I_1\cup I_2}f\leq\int_{[a_1, b_1]\cup[b_1, b_1+b_2-a_2]}f=\int_{[a_1, b_1+b_2-a_2]}f\leq |b_1-a_1+b_2-a_2|^{1/q}=|I_1\cup I_2|^{1/q}.$$

• Now, For any measurable set $A\subset [0,1]$ and any $\varepsilon>0$, there exists a countable union of disjoint intervals $\{I_j\}_{j\geq1}$ such that $A\subset \cup_{j\geq1}I_j$ and $\sum_j|I_j|\leq |A|+\varepsilon$. Let $E_n=\cup_{j=1}^nI_j$. For any $n\geq 1$, we have $$\int_{E_n} f\leq |E_n|^{1/q}.$$ Letting $n\to\infty$, the monotone convergence theorem implies that $$\int_A f\leq\int_{\cup_{j\geq1}I_j}f\leq |\cup_{j=1}^\infty I_j|^{1/q}\leq (|A|+\varepsilon)^{1/q}.$$ Letting $\varepsilon\to0+$ implies $(*)$.

Answer 2

Suppose that $f\colon [0,1]\to\mathbb R$ is a non-negative function which satisfies $\int_A f \leqslant \lambda(A)^{1/q}$. Then for each $t>0$, denoting $A_t:=\{x\in [0,1], f(x)\geqslant t\}$, $$ t^{q/(q-1)}\lambda\left(A_t\right)\leqslant t^{q/(q-1)-1}\int_{A_t}f(x)dx\leqslant t^{q/(q-1)-1}\left(\lambda\left(A_t\right)\right)^{1/q} $$ hence $$ t\left(\lambda(A_t)\right)^{(q-1)/q}\leqslant 1 $$ and it follows that $$ \sup_{t>0}t\left(\lambda(A_t)\right)^{(q-1)/q}\leqslant 1.\tag{*} $$ Conversely, assume that $(*)$ holds. For a set $A$, one writes $$ \int_A f =\int_0^\infty \lambda(A\cap A_t)dt\leqslant \int_0^\infty\min\{\lambda(A),t^{-(q-1)/q}\}dt\leqslant \lambda(A_t). $$ Note that $(*)$ is equivalent to $$ \sup_{n\geqslant 1}2^{nq/(q-1)}\lambda(A_{2^n})\leqslant 1 $$ while $f$ belongs to $\mathbb L^{q/(q-1)}$ to $$ \sum_{n\geqslant 1}2^{nq/(q-1)}\lambda(A_{2^n})<\infty. $$ Moreover, these considerations help to see in an other way that the example given by Feng works.