Let $x \in \mathbb{R}^n$ and $f:\mathbb{R^n} \to \mathbb{R}$ be a non-negative function such that $f(x)=0$. Is it true that (assuming $\alpha,\beta>0$) $$\limsup_{r \to 0} r^{-\alpha \beta}\frac{1}{|B_{r}(x)|}\int_{B_{r}(x)} f(y)^\alpha dy < \infty$$ implies $$|f(z)| \le C|z-x|^\beta,$$ for $z$ close enough to $x$ and for some constant $C>0$?
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$\begingroup$ What is the range of $f$ (if $f$ may take arbitrary negative values, the expression $f(y)^\alpha$ could get you into trouble)? Also, is $|B_r (x)|$ the volume of the $n$-dimensional ball in $\mathbb R^n$? $\endgroup$– Alex M.Commented Mar 31, 2018 at 12:35
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$\begingroup$ @AlexM. We may take $f$ non-negative. Yes, it is the volume of the ball. $\endgroup$– RikuCommented Mar 31, 2018 at 12:37
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$\begingroup$ Do we know anything about $f$? Is it continuous, integrable, etc.? $\endgroup$– Alex M.Commented Mar 31, 2018 at 12:50
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$\begingroup$ @AlexM. I'm mostly interested in a sufficiently weak assumption that makes the result hold true. $\endgroup$– RikuCommented Mar 31, 2018 at 13:11
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No. Take $n=2$, $x=0$ (and $\alpha=1$, wlog). Define $f$ in polar coordinates as $f(r,\theta)=r^\beta g_r(\theta)$ where $\int_0^{2\pi} g_r(\theta)\ d\theta=1$ and $\lim_{r\to 0}g_r(0)=\infty$, which is clearly possible while keeping $f$ continuous. (For example, $g_r$ could be piecewise linear on $[-r^\gamma,+r^\gamma]$ and vanish elsewhere, with $g_r(0)=r^{-\gamma}$, $0<\gamma<\beta$).
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$\begingroup$ Thanks. And what is an assumption on $f$ that makes the result hold true? $\endgroup$– RikuCommented Mar 31, 2018 at 17:57
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1$\begingroup$ Maybe $f$ Hölder continuous of order $\beta$ ?? (Sorry, we're April 1st...). $\endgroup$ Commented Apr 1, 2018 at 15:46