Using scaling arguments to determine relationships between Sobolev spaces? I was looking up how to find relationships between Sobolev spaces and I came across this post on MO in which the first comment talks about a scaling procedure for understanding the relationships:

To find the right values of k,p,r,a, I was taught to use scaling
  arguments. Take a nice enough singularity for the function inside the
  domain, and make it worse and worse, and compare how the norms change.
  This usually lets you solve for the appropriate parameter (and if it
  doesn't it tells you there is something special about that embedding).
  For unbounded domains you can do the same thing, and this often shows
  why certain embeddings cannot exist (handling singularities both
  inside and at infinity is hard).

Other answers also mentioned this scaling approach, which I have never heard of, with one comment saying:

Comparing the behavior of the different Sobolev norms for a compactly
  supported family of functions converging to a Dirac delta function is
  indeed the simplest way to figure out these inclusions. In general,
  understanding how things behave under rescalings is extremely useful
  but, as far as I know, rarely mentioned in print. It seems that you
  have to learn about it by word of mouth or stumble onto it yourself.
  Physicists and chemists also use this often and call it "unit
  analysis".

So does anyone here know of this scaling approach? If so could you give a worked example for some specific function, so somebody (like myself) who is new to Sobolev spaces can learn how to apply it and gain a better understanding of these spaces?
 A: Yet another term: I think of these as homogeneity arguments.
A simple example (has to be simple, I don't know much about Sobolev spaces): You know that if $f$ is a function on the line and $f'\in L^2$ then $f\in Lip_{1/2}$. Are there any other $Lip_\alpha$ spaces for which this is true? No.
Suppose $$\sup_{x\ne y}\frac{|f(x)-f(y)|}{|x-y|^\alpha}\le c||f'||_2$$(that's the $L^2$ norm). For $\lambda>0$ let $$g(x)=f(\lambda x).$$Replacing $f$ by $g$ in that inequality leads to $$\lambda^\alpha\sup_{x\ne y}\frac{|f(x)-f(y)|}{|x-y|^\alpha}\le c\lambda^{1/2}||f'||_2,$$or $$\sup_{x\ne y}\frac{|f(x)-f(y)|}{|x-y|^\alpha}\le c\lambda^{1/2-\alpha}||f'||_2.$$If $\alpha<1/2$ you get a contradiction by letting $\lambda\to0$, wihle if $\alpha>1/2$ you get a contradiction by letting $\lambda\to\infty$.
Bonus You can also sometimes use homogeneityy arguments to improve inequalities. My favorite example: It's a fact that if $f,f'\in L^2(\Bbb R)$ then $\hat f\in L^1(\Bbb R)$. (Where $\hat f$ is the Fourier transform.) Which is to say $$||\hat f||_1\le  c(||f||_2+||f'||_2).$$
Now replace $f$ by $f(\lambda x)$, look at how all three quantities transform, divide by the power of $\lambda$ you get on the LHS and then choose $\lambda$ to minimize the RHS: You get $$||\hat f||_1\le c\left(||f||_2||f'||_2\right)^{1/2}.$$That's the "homogenized" version of the first inequality.
