Let $u \in C^{\alpha}(\Omega)$ and $B_R(x) \subset \Omega$. How can I see that \begin{equation} |u(x) - u_{B_R(x)}| \le (2R)^{\alpha} \| u\|_{C^\alpha(\Omega)}. \end{equation} where $u_{B_R(x)}= \dfrac{1}{|B_R(x)|}|u|$.
$u \in C^{\alpha}(\Omega) \Rightarrow |u(x) - u_{B_R(x)}| \le (2R)^{\alpha} \| u\|_{C^\alpha(\Omega)}.$
0
$\begingroup$
analysis
-
0I think you are missing an integral sign in the definition of $u_{B_R(x)}$. Also, why do you write $B_R(x) \subset B_R(x)$? Should these be two different balls? And what would the role of the second ball be? – 2012-10-25
-
0What is $\Omega$? – 2012-10-25
-
0This is the first line of proof of theorem 9.1.4 (Campanato characterizing H¨older continuity in terms of Lp-approximability by means on balls:) of Partial Diffeerential Equations of Jürgen Jost. – 2012-10-25
-
0$ \Omega$ is a subset of a $\mathbb{R}^n$. – 2012-10-25
-
0I made a mistake and I corrected. Please see. – 2012-10-25
-
0Now is $\alpha \in (0,1)$ and $C^\alpha$ the set of $\alpha$-Hölder functions? And is $\|u\|_{C^\alpha(\Omega)}$ the best constant in $|u(x)-u(y)|\le C|x-y|^\alpha$? – 2012-10-25
-
0And how are $r$ and $R$ linked? Furthermore, it would be better if you include information in the OP, not on the comments (but just use these to inform us you have done the edit). – 2012-10-25
-
0Sorry, again I made a mistake. – 2012-10-25
-
1It is still not correct. As I said before, you are missing an integral in the definition of $u_{B_R(x)}$, and you have not defined the meaning of $C^\alpha$, the $C^\alpha$-(semi)norm and the range of $\alpha$. Anyway, assuming my interpretation above is correct, you should get the result by simply integrating the Hölder inequality over the ball. – 2012-10-25