Motivation/Intuition behind Lorentz spaces

Solution 1:

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 $p$.