Let $\Omega$ be a bounded, connected open subset of $\mathbb R^n$ with smooth boundary. For any $f \in L^1 (\Omega)$, is it true that $f \in L^\infty (\Omega)$ if and only if the following condition holds?

For every $\delta > 0$, there exists some $C > 0$ such that for almost every $x \in \Omega$, and every open ball $B \subset \Omega$ centered at $x$ with radius at least $\delta$ we have

$$|f(x)| \leq C \def\avint{\mathop{\,\rlap{-}\!\!\int}\nolimits} \avint_B |f|.$$

Note: Here $\def \avint{\mathop{\,\rlap{-}\!\int}\nolimits} \avint_B |f|$ denotes the average integral of $|f|$ over $B$.

Edit: The original question asked about the case where the balls $B$ are not necessarily centered at $x$. This has been solved in the comments - the "if" direction holds, but "only if" does not.

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    $\begingroup$ Concerning the "only if" part: the value of $f(x)$ is not even defined at any $x$. $\endgroup$ Oct 30, 2023 at 15:42
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    $\begingroup$ I think convex unbounded functions provide counterexamples when $d = 1$. $\endgroup$
    – Zarrax
    Oct 30, 2023 at 15:48
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    $\begingroup$ But then fix $\delta=1$ and get $|f(x)| \leq C \int_B |f| \leq C\|f\|_1$..or I misunderstood? $\endgroup$ Jan 17 at 18:23
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    $\begingroup$ @FedorPetrov Yes indeed, the “only if” direction fails horribly. Although, see my comment directly before this one. $\endgroup$
    – Nate River
    Jan 18 at 0:40
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    $\begingroup$ Yes, for centered balls there is $C$ such that this works for all Lebesgue points of $f$ $\endgroup$ Jan 18 at 6:14

2 Answers 2


Let me prove that for $f\in L^\infty$ and fixed $\delta>0$ (and centered balls) such $C$ exists. The idea is to find $C$ which works for all Lebesgue points of $f$. Assume that there is no such $C$, then for $C=n$ there exist violating Lebesgue points $x_n$ and $B_n$ with center $x_n$ and radius $r_n\geqslant \delta$ and $$|f(x_n)|> n\def\avint{\mathop{\,\rlap{-}\!\!\int}\nolimits} \avint_{B_n}|f|.\tag{$\heartsuit$}$$ Without loss of generality, $x_n$ converge to a point $x_0$ (it lies in $\Omega$, since all $x_n$ are at distance at least $\delta$ from $\mathbb{R}^n\setminus \Omega$, thus so does $x_0$). Then for the ball $B(x_0,\delta)$ we have $\int_{B(x_0,\delta)}f=0$: otherwise the integrals over $B_n$ would be bounded from below, and RHS's of $(\heartsuit)$ would tend to infinity, contradicting to LHS's being bounded from above by $\|f\|_\infty$. But for large $n$ we have $x_n\in B(x_0,\delta)$, and, since by $(\heartsuit)$ we have $f(x_n)\ne 0$ and $x_n$ is a Lebesgue point of $f$, we conclude that $f$ is non-zero on some set of positive measure inside $B(x_0,\delta)$. A contradiction.

For other direction, for centered balls this trivially fails for example for $f(x)=x^{-1/2}$, $\Omega=(0,1)$ (for $x$ near the singularity there are simply no centered balls of radius $\geqslant \delta$ contained in $\Omega$). For not centered balls it is However true, and settled in comments, the key word is uniform interior sphere.


Too long for a comment. For $d=1$ should be true. Let me simplify a bit (hopefully without loss of generality). If $\Omega=(-1,1), f\in C(-1,1)$. Then what you ask implies that \begin{equation} f(x) \leq A + B \int_0^x f(t) dt \leq A + B \int_0^x f(t)dt, \,\,\, 0\leq x <1. \end{equation} for some $A,B>0$ Then, for example by Gronwall's inequality you have \begin{equation*} f(x) \leq A e^{xB} \leq Ae^B < +\infty.\end{equation*} For $0\leq x <1$. Similarly for $-1<x\leq 0 $.

  • $\begingroup$ Ah, Gronwall’s inequality - very nice. $\endgroup$
    – Nate River
    Oct 30, 2023 at 17:28

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