Lorentz Space
The purpose of this note is to give a few equivalent definitions, \eqref{def1}, \eqref{def2}, \eqref{def3} and \eqref{def4} for Lorentz norm $L ^{p, q}$.
Distribution function and Decreasing Rearrangement
Similar as Lebesgue space, Lorentz space only cares about the height of a function and the space it occupies at each height. Let $f \in \mathcal L ^0 \pth{X, \d \mu}$ be any complex-valued measurable function. Any definition below that works for $f \ge 0$ can be generalized for complex-valued (and in general Banach space-valued) $f$ by replacing $f$ by $\abs{f}$.
Definition. The distribution function of $f$ is defined as the measure of the superlevel set, that is
\begin{align} d _f (\alpha) := \mu \pth{\set{x \in X: |f (x)| > \alpha}}, \qquad \alpha \ge 0. \end{align}
This definition is the opposite to the definition of cdf (cumulative distribution function) for a random variable that we are used to in probability. So in contrast to cdf, $d _f$ is left/lower semicontinuous, and is a decreasing function.
Definition. The decreasing rearrangement of $f$ is defined on $[0, \infty)$ by
\begin{align} f ^* (\lambda) := \inf \set{t > 0: d _f (t) \le \lambda} \end{align}
This is almost like the “inverse function” of the distribution function. It is also left/lower semicontinuous and decreasing. Notice that the superlevel set of $f ^*$ has the same measure as $f$.
One can take the following example to understand the construction of the distribution function and decreasing rearrangement. Assume the following is the graph of $f = \ind{S _1} + 4 \ind{S _2} + 2 \ind{S _3}$ where $\mu (S _1) = \mu (S _2) = \mu (S _3) = 1$.
To find its distribution function, rotate by 90 degrees and then find the measure of superlevel set at each level $\alpha$.
Finally, rotate back and flip the distribution function to obtain decreasing rearrangement.
Compare $f$ and $f ^*$, it looks like just push all the level sets to the left onto the $\alpha$ axis, and end up with something decreasing.
Layer Cake Representation and $L ^p$ Norm
The layer cake representation states the following. For $f \ge 0$,
\begin{align} \int _X f \d \mu = \int _0 ^\infty d _f (\alpha) \d \alpha. \end{align}
This comes from thinking integration over the hypograph,
\begin{align} \Omega _f = \set{(x, \alpha) \in X \times \R _+: f (x) > \alpha}. \end{align}
Therefore
\[\begin{align} \int _X f \d \mu &= \int _X \int _0 ^{f(x)} \d \alpha \d \mu \notag\\\ &= \int _{\Omega _f} \d \alpha \d \mu \notag\\\ &= \int _0 ^\infty \int _X \ind{\set{f(x) > \alpha}} \d \mu \d \alpha \notag \\\ &= \int _0 ^\infty \mu (f > \alpha) \d \alpha \notag \\\ &= \int _0 ^\infty d _f (\alpha) \d \alpha. \notag \end{align}\]Using decreasing rearrangement, we can also think
\begin{align} \int _X f \d \mu = \int _0 ^\infty f ^* \d \lambda = \int _{\Omega _{f ^*}} \d \alpha \d \lambda. \notag \end{align}
Here $\Omega _{f ^*}$ is the hypograph of $f ^*$ (and also the hypograph of $d _f$)
\begin{align} \Omega _{f ^*} = \set{(\lambda, \alpha) \in \R _+ ^2: f ^* (\lambda) > \alpha} = \set{(\lambda, \alpha) \in \R _+ ^2: d _f (\alpha) > \lambda}. \notag \end{align}
Similarly, for $L ^p$ norm,
\begin{align} \int _X f ^p \d \mu &= \int _{\Omega _{f ^*}} \d (\alpha ^p) \d \lambda = \int _0 ^\infty d _f (\alpha) \d (\alpha ^p) = \int _0 ^\infty p \alpha ^{p - 1} d _f (\alpha) \d \alpha. \notag \end{align}
So the integral of $f ^p$ can be just thought as the $\d(\alpha ^p) \tensor \d \lambda$ measure of the hypograph.
