Control level sets

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Control level sets

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[QUICK DESCRIPTION]

From Fubini's theorem one has the identity

[math fub] \int_E f(x)\ d\mu(x) = \int_0^\infty \mu( \{ x \in E: |f(x)| > t\} )\ dt[/math]

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

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

[math pdist] \int_E |f(x)|^p\ d\mu(x) = p \int_0^\infty \mu( \{ x \in E: |f(x)| > t\} ) t^{p-1}\ dt.[/math]

As another variant, one can view [eqref fub] as a means to decompose any non-negative function $f$ as a superposition of indicator functions:

[math] f = \int_0^\infty 1_{f > t}\ dt.[/math]

[PREREQUISITES]

Measure theory

[EXAMPLE]

Let $X$ be a finite set, and let $f: X \to \C$ be a function.  Show that
[math] \|f\|_{\ell^p(X)}^p \leq C_p \log(2+|X|) \|f\|_{\ell^{p,\infty}(X)}^p[/math]
for any $0 < p < \infty$.

Solution: we may as well normalize $\|f\|_{\ell^{p,\infty}(X)} = 1$, thus
[math] \# \{ x \in X: |f(x)| \geq t \} \leq t^{-p}.[/math]

If we insert this bound directly into [eqref pdist] (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 [eqref pdist] into the regions $t > 1$, $t < |X|^{-1/p}$, and $|X|^{-1/p} \leq t \leq 1$, one obtains the claim.

[EXAMPLE]

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 [eqref pdist] of the $L^p$ norm of a function. Indeed we can write

[math] \begin{align}\int_X |f(x)|^p d\mu(x)&= p \int_0^\infty \mu( \{ x \in X: |f(x)| > t\} ) t^{p-1}\ dt\leq \\ &\leq p\int_0 ^1 \mu( \{ x \in X: |f(x)| > t\} ) t^{p-1}\ dt+ p\int_1 ^\infty  \mu( \{ x \in X: |f(x)| > t\} ) t^{p-1}\ dt.\end{align}[/math]

For the first term (which corresponds to the set where $f$ is small) we use the weak $L^{p_0}$ estimate
[math] \int_0 ^1 \mu( \{ x \in X: |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.[/math]

Similarly we get for the second term
[math] \int_0 ^1 \mu( \{ x \in X: |f(x)| > t\} ) t^{p-1}\ dt\leq  \frac{\|f\|^{p_1} _{L^{p_1,\infty}}}{p_1-p}<+\infty.[/math]

Thus we have

[math]\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,[/math]

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 [[To optimize a sum try making the terms roughly equal in size|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 $\sum_{n \in \Z} 2^{np} \mu(\{|f| \geq 2^n\})$.

The [[w:Marcinkiewicz interpolation theorem]] relies heavily on these sorts of level set decompositions.

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