Hölder estimation for weak solution of Laplace equation

Question

Let $B_{R_0}(x_0)=\{y\in \mathbb R^n : |y-x_0|\le R_0\}$, $u\in H^1_{loc}(\mathbb R^n)$. $\alpha\in (0,1)$.

If for any $0<R\le R_0$, we have $$ \sup_{B_{R}~(x_0)} u -\inf_{B_{R}~(x_0)} u \le C_1 R^\alpha $$ where $C_1$ is a positive constant, then how to show there is positive constant $C_2>0$ and $\theta\in (0,1)$ such that $$ [u]_{\alpha, B_{\theta R_{~0}}(x_0)} \le C_2 $$ where $[\cdot]_{\alpha, B_{\theta R_{~0}}(x_0)} $ is Holder seminorm.

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For makeing the problem more complete, I add the proof which the problem origin from. In fact, I have two problem in this proof. Firstly, how to get (1) by Lemma 5.1.5 ? Second, why (1) means (2) ?

Lemma 5.1.5: $\omega$ is nonnegative undecreasing function on $[0, R_0]$. If exiting $0<\theta, \eta<1, 0<\gamma\le 1, K\ge 0$ such that $$ \omega(\theta R) \le \eta \omega(R)+KR^\gamma, ~~~~~~0<R\le R_0, $$ then, there are constants $\alpha\in (0, \gamma), C>0$ depending on $\theta,\eta,\gamma$ such that $$ \omega(R)\le C(\frac{R}{R_0})^\alpha(\omega(R_0)+ KR_0^\gamma), ~~~~~~0<R\le R_0. $$

Theorem 5.1.6: $u\in H_{loc}^1(R^n)$ is bounded weak solution of $$ -\Delta u=0 $$ then, there is $\alpha\in (0,1)$ such that for any bounded domain $\Omega\subset R^n$, there is $$ [u]_{~\alpha;~\Omega} \le C $$ where $C$ depending on $n, \Omega$.

Proof of Theorem 5.1.6: For any fixed $x^0\in R^n$，$R>0$, let $$ m(R)=\inf_{B_R} u,~~~ M(R)=\sup_{B_R} u $$ where $B_R=B_R(x^0)$ is ball. Let $$ v(x)= u(x)- m(R),~~~ \omega(x)= M(R)-u(x). $$ Then $v,\omega\in H^1_{loc}(R^n)$ are nonnegative bounded. And in the sense of weak, we have $$ -\Delta v = -\Delta \omega =0. $$ By the Harnack inequation, we have $$ \sup_{B_{R/2}} v\le C\inf_{B_{R/2}} v,~~~~~~\sup_{B_{R/2}} \omega\le C\inf_{B_{R/2}} \omega $$ Namely, $$ M(R/2)-m(R)\le C(m(R/2)-m(R))\\ M(R)-m(R/2)\le C(M(R)-M(R/2)) $$ we can assume $C>1$. Then, by the above two inequation, we have $$ M(R/2)-m(R/2)\le \frac{C-1}{C+1}(M(R)-m(R)) $$ Letting $f(R)=M(R)-m(R), \eta=\frac{C-1}{C+1}$, $f$ is nonnegative nondecreasing, and $$ f(R/2)\le \eta f(R). $$ By Lemma 5.1.5, there is $\alpha\in(0,1)$ such that $$ f(R)\le CR^\alpha \tag{1} $$ namely $$ [u]_{\alpha; B_{R}(x^0)} \le C \tag{2} $$ Finally, just using finite covering for $\overline \Omega$.

Answer 1

There is a bit of an abuse of notation in the text you quoted. They're proving that for all $x_0\in \mathbb{R}^n$ and all $R>0$ it holds, $$ f(R/2)\leq \theta f(R), $$ where $f(R)= M(R)-m(R)$ and $\eta=(C-1)/(C+1)$. Now fix any $R_0>0$ and apply the Lemma with $f=\omega$, $\theta=1/2$, and $K=0$ to obtain $$ f(R)\leq C\left( \dfrac{R}{R_0}\right)^\alpha f(R_0), \qquad 0<R<R_0, $$ for some $\alpha>0$.

From here tha Hölder continuity follows: Fix $x_0 \in \mathbb{R}^n$ and $R_0>0$. Let $x,y\in B_{R_0/2}(x_0)$. Without loss of generality $u(x)>u(y)$ and we have two cases.

Case 1. $|x-y|>R_0/4$. Here we simply estimate $$ \begin{split} |u(x)-u(y)| &= u(x)-u(y)\leq \sup_{B_{R_0/2}(x_0)} u - \inf_{B_{R_0/2}(x_0)}u = f(R_0/2) \\ &\leq C\left(\dfrac{R_0/2}{R_0}\right)^\alpha f(R_0) \\ & \leq \dfrac{Cf(R_0)}{R_0^\alpha} |x-y|^\alpha. \end{split} $$ This gives the result in this case with $C_2= Cf(R_0)/R_0^\alpha$.

Case 2. $|x-y|\leq R_0/4$. Here note that, if $s=|x-y|$, then $B_s(x)\subset B_{R_0}(x_0)$ and so, estimating as before, $$ |u(x)-u(y)| \leq \sup_{B_s(x)} u - \inf_{B_s(x)} u = f(s, x) \leq C\left( \dfrac{s}{R_0/2}\right)^\alpha f(R_0/2, x), $$ where $f(s, x)$ is the same as $f$ but with the balls now centered at $x$ instead of $x_0$ (this is why it's important that we proved the estimate for all $x_0$, in particular it holds for $x$). Since $f(R_0/2, x)\leq f(R_0)$ we arrive at $$ |u(x)-u(y)|\leq \dfrac{2^\alpha C f(R_0)}{R_0^\alpha} |x-y|^\alpha. $$ This is the claim with $C_2= 2^\alpha Cf(R_0)/R_0^\alpha$.

Remark. Notice that the appearance of $f(R_0)/R_0^\alpha$ is natural, simply because of the way the quantity $[u]_\alpha$ behaves under the scaling $u(x)\mapsto u(\lambda x)$.