Lorenze Space
Definition. For $p, q > 0$, we define the Lorenze norm of $f \ge 0$ by
\begin{align} \label{def1} \nor{f} _{L ^{p, q}} ^q = \int _{\Omega _{f ^*}} \d (\alpha ^q) \d \pth{\frac pq \lambda ^{\frac qp}} = \int _{\Omega _{f ^*}} \lambda ^{\frac qp - 1} \d (\alpha ^q) \d \lambda. \end{align}
That is, we change the measure we put on domain, and put $L ^q$-typed measure on range. If we integrate $\lambda$ first, we end up with
\[\begin{align} \nor{f} _{L ^{p, q}} ^q &= \int _0 ^\infty \frac pq \bkt{d _f (\alpha)} ^{\frac qp} q \alpha ^{q - 1} \d \alpha \notag \\\ &= p \int _0 ^\infty \bkt{d _f (\alpha)} ^{\frac qp} \alpha ^{q} \frac{\d \alpha}\alpha \label{split} ,\\\ \nor f _{L ^{p,q}} &= p ^\frac1q \nor{\alpha [d _f (\alpha)] ^\frac1p} _{L ^q \pth{\R _+, \frac{\d \alpha}\alpha}} . \label{def2} \end{align}\]If we integrate $\alpha$ first, we end up with
\[\begin{align} \nor f _{L ^{p, q}} ^q &= \int _0 ^\infty [f ^* (\lambda)] ^q \lambda ^{\frac qp - 1} \d \lambda \notag \\ &= \int _0 ^\infty \lambda ^{\frac qp} [f ^* (\lambda)] ^q\frac{\d \lambda}\lambda \notag ,\\ \nor f _{L ^{p, q}} &= \nor{\lambda ^{\frac1p} f ^* (\lambda)} _{L ^q \pth{\R _+, \frac{\d \lambda}\lambda}}. \label{def3} \end{align}\]Dyadic Decomposition in Range
We decompose $f \ge 0$ by range as the following. Let $E _k = \set{f > 2 ^k}$ denote a sequence of nested dyadic level sets, and define
\[\begin{align*} f _k &= 2 ^{k-1} \ind{E _{k + 1}} + (f - 2^{k - 1})(\ind{E _k} - \ind{E _{k + 1}}) \\\ &= f(\ind{E _k} - \ind{E _{k + 1}}) + 2 ^k \ind{E _{k + 1}} - 2^{k - 1} \ind{E _k} . \end{align*}\]Then $f = \sum _{k \in \mathbb Z} f _k$, $f _k$ is supported in $E _k$, and $f _k$ is comparible to $2 ^k \ind{E _k}$:
\[\begin{align*} \frac12 2 ^k \ind{E _k} \le f _k \le \frac32 2 ^k \ind{E _k}. \end{align*}\]
Now we also split the integral \eqref{split} dyadically,
\[\begin{align*} \frac1p \nor{f} _{L ^{p, q}} ^q &= \int _0 ^\infty \bkt{d _f (\alpha)} ^{\frac qp} \alpha ^{q} \frac{\d \alpha}\alpha \notag \\\ &= \sum _{k \in \Z} \int _{2 ^k} ^{2 ^{k + 1}} \bkt{d _f (\alpha)} ^{\frac qp} \alpha ^{q} \frac{\d \alpha}\alpha .\notag \\\ \end{align*}\]Since $d _f$ is decreasing,
\[\begin{align*} 2 ^{kq} \bkt{d _f \pth{2 ^{k + 1}}} ^{\frac qp} \lesssim \int _{2 ^k} ^{2 ^{k + 1}} \bkt{d _f (\alpha)} ^{\frac qp} \alpha ^{q} \frac{\d \alpha}\alpha \lesssim 2 ^{kq} \bkt{d _f \pth{2 ^{k}}} ^{\frac qp}. \notag \end{align*}\]By definition of $E _k$,
\[\begin{align*} 2 ^{kq} \mu\pth{E _{k + 1}} ^{\frac qp} \lesssim \int _{2 ^k} ^{2 ^{k + 1}} \bkt{d _f (\alpha)} ^{\frac qp} \alpha ^{q} \frac{\d \alpha}\alpha \lesssim 2 ^{kq} \mu\pth{E _{k}} ^{\frac qp}. \notag \end{align*}\]Since $\nor{f _k} _{L ^p} \sim \nor{2 ^k \ind {E _k}} _{L ^p} = 2 ^k \mu \pth{E _k} ^{\frac 1p}$,
\[\begin{align*} \nor{f _{k + 1}} _{L ^p} ^q \lesssim \int _{2 ^k} ^{2 ^{k + 1}} \bkt{d _f (\alpha)} ^{\frac qp} \alpha ^{q} \frac{\d \alpha}\alpha \lesssim \nor{f _k} _{L ^p} ^q. \notag \end{align*}\]By taking the summation over $k$, we obtain an equivalent way of defining $L ^{p, q}$ norm,
\[\begin{align} \label{def4} \nor f _{L ^{p, q}} \sim _{p,q} \nor{ \set{\nor{f _k} _{L ^p}} _{k \in \Z}} _{\ell ^q}. \end{align}\]