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 $C>0$?



Morrey condition (integral condition) and (local) Holder condition
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 $g\_r(0)=r^{-\gamma}$, $0<\gamma<\beta$).


