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I want to prove the following "duality" and don't know how to get started... Does it have anything to with covering theorems?

$\limsup_{r\rightarrow0} \sup_{y \in B_r (x)} V^{-1} \int_{B_r (y)}f \mathrm{d}m^d = \liminf_{r\rightarrow0} \inf_{y \in B_r (x)} V^{-1} \int_{B_r (y)}f \mathrm{d}m^d$

where $V$ denotes the volume of $d$-ball with Radius $r$ and $m^d$ the $d$-dimensional Lebesgue measure and $f$ is in $L^1(\mathbb{R}^d;\mathbb{R})$.

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Not true. Take $f(x) = 1, 100 > x > 0, f(x) = 0$ otherwise. With x=0, for fixed r $ \inf_{y \in B_r (x)} V^{-1} \int_{B_r (y)}f \mathrm{d}m^d = 0$ (it's achieved at $y = -1$) and the sup on the other side is always 1.

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  • $\begingroup$ mhmmm... does it change anything writing for almost all $x$? $\endgroup$
    – tubmaster
    Dec 8 '16 at 17:01
  • $\begingroup$ I think that is true but don't have an argument off hand. $\endgroup$
    – user83457
    Dec 8 '16 at 17:02
  • $\begingroup$ It must be true ;) A collegue of mine said that it should have somthing to do with covering theorems (perhaps Besicovitch or something)... $\endgroup$
    – tubmaster
    Dec 8 '16 at 17:04
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It is true for almost all $x$, in particular for Lebesgue points of $f$.

Suppose $x$ is a Lebesgue point of $f$. Given $\varepsilon > 0$, for $r$ sufficiently small we have $$\int_{B_{2r}(x)} |f(z) - f(x)| \; dm^d(z) < \varepsilon V(2r) = 2^d \varepsilon V(r)$$ (where $V(r)$ is the volume of $B_r(x)$). If $y \in B_{r}(x)$ we have $B_{r}(y) \subseteq B_{2r}(x)$, so that

$$ \int_{B_{r}(y)} |f(z) - f(x)| \; dm^d(z) < 2^d \varepsilon V(r)$$ and then $$ f(x) - 2^d \varepsilon < \dfrac{1}{V(r)} \int_{B_r(y)} f(z)\; dm^d(z) < f(x) + 2^d \varepsilon$$

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