Mean Oscillation Over Cubes and Hölder Continuity

$\mathbf{Problem:}$ Suppose $f\in L^1_{Loc}(\mathbb{R}^n)$ is hölder $\alpha-$continuous, where $0<\alpha\leq 1$. It is equivalent to say for any ball $B$ centered at $x\in \mathbb{R}^n$ we have

$\displaystyle \frac{1}{|B|}\int_B\left|f(y)-\frac{1}{|B|}\int_B f(z)dz\right|dy\leq C|B|^{\alpha/n}$

where $C$ is independent from $x$ and the radius of the ball.

$\mathbf{Proof:}$ “necessity” If $f$ is hölder $\alpha-$continuous, then

$\displaystyle \frac{1}{|B|}\int_B\left|f(y)-\frac{1}{|B|}\int_B f(z)dz\right|dy\leq \frac{1}{|B|^2}\int_B\int_B|f(y)-f(z)|dydz$

$\displaystyle \leq \frac{C}{|B|^2}\int_B\int_B |y-z|^\alpha dydz\leq \frac{C}{|B|^2}\int_B\int_B(2r)^\alpha dydz=Cr^\alpha=C|B|^{\alpha/n}$

here $r$ is the radius of $B$.

“sufficiency” Let $B_k=B(x_0,2^{-k}r)$ is the ball centered at $x_0$ with radius $2^{-k}r$. Denote $\displaystyle f_B=\frac{1}{|B|}\int_Bf(y)dy$ then $\displaystyle\frac{1}{|B|}\int_B|f(y)-f_B|dy\leq C|B|^{\alpha/n}$.

Since $\displaystyle |f_{B_k}-f_{B_{k+1}}|\leq |f(y)-f_{B_k}|+|f(y)-f_{B_{k+1}}|$, $\forall\, y\in B_{k+1}$

Integrating this over $B_{k+1}$, we get

$\displaystyle |f_{B_k}-f_{B_{k+1}}|\leq \frac{1}{|B_{k+1}|}\int_{B_{k+1}}|f(y)-f_{B_{k+1}}|dy+\frac{|B_k|}{|B_{k+1}|}\frac{1}{|B_k|}\int_{B_k}|f(y)-f_{B_k}|dy\leq C|B_{k+1}|^{\alpha/n}+C2^n|B_k|^{\alpha/n}\leq C(n)2^{-k\alpha}r^\alpha$

So $\displaystyle |f_{B_{k}}-f_{B_{k+1}}|\leq \sum\limits_{i=k}^{k+l-1}|f_{B_i}-f_{B_{i+1}}|\leq \sum\limits_{i=k}^{k+l-1}C(n)2^{-i\alpha}r^\alpha\leq C(n)r^\alpha 2^{-(k-1)\alpha}$.

Since $f\in L^1_{Loc}(\mathbb{R}^n)$, from lebesgue differentiation theorem $\lim\limits_{n\to\infty}f_{B_{m}}=f(x)$ a.e. Let $l\to \infty$, we know $|f_{B_k}-f_{x_0}|\leq C(n)2^\alpha r^\alpha$ a.e. $k\geq 0$.

So when $k=0$, $|f_{B_0}-f(x_0)|\leq C(n)2^\alpha r^\alpha$.

Set $r=|x_0-y_0|$, $B'_0=B(y_0, r)$, then

$\displaystyle |f(x_0)-f(y_0)|\leq |f(x_0)-f_{B_0}|+|f(y_0)-f_{B'_0}|+|f_{B_0}-f_{B'_0}|\leq C(n)2^{\alpha+1}r^\alpha+|f_{B_0}-f_{B'_0}|$ holds almost everywhere.

Since we have $|f_{B_0}-f_{B'_0}|\leq |f(z)-f_{B_0}|+|f(z)-f_{B'_0}|$, $z\in B_0\cap B'_0$

Integrating this on $B_0\cap B'_0$, we get

$\displaystyle |f_{B_0}-f_{B'_0}|\leq \frac{|B_0|}{|B_0\cap B'_0|}\frac{1}{|B_0|}\int |f(z)-f_{B_0}|dz+\frac{|B'_0|}{|B_0\cap B'_0|}\frac{1}{|B'_0|}\int|f(z)-f_{B'_0}|dz\leq C(n)|B_0|^{\alpha/n}+C(n)|B'_0|^{\alpha/n}=C(n)r^\alpha$

So $|f(x_0)-f(y_0)|\leq C(n,\alpha)r^\alpha=C(n,\alpha)|x_0-y_0|^\alpha$ a.e. So $f$ is a hölder $\alpha-$continuous.

$\text{Q.E.D}\hfill \square$$\mathbf{Remark:}$ Fanghua Lin, Qing Han. Elliptic partial differential equations.