TITLE: Motivation/Intuition behind Lorentz spaces
QUESTION [12 upvotes]: My current understanding is that the Lorentz spaces $L^{p,q}$ arise naturally as interpolation spaces between $L^1$ and $L^\infty$, but then people often describe them heuristically by saying something along the lines of "Lorentz spaces provide a finer control than $L^p$ spaces", and this is where I'm lost - what does that really mean?
It certainly seems like a reasonable claim, if only because you now have an extra parameter to tweak, and since $L^{p,p}=L^p$, well the Lorentz spaces are simply a larger class of spaces amongst which your classical $L^p$ spaces live, so sure, they are "better" because there's more of them, so I can give more nuanced descriptions, but I don't really understand where the nuance lies, I don't understand what extra control the Lorentz spaces provide you that the usual $L^p$ spaces do not.
I feel like my question is very vague overall, so feel free to ask for clarifications. As an example of the type of answer that I think there might be to what I am asking is the following cryptic (to me anyway) comment on the wikipedia page for "Lorentz spaces": "The Lorentz norms provide tighter control over both qualities than the $L^{p}$ norms, by exponentially rescaling the measure in both the range (p) and the domain (q)". I have no idea what that means, if anyone does, please let me know, but it seems like, after clarification, it would provide a nice intuitive explanation for precisely how Lorentz spaces provide finer control than $L^p$ spaces do.

REPLY [2 votes]: Note that if $F$ is the distribution function of $f$, and $f = H \cdot \mathbf{I}_E$, where $E$ is a measurable set with $|E| = W$, then $F = W \cdot \mathbf{I}_{[0,H]}$. Thus, in some sense, the distribution function switches the domain and range of a function, so that the 'range' of $f$, is the 'domain' of $F$, and vice versa. In particular, the $L^p$ norms
$$ \| f \|_p = \left( \int |f(x)|^p\; dx \right)^{1/p} \sim \left( \int_0^\infty F(t) t^p \frac{dt}{t} \right)^{1/p} $$
try to understand the distribution of $f$ by scaling it's range, or by scaling the `domain' of $F$ (changing the power of $t$ in the equation). Conversely, the Lorentz norm
$$ \| f \|_{p,q} \sim \left( \int_0^\infty (tF(t)^{1/p})^q \frac{dt}{t} \right)^{1/q} $$
have two separate powers $p$ and $q$. Here $p$ scales the domain of $f$, and $q$ scales the domain and range of $f$ simultaneously. We changed $F(t)$ from being linear to being a power of $1/p$, but this is only slightly confusing because
$$ \left( \int_0^\infty (t^pF(t))^{1/q} \frac{dt}{t} \right)^{1/q} \sim \| f \|_{p,q/p} $$
The reason that $q$ needs to scale the domain and range simultaneously is so that it acts as a second order parameter for the family of quasinorms, when compared to the primary exponent, which is $p$.