User:Addemf/sandbox/Markov and Hardy-Littlewood

Markov
As mentioned in the previous lesson, when we proceed to the proof of length-measure differentiation, we will find Markov's inequality useful. This theorem states, roughly, that an integrable function cannot have too many points where it is extremely large.

More formally, let $$f:\Bbb R\to\Bbb R$$ be an integrable function and $$c\in\Bbb R^+$$.

Define $$E_c = \{x\in\Bbb R:c\le|f(x)|\}$$. (Think of this as the set of points at which f is large.)

Then


 * $$\lambda(E_c)\le\frac 1 c\int f$$

Exercise 1. Prove Markov's
Prove Markov's inequality by construing $$\lambda(E_c)$$ as the integral of $$\mathbf 1_{E_c}$$, and then apply the ML bound.

(Would that it were so simple to prove the Hardy-Littlewood inequality.)

Hardy-Littlewood
The Hardy-Littlewood theorem asserts, much like the Markov inequality, that if f is integrable, then there cannot be too many points at which the maximal average is too large.

More formally, let $$E_c=\{x\in\Bbb R: c< f^*(x)\}$$ and the theorem states


 * $$\lambda(E_c) < \frac 3 c \int f$$

To start on the proof, we first observe that if $$x\in E_c$$ is arbitrary then there is some $$t\in\Bbb R^+$$ such that $$\frac 1{2t}\int_{(x-t,x+t)}|f| > c$$.

This allows us to define, for each x, the corresponding open interval $$(x-t,x+t)$$ with t as above. However, that would be senseless if we made the cover for all of $$E_c$$.

It is preferable if we could cover some compact set instead. We would thereby be guaranteed a finite subcover of open intervals.

Whence we let $$F\subseteq E_c$$ be any compact subset. The idea behind what we will do for the remainder of this proof, is to show that $$\lambda(F)\le\frac 3 c \int |f|$$. It will then follow that $$\lambda(E_c)\le\frac 3 c \int|f|$$.

Now for each $$x\in F$$ we define the corresponding open interval $$I_x=(x-t,x+t)$$ such that $$\frac 1{2t}\int_{(x-t,x+t)}f^* > c$$. By compactness there is a finite subcover, which we will choose to call $$J_1,\dots,J_n$$.

We would like to reason as follows (although, of course, I would only phrase it this way if there is an obstacle coming):


 * $$\begin{aligned}

\lambda(F) &\le \sum_{i=1}^n\lambda(J_i) \\ &=\sum_{i=1}^n 2t_i\quad \text{ where } t_i \text{ is the width of interval } J_i \\ &<\sum_{i=1}^n\frac{1}{c}\int_{J_i}|f| \\ &\le \frac 1 c\int|f| \end{aligned}$$

If this argument were correct then we could get a smaller bound on $$\lambda(F)$$, using $$\frac 1 c\int|f|$$ rather than $$\frac 3 c \int |f|$$.

However, the last inequality is not justified because the intervals $$J_1,\dots,J_n$$ may overlap.

Exercise 2. Justify the Hardy-Littlewood Steps
In this exercise you will justify each of the steps in


 * $$\begin{aligned}

\lambda(F) &\le \sum_{i=1}^n\lambda(J_i) \\ &=\sum_{i=1}^n 2t_i\quad \text{ where } t_i \text{ is the width of interval } J_i \\ &<\sum_{i=1}^n\frac{1}{c}\int_{J_i}|f| \\ &\le \frac 1 c\int|f| \end{aligned}$$

except for the last one, which we have observed is actually invalid.

1. Explain why $$\lambda(F)\le\sum_{i=1}^n\lambda(J_i)$$.

2. Explain why $$\lambda(J_i)=2t_i$$.

3. Explain why $$2t_i < \frac 1 c \int_{J_i}|f|$$. Hint: Recall the defining property of $$J_i$$.

The Vitali Covering Lemma
In order to overcome the obstacle above, that $$J_1,\dots,J_n$$ may fail to be disjoint, we will try to relate this collection to some other collection which is disjoint.

One strategy would be to simply merge overlapping intervals. For instance, if $$J_1,J_2$$ overlap each other, we could replace the pair with $$J'=J_1\cup J_2$$. Repeating the procedure finitely many times would produce a new family which is now composed of disjoint intervals.

However, notice that if we did so, the we would no longer be able to say $$ \lambda(J_i)=2t_i$$.

So we have two competing needs: The need for the intervals to be disjoint but also the need for the intervals to maintain their size.

The easiest resolution is to take the interval in $$J_1,\dots,J_n$$ with the greatest length (ties may be broken arbitrarily), assume without loss of generality that this is $$J_1$$. If this intersects any other interval then we simply remove those intersecting intervals.

Now define $$3*J_1$$ to be the interval with the same center as $$J_1$$, but with three times its length.

Exercise 3. Show that 3 Is Enough
Let $$I=(a,b), \ J=(c,d)$$ be two open bounded intervals. Assume that $$\ell(I)<\ell(J)$$ and that they intersect, $$I\cap J\ne\emptyset$$.

Prove that $$I\subseteq 3*J$$.

Exercise 4. Show that Vitali Covering Continues
Let $$I_1,\dots,I_m$$ be any family of open bounded intervals. Show that there exists a sub-family $$I_{k_1},\dots,I_{k_n}$$ with the properties


 * the family is pairwise disjoint, and
 * $$\bigcup_{i=1}^m I_i \subseteq \bigcup_{i=1}^n 3*I_{k_i}$$

This sub-family is called the Vitali-covering for the family $$I_1,\dots,I_m$$

Exercise 5. Complete the Hardy-Littlewood Proof
Using the Vitali-covering for the family $$J_1,\dots,J_n$$ as in the initial "false proof", correct the false proof to obtain a correct proof that


 * $$\lambda(F)\le \frac 3 c \int|f|$$

and then conclude the theorem.