Intuition regrading the weak $L^p$ functions

Question

In this set of notes in harmonic analysis by Tao, the following remark is made:

On a Enclidean space ${\bf R}^d$, the power function $|x|^{-\alpha}$ lies in weak $L^p$ if and only if $\alpha=d/p$. Indeed one can think of a weak $L^p$ function as a function which is pointwise dominated in magnitude by a rearrangement of (a multiple of) $|x|^{-d/\alpha}$.

It is not very difficult to follow the definition to show that $|x|^{-\alpha}\in L^{p,\infty}({\bf R}^d)$ iff $\alpha=d/p$.

Question: How should one understand the second sentence:

"Indeed one can think of a weak $L^p$ function as a function which is pointwise dominated in magnitude by a rearrangement of (a multiple of) $|x|^{-d/\alpha}$"?

If this is made as a precise statement, what would it be?

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It seems to me that this literally means $f\in L^{p,\infty}({\bf R}^d)$ if and only if $|f(x)|\leq C |x|^{-d/\alpha}$ almost everywhere. But I don't see how it relates to the first sentence in the remark.

Answer 1

The weak $L^p$ spaces are rearrangement invariant. Indeed this is due to their use of the distribution function $d_f$ in their very definition,

$$f\in L^{p,\infty} ⇔ d_f(t) := \mu(|f|> t) ≤ \frac{[f]^p_{L^{p,\infty}}}{t^p}$$

There are many different functions $g$ with $d_f = d_g $, (for instance $d_{|f|} = d_f$) so it is not the specific shape of $f$ that matters, only that of $d_f$, and one such function is the symmetric decreasing rearrangement of $f$, normally written $f^*$. In addition, for radial decreasing nonnegative functions $f,g$ with $f(0)=g(0)$, $d_f ≤ d_g$ iff $f≤g$.

The remark is clear once we shift attention to $f^*$: If $\alpha = d/p$ and $|f^*(x)| = |f^*(|x|)| ≤ C |x|^{-\alpha}$, then since $$d_f(t) = d_{f^*}(t) = \mu(|f^*|>t) ≤ \mu(C|x|^{-\alpha} > t) = d_{C|\bullet|^{-\alpha}}(t)$$ we have $f\in L^{p,\infty}$. Also, since $d_{|\bullet|^{-\alpha}}(t)$ is actually equal to $\frac{\tilde C^p}{t^p}$, the reverse is also true: a function in $L^{p,\infty}$ is, up to a rearrangement, dominated in magnitude by a multiple of $|x|^{-\alpha}.$

Answer 2

By definition we have $$f^*(x)=\inf\{s>0:|\{t:|f(t)|>s\}|\leq x\}$$

Let $k(x) = |x|^{-d/p}$

we have $$|\{t:|f(t)|>s\}| =|\{t:|t|<s^{-p/d}\}| =|B(0,s^{-p/d} )| =c_ds^{-p} $$ where $c_d =|B(0,1)|$ . Whence, $$k^*(x)=\inf\{s>0:|\{t:|f(t)|>s\}|\leq x\} = \inf\{s>0:c_ds^{-p} \leq x\} \\=\inf\{s>0: s^{} \geq c_d^{1/p}x^{-1/p}\} =\color{red}{c_d^{1/p}x^{-1/p}.}$$

we have $$ \color{red}{[k^*(x)]^p=c_d x^{-1}.}$$

On the other hand, the magnitude function (or distribution function) of $f$ is

$$\color{red}{s\mapsto |\{x: |f(x)|>s\}| :=d_f(s)}$$

By definition of the weak $L^p$-norm we have $$f\in L^{p,\infty} \Longleftrightarrow \exists~~ C>0: \text{such that } \|f\|_{L^{p,\infty}} =\sup_{t>0} td_f(t)^{1/p} = \color{red}{c_d^{}}\sup_{t>0} \color{red}{c_d^{-1}}td_f(t)^{1/p} \\ =\color{red}{c_d^{}}\sup_{t>0} \color{red}{k^*(t)}^{-p}d_f(t)^{1/p}<C$$

That is, up to relabeling the constant $C>0$ we get: $$f\in L^{p,\infty} \Longleftrightarrow \exists~~ C>0: \text{for all $t>0$}~~\color{red}{k^*(t)}^{-p} d_f(t)^{1/p} := \color{red}{k^*(t)}^{-p}|\{x: |f(x)|>s\}|^{1/p} ≤ C $$

that is $$f\in L^{p,\infty} \Longleftrightarrow \exists~~ C>0: \text{for all $t>0$}~~~ \color{blue}{d_f(t) := |\{x: |f(x)|>s\}| \le} C\color{red}{k^*(t)^{p^2}}$$

This plainly justify the following statement

"Indeed one can think of a weak $L^p$ function as a function which is pointwise dominated in magnitude by a rearrangement of (a multiple of) $|x|^{-d/\alpha}$".