Measure Theory/Markov and Hardy-Littlewood

Markov's Inequality

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We've already had enough preamble for this theorem. So let's just jump in and state, then prove, the following theorem.

Theorem: Let   be an integrable function and  .

Define  . (Think of this as the set of points at which f is large.)

Then

 


Exercise 1. Prove Markov's


Prove Markov's inequality by construing   as the integral of  , and then apply the ML bound.

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

Hardy-Littlewood

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Let   and we want to show that

 

To start on the proof, we first observe that if   is arbitrary then there is some   such that  .

This allows us to define, for each x, the corresponding open interval   with t as above. This looks like, perhaps, a cover of   by open intervals! That sounds provocative and familiar.

However, it would be senseless to cover for all of  , since there is no guarantee that   is compact. Compactness is precisely what would make an open interval cover useful.

Whence we let   be any compact subset. The idea behind what we will do for the remainder of this proof, is to show that  . It will then follow that  , which you will prove in an exercise below.

Now for each   we define the corresponding open interval   such that  . By the compactness of F, there is a finite subcover, which we will choose to call  .

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

 

If this argument were correct then we could get a smaller bound on  , using   rather than  .

However, the last inequality is not justified because the intervals   may overlap.


Exercise 2. Justify the Hardy-Littlewood Steps


In this exercise you will justify each of the steps in

 

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

1. Explain why  .

2. Explain why  .

3. Explain why  . Hint: Recall the defining property of  .

The Vitali Covering Lemma

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In order to overcome the obstacle above, that   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   overlap each other, we could replace the pair with  . 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  .

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   with the greatest length (ties may be broken arbitrarily), assume without loss of generality that this is  . If this intersects any other interval then we simply remove those intersecting intervals.

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


Exercise 3. Show that 3 Is Enough


Let   be two open bounded intervals. Assume that   and that they intersect,  .

Prove that  .


Exercise 4. Show that Vitali Covering Continues


Let   be any family of open bounded intervals. Show that there exists a sub-family   with the properties

  • the family is pairwise disjoint, and
  •  

This sub-family is called the Vitali-covering for the family  

The End

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Exercise 5. Complete the Hardy-Littlewood Proof


Using the Vitali-covering for the family   as in the initial "false proof", correct the false proof to obtain a correct proof that

 

and then conclude the theorem.