I am trying to solve this problem. Let $f:\mathbb{R}^n\rightarrow \mathbb{R}$ be such that for $x\in \mathbb{R}^n$ and $r\in(0,1)$, $$\frac{1}{\text{Vol}(B(x,r))}\int_{B(x,r)}|f(y)-f_{x,r}|dy<\kappa r^{\alpha}$$ where $$f_{x,r}=\frac{1}{\text{Vol}(B(x,r))}\int_{B(x,r)}f(y)dy,$$ $\kappa>0$ and $\alpha\in(0,1)$. Show that there exists a constant $C_0=C(n,\kappa)$ such that for all $x,y\in\mathbb{R}^n$ with $|x-y|\leq1$, $$|f(x)-f(y)|\leq C_0|x-y|^{\alpha}.$$ My first thought was to take $r=|x-y|$ when $|x-y|<1$ and consider two balls $B(x,r)$ and $B(y,r)$, then proceed like one proves Morrey's inequality, say let $z\in B(x,r)\cap B(y,r)$ then $$|u(x)-u(y)|\leq \frac{C_n}{\text{Vol}(B(r))}\left(\int_{B(x,r)}|u(x)-u(z)|dz+\int_{B(y,r)}|u(y)-u(z)|dz\right)$$ where $C_n$ is the volume ratio. But this doesn't quite work, did I miss something or can anybody suggest an another thought on this? When $\alpha$ approaches $0$, does this inequality provide some information of BMO space, or does it simply collapse? The left hand side of the given condition also looks like that of a poincare inequality, is there any connection?

I found this problem when I was looking for some resources of geometric measure theory.

  • $\begingroup$ It does not hold for the all zero function modified by adding one point of discontinuity (which does not change any integrals). $\endgroup$ – Michael May 24 at 14:29
  • $\begingroup$ You are right, what if one add continuity to f? Although this condition is not in the original problem as I just double checked. $\endgroup$ – WhiteDwarf May 24 at 21:53

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