Scale invariant definition of the Sobolev norm $\|\|_{m,\Omega}$ for $H^m(\Omega)$

Question

I learned the following from Constantin and Foias's Navier-Stokes Equations (Chapter 4):

We say that a function of a bounded open set $\Omega\subset\mathbb{R}^n$, $c(\Omega)$, is scale invariant if $c(\Omega)=c(\Omega')$ for all $\Omega'$ obtained from $\Omega$ by a rigid transformation and a dilation $x\mapsto\delta x$.

Denote by $T_\delta$ the operation mapping functions defined on $\Omega$ to function defined on $\Omega_\delta:=\{\delta x\mid x\in\Omega\}$: $$ (T_\delta f)(y)=f(\frac{y}{\delta}). $$ When we refer to the way something scales we mean under the dilations $\delta$ and operations $T_\delta$. One can modify the definition of the norms $\|\cdot\|_{m,\Omega}$ of $H^m(\Omega)$ in such a way that they scale as the pure $m$-th order derivatives do. We shall denote by $ |\Omega|=\int_\Omega 1\ dx $ and by $ L(\Omega)=L=|\Omega|^{1/n} $ the linear size of $\Omega\subset\mathbb{R}^n$.

Define $\|\|_{m,\Omega}$ with $$ \|f\|_{m,\Omega}^2=\sum_{|\alpha|\leq m}\color{blue}{\big|\Omega\big|^{\dfrac{2(|\alpha|-m)}{n}}} \int_{\Omega}|D^\alpha f|^2\ dx $$ With this definition the quantity $\|f\|_{m,\Omega}$ scales like $L^{\frac{n}{2}-m}$, i.e., $$ C(\Omega):=\dfrac{1}{\big|\Omega\big|^{\dfrac{1}{2}-\dfrac{m}{n}}}\cdot \|f\|_{m,\Omega} $$ is scale invariant.

Here are my questions:

• Would anybody find a heuristic way to cook up the exponent in the term $$ \displaystyle \color{blue}{\big|\Omega\big|^{\dfrac{2(|\alpha|-m)}{n}}}? $$
• A direct simplification gives $$ C(\Omega)=\sqrt{\sum_{|\alpha|\leq m} \big|\Omega\big|^{\dfrac{2|\alpha|}{n}-1}\int_\Omega|D^\alpha f|^2\ dx}. $$ Would anybody show me why this is scale invariant?

Answer 1

The related question

Scalar property of $ C(\Omega)=\sum_{|\alpha|\leq m}\color{blue}{\big|\Omega\big|^{\dfrac{2|\alpha|-n}{n}}} \int_{\Omega}|D^\alpha f|^2\ dx $

and its answer gives a direct answer for the second question.

For the first question, one can use the calculation of $\|T_\delta f\|_{m,\delta\Omega}$ to find the correct exponents.

Answer 2

The following is an answer by Terry Tao, which is similar to Winther's comment.

If $f$ is dimensionless, then $D^\alpha f$ has the units of $L^{-|\alpha|}$, and so $\int_\Omega |D^\alpha f|^2\ dx$ has the units of $L^n \times (L^{-\alpha})^2$. As $\Omega$ has units of $L^n$, each summand in $\|f\|_{m,\Omega}^2$ has the units of $(L^n)^{\frac{2(|\alpha|-m)}{n}} \times L^n \times (L^{-\alpha})^2 = L^{n-2m}$, and hence $\|f\|_{m,\Omega}$ scales as $L^{\frac{n}{2}-m}$ as claimed.