In the lecture I saw the Campanato theorem, that embeds the Campanato space in the Hölder space, more precisely see Theorem below. I came across something which I did not understand in the proof. Let me mention it after I give the theorem. But let me first start with some noatations and assumptions:
- Let $\Omega \subset \mathbb{R}^n$ be open always be open. For $x_0 \in \Omega, r > 0$ we set$$\Omega_{r}(x_0)=B_r(x_0) \cap \Omega,$$ where $B_r(x_0)$ is the open ball of radius $r$ centered at $x_0$.
- For $1 \leq p < \infty, \lambda > 0$ the campanato space is defined as $$\mathcal{L}^{p,\lambda}(\Omega)=\{u \in L^p(\Omega): [u]_{\mathcal{L}^{p,\lambda}(\Omega)}:=\sup_{x_0 \in \Omega, 0 < r < r_0} (r^{-\lambda} \int_{\Omega_r(x_0) }|u-u_{x_0,r}|^pdx)^{1/p}\},$$ where $$u_{x_0,r}:=\frac{1}{\mathcal{L}^n(\Omega_r(x_0))} \int_{\Omega_r(x_0)}u dx,$$ with norm $$||u||_{\mathcal{L}^{p,\lambda}}:= ||u||_{L^p}+[u]_{\mathcal{L}^{p,\lambda}}$$
- A map $u : \Omega \rightarrow \mathbb{R}$ is called Hölder continuous with exponent $\alpha \in (0,1]$, if $$\forall x,y \in \Omega: |u(x)- u(y)| \leq C |x-y|^\alpha.$$ The Hölder space is defined as $$C^{0,\alpha}(\overline{\Omega})=\{u \in C^0(\overline{\Omega}): u \text{ is Hölder continuous with exponent } \alpha\},$$ with semi-norm given by $$[u]_{C^{0,\alpha}}:= \sup_{x,y \in \Omega,x ≠ y} \frac{|u(x)-u(y)|}{|x-y|^\alpha}$$
Theorem: Let $\Omega$ be open and of type $A$, i.e. $\mathcal{L}^n(\Omega_r(x_0)) \geq Ar^n$ for all $x_0 \in \Omega$ and $0< r< r_0(\Omega):=\min\{1,diam(\Omega)\}$. Let $\lambda > n$, $1 \leq p < \infty$ and $\alpha=\frac{\lambda-n}{p}$. Then $\mathcal{L}^{p,\lambda}(\Omega)$ can be embedded into $C^{0,\alpha}(\overline{\Omega})$ and $$\forall u \in \mathcal{L}^{p,\lambda}(\Omega): ||u||_{C^{0,\alpha}} \leq C ||u||_{\mathcal{L}^{p,\lambda}(\Omega)} $$ for a constant $C > 0$ independent of $u$.
In the proof we constructed a continuous representative $\overline{u}$ of $u$ and proved
Claim: $\overline{u} \in C^{0,\alpha}(\overline{\Omega})$ and for $x_0 \neq y_0 \in \Omega$ with $0 < |x_0-y_0| =r < \frac{r_0}{2}$ we have $$|\overline{u}(x_0)-\overline{u}(y_0)| \leq C r^\alpha [u]_{\mathcal{L}^{p,\lambda}}$$
So my question: I understood proving the estimate in the claim, but I cannot understand why we can then see that $\overline{u}$ is Hölder continuous, since the estimate only holds for $r < \frac{r_0(\Omega)}{2}$?
I would also appreciate if anyone could maybe cite another source for a prove, which contains a lot of details. Thanks a lot in advance!
The large-distance estimate is obtained by a rough bound on $\sup |u|$ as follows.
Lemma. Suppose $u\in L^p(\Omega)$ and there exist constants $C,\alpha,r_0$ such that $|u(x)-u(y)|\le C|x-y|^\alpha$ whenever $x, y\in\Omega$ and $|x-y|< r_0$. Then $u\in C^{0, \alpha}(\overline{\Omega})$ and $[u]_{C^{0, \alpha}}\le C'$ where $C'$ depends only on $C, \alpha, r_0, \|u\|_{L^p}$, and $A$ (the constant in the definition of type $A$).
Proof. For $|x-y|\ge r_0$ we have $$ \frac{|u(x)-u(y)|}{|x-y|^\alpha} \le \frac{2\sup_{\overline{\Omega}}|u|}{r_0^\alpha} $$ so it suffices to give an upper bound for $|u|$. Suppose $|u(x_0)|=M$ where $M>2Cr_0^\alpha$. Then for all $y\in \Omega_{r_0}(x_0)$ we have $|u(y)|\ge M - Cr_0^\alpha > M/2$, hence $$ \int_\Omega |u|^p \ge (M/2)^p \mathcal L^n(\Omega_{r_0}(x_0)) \ge (M/2)^p A r_0^n $$ hence $M\le 2\|u\|_{L^p}A^{-1/p}r_0^{-n/p}$. Conclusion: $$ \sup |u| \le \max(2Cr_0^\alpha, 2\|u\|_{L^p}A^{-1/p}r_0^{-n/p}) $$ which proves the lemma.