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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\}).

The Marcinkiewicz interpolation theorem relies heavily on these sorts of level set decompositions.