estimate Holder norm via integral inequality

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.

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