# Weighted Holder Norms

I am reading chapter 4 & 6 in the book " Elliptic Partial Differential Equation of Second Order" by Gilbarg and Trudinger. Instead of using the standard Holder norms, the authors use some sort of weighted Holder norms. Let $\Omega$ be an open set. For any $x\in\Omega$, let $d_x=distance (x,\partial\Omega), d_{x,y}=min\{d_x,d_y\}$. They defined for any function $u$ in $C^{2,\alpha}(\Omega)$

\begin{align*} [u]^*_{k,0;\Omega}&=[u]^*_{k;\Omega}=\sup_{|\beta|=k}\sup_{x\in\Omega}(d_{x})^{k}|D^{\beta}u(x)|,\ \ k=0,1,\ldots;\\ [u]_{k,\alpha;\Omega}^*&=\sup_{|\beta|=k}\sup_{x,y\in\Omega}(d_{x,y})^{k+\alpha}\frac{|D^{\beta}u(x)-D^{\beta}u(y)|}{|x-y|^{\alpha}},\ \ 0<\alpha\le 1;\\ \|u\|^*_{k;\Omega}&=\|u\|^*_{k,0;\Omega}=\sum_{j=k}^{k}[u]_{j;\Omega}^*;\\ \|u\|^*_{k,\alpha;\Omega}&=\|u\|^*_{k;\Omega}+[u]_{k,\alpha;\Omega}^*. \end{align*} It turns out that these norms help the proof of the Schauder estimates easier and sharper. Is there any one know what is the idea behind these norms ? Thank you very much.

• The obvious question: what is $d_{x}^k$? Dec 2, 2016 at 5:48
• Oh! Sorry. I forgot to define $d_x$. Dec 2, 2016 at 5:57

Here are a couple (related) ways to motivate this choice. The first is a physics-based argument related to units. If we think of the function $u$ as having units (say meters or seconds or something) $U$ then $\partial^\alpha u$ has units $U/D^{|\alpha|}$ for every $\alpha$, where $D$ is units of distance. In other words, when we take derivatives we change the units of the quantity under consideration. The same thing happens when we consider $(\partial^\alpha u(x) - \partial^\alpha u(y))/|x-y|^\beta$: it has units $U/ D^{|\alpha| + \beta}$ By multiplying by $d_{xy}^{\beta + |\alpha|}$ we restore the units back to just $U$. This means that all of the terms appearing in the norms have the same units.
Why should we care about this? Well, one reason is that it makes things scale in a better way. To show what I mean consider $\Omega =\{x \in \mathbb{R}^n : x_n >0\}$. Given a function $u : \Omega \to \mathbb{R}$ we can consider the rescalings of $u$ given by $u_\lambda(x) = u(\lambda x)$ for $\lambda >0$. I'll leave it to you to check that $$\Vert u_\lambda \Vert_{k,\alpha}^\ast = \Vert u \Vert_{k,\alpha}^\ast.$$ This shows that the norm is scaling-invariant. If we drop the $\ast$ and return to the norm without weights, then this scaling invariance is broken.