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Measure Theory/Integral Product Bound and Triangle Inequality

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The Lp Product Bound

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Now recall that the point of obtaining the real product-to-sum bound from the previous lesson,

for any and and

was in the hope of obtaining a bound on an integral of the form .


Exercise 1. Integral Product Bound


The following result is often called Hölder's inequality.

Let and and then let and .

Prove that .

Hint: In many cases of a normed vector space, it is easier to prove a claim first for vectors of unit norm, and then relate this result to the space of all vectors. In this case, that approach can help to simplify the proof. So start by assuming that , in which case your goal is to prove

.

Now apply the sum-to-product bound.

When you have finished that, you now need to let have any norm at all. Consider first the case that either f or g has norm equal to zero, which should be straight-forward.

Next consider that their norms are both not zero and consider the normalized vectors . Argue that these two vectors now each have norm equal to 1 and then apply your result from before.

Then with a little bit of manipulation, infer the desired result.


Exercise 2. The Conjugate Function


Let and let . Define to be the function which "complements" f with respect to the norm. What I mean by that is that we take to be a function which makes the equation

come out correct. The function which does so, is sometimes called the conjugate of f.

1. Prove that .
2. Prove that .
3. Show that .

The Triangle Inequality

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Finally we are in a position to prove the triangle inequality for .


Exercise 3. The Lp Triangle Inequlity


This theorem is sometimes called Minkowski's inequality.

For prove that .

Hint: From split off a single factor of and then use this to distribute the multiplication and then the integration. What is left should be a conjugate function for . Continue from there.


Exercise 4. Lp Is Complete


Recall that one of the most important properties of which we proved, was its completeness. The proof of its completeness was by way of a theorem for normed vector spaces which applies just as well to at this point.

Show that the proof for completeness is exactly the same proof for completeness, mutatis mutandis.