better than Hölder leads to regularity?

Question

Consider a $L^2$-function $f:\Omega\subset \mathbb{R}^n\to[0,\infty)$ such that for any ball $B(x,r)\subset \Omega$, we have better than Hölder inequality : $$ \int_{B(x,r/2)} |f| \leq \epsilon |B(x,r/2)|^\frac{1}{2} \left( \int_{B(x,r)} |f|^2\right)^\frac{1}{2} $$ for some $\epsilon\in(0,1)$. What can we say about the regularity of $f$ ? Is there an $\epsilon>0$ (depending only on the dimension or the domain) such that every fonction satisfying these inequalities is continuous or anything better than just $L^2$ ?
I have noticed that if we forget the $1/2$, and if we choose $x$ as a lebesgue point of $f$ and $f^2$, then one can divide by $|B(x,r)|$ and pass to the limit $r\to 0$ to get that $f=0$.
But if we consider the $1/2$, then a lot of nonzero functions satisfy this kind inequality (for instance if $f$ is harmonic). But in order to prove regularity, all proofs that i have seen need to have the same exponent on both side of the inequality : something of the form $$ \int_{B(x,r/2)} |f|^p \leq \frac{1}{2}\int_{B(x,r)} |f|^p $$ in order to iterate it and get an inequality of the form $$ \sup_{B(x,r)\subset B(0,\rho)} \frac{1}{r^{n-p}} \int_{B(x,r)}|f|^p \leq C \rho^\alpha $$ for some $C,\alpha >0$, which is a Morrey space. In the case where $f=\nabla u$ for some $u$, this gives the Hölder continuity of $u$.
The question is : without the same exponent, can we still conclude to some regularity better than $L^2$ ?