# Does this integral condition characterize $L^\infty$?

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.

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

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 $$$$f(x) \leq A + B \int_0^x f(t) dt \leq A + B \int_0^x f(t)dt, \,\,\, 0\leq x <1.$$$$ for some $$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.