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Let $\Omega$ an open bounded domain in $R^n$ with smooth boundary and consider a function $u \in W^{1,p}(\Omega) (2<p < +\infty)$ . Suppose that for every ball $B \subset \subset \Omega$ exists a constant $A$ that does not depend on $B$ such that

$$\int_{B} |\nabla u - (\nabla u )_{B}|^p \leq A (r_{B})^{n},$$ where $r_B$ is the radius of the ball $B$ and

$$ (\nabla u )_{B}:= \left( \frac{1}{|B|} \int_{B} \frac{\partial u}{ \partial x_1},..., \frac{1}{|B|} \int_{B} \frac{\partial u}{\partial x_n}\right)$$

I am reading a paper and the author says that if the function satisfy the things above, then $u$ is locally log - lipschitz, that is, for every ball $B \subset \subset \Omega$ the is a constant K = K(B) such that

$$ |u(x) - u(y)| \leq K |x-y||log|x-y||.$$

The author says that this is true by theorem 3 of this paper, but I am not seeing how to apply such theorem ...

Someone could help me to understand the justification of the author or justify the affirmation by other way?

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  • $\begingroup$ So I don't have search for the remark, where in the paper does the author make the claim with which you are having trouble? $\endgroup$ Oct 14, 2015 at 15:52

2 Answers 2

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The John-Nirenberg inequality implies that every BMO function $f$ is in exponentially integrable (on domains of suitable shape: cubes or balls), meaning there is $\lambda>0$ such that $\exp(f/\lambda)\in L^1$.

The exponential function is an Orlicz function with its conjugate being $\psi(t)=t\log t$. Theorem 3 in Cianchi's paper yields the modulus of continuity $C\psi^{-1}(|x-y|^{-n})$. Since $\psi^{-1}(t)\sim t/\log t$ when $t \to\infty$, this is exactly the log-Lipschitz continuity.

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There is one fundamental proof for the inequality when |x-y|<1/2. (Notation: $(f)_S =\int_S f/|S|$, denotes the average of $f$ on the set $S$.) WLOG, only need to consider the case when $x,y\in B_1$.

  1. when $r<1$, $|(|Du|)_{B_r} - (|Du|)_{B_{r/2}}| \leq C[Du]_{BMO}$. Then, it implies $|(Du)_{B_r}| \leq -C\log r$.
  2. By Poincare, $\int_{B_r}\int_{B_r}|u(x)-u(y)|/r^{2n}dxdy\leq Cr\int_{B_r}|Du|/r^ndx\leq Cr(-\log r)$.
  3. $|(u)_{B_r} - (u)_{B_{r/2}}|\leq C\int_{B_r}\int_{B_r}|u(x)-u(y)|/r^{2n}dxdy\leq Cr(-\log r)$.
  4. Let $r=|x-y|$, then $|u(x)-(u)_{B_r(x)}| = \sum_{j=1}^\infty |(u)_{B_{r/2^j}(x)} - (u)_{B_{r/2^{j-1}}(x)}|\leq -\sum_{j=1}^\infty C\frac{r}{2^{j-1}}\log\frac{r}{2^{j-1}} = -Cr\log r$. Similar result holds for $y$.
  5. Set $z=(x+y)/2$, then $B_r(x), B_r(y)$ are both contained in $B_{3r/2}(z)$. Combine step 2 and 4 to draw the conclusion: \begin{equation*} \begin{split} |u(x)-u(y)|\leq & |u(x)-(u)_{B_r(x)}| + |u(y)-(u)_{B_r(y)}| \\ &+ C\int_{B_{3r/2}(z)}\int_{B_{3r/2}(z)} |u(x')-u(y')|/r^{2n}dx'dy'\\ \leq &-Cr\log r. \end{split} \end{equation*}
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