Control level sets

Quick description

From Fubini's theorem one has the identity

$|f(x)| > t\} )\ dt$ (1)

for non-negative $f$ . Thus, to estimate an integral, one way to do this is to control the measure of the level sets $|f(x)| > t\}$ for different values of $t$ .

A slight variant: the $L^p$ norm of a function $f$ can be expressed by the formula

$|f(x)| > t\} ) t^{p-1}\ dt.$ (2)

As another variant, one can view (1) as a means to decompose any non-negative function $f$ as a superposition of indicator functions:

$f = \int_0^\infty 1_{f > t}\ dt.$

Prerequisites

Measure theory

Example 1

Let $X$ be a finite set, and let $X \to \C$ be a function. Show that

$\|f\|_{\ell^p(X)}^p \leq C_p \log(2+|X|) \|f\|_{\ell^{p,\infty}(X)}^p$

for any $0 < p < \infty$ .

Solution: we may as well normalize $\|f\|_{\ell^{p,\infty}(X)} = 1$ , thus

$\# \{ x \in X: |f(x)| \geq t \} \leq t^{-p}.$

If we insert this bound directly into (2) (using counting measure $\#$ instead of $\mu$ ) we obtain a logarithmic divergence. But we can improve the bound in two ways. Firstly, when $t > 1$ , then $\# \{ x \in X: |f(x)| \geq t \} = 0$ , since there is no other non-negative integer less than $t^{-p}$ . Secondly, we also have the trivial upper bound of $|X|$ , which is superior when $t < |X|^{-1/p}$ . If we then use "divide and conquer" and partition the integral on the right-hand side of (2) into the regions $t > 1$ , $t < |X|^{-1/p}$ , and $|X|^{-1/p} \leq t \leq 1$ , one obtains the claim.

Example 2

Show that if a function $f$ lies in the weak $L^p$ spaces $L^{p_0,\infty}(X)$ and $L^{p_1,\infty}(X)$ for some measure space $(X,\mu)$ and some exponents $0 < p_0 < p_1 < \infty$ , then it lies in the strong $L^p$ spaces $L^p(X)$ for all $p_0 < p < p_1$ .

We need to use "divide and conquer" efficiently in order to 'interpolate' the information we have at the endpoint weak $L^p$ spaces. The idea is that when the function $f$ is large, say $|f|>1$ , then the $L^p$ norms increase when $p$ increases which is the same as saying that $t^p<t^{p_1}$ whenever $1<t$ and $p<p_1$ . This indicates that we should split the integral $\int_X |f(x)|^pd\mu(x)$ into two parts; one integral over the set where $f$ is large and one over its complement. Then the integral over the region where $f$ is large is controlled by the weak $L^{p_1}$ norm of $f$ and the integral where $f$ is small is controlled by the weak $L^{p_0}$ norm of $f$ .

This can be done in a very elegant fashion by using the description (2) of the $L^p$ norm of a function. Indeed we can write

$|f(x)| > t\} ) t^{p-1}\ dt.\end{align}$

For the first term (which corresponds to the set where $f$ is small) we use the weak $L^{p_0}$ estimate

$|f(x)| > t\} ) t^{p-1}\ dt\leq \|f\|_{L^{p_0,\infty}}\int_0 ^1 t^{p-p_0-1}dt=\frac{\|f\|^{p_0} _{L^{p_0,\infty}}}{p-p_0}<+\infty.$

Similarly we get for the second term

$|f(x)| > t\} ) t^{p-1}\ dt\leq \frac{\|f\|^{p_1} _{L^{p_1,\infty}}}{p_1-p}<+\infty.$

Thus we have

$\int_X |f(x)|^p d\mu(x) \leq p \bigg(\frac{\|f\|^{p_0} _{L^{p_0,\infty}}}{p-p_0}+\frac{\|f\|^{p_1} _{L^{p_1,\infty}}}{p_1-p}\bigg)<+\infty,$

and in particular $f\in L^p(X).$

More efficiently, one could split the integral at some point $\lambda$ instead of the point $1$ and then optimize in the parameter $\lambda$ .

General discussion

This method is closely related to double counting and "interchange integrals or sums".

The method also combines well with dyadic decomposition. Indeed, one easily verifies that $\int_E |f(x)|\ d\mu(x)$ is comparable to $\sum_{n \in \Z} 2^n \mu(\{ |f| \geq 2^n \} )$ , and more generally $\int_E |f(x)|^p\ d\mu(x)$ is comparable (up to constants depending on $p$ ) to $\sum_{n \in \Z} 2^{np} \mu(\{|f| \geq 2^n\})$ .