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Several ways to prove Hardy's inequality

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Several ways to prove Hardy's inequality

Suppose and (with respect to Lebesgue measure of course) is a nonnegative function, take we have Hardy's inequality where of course.

There are several ways to prove it. I think there are several good reasons to write them down thoroughly since that may be why you find this page. Maybe you are burnt out since it's left as exercise. You are assumed to have enough knowledge of Lebesgue measure and integration.

Minkowski's integral inequality

Let be two measurable set, suppose is measurable, then A proof can be found at here by turning to Example A9. You may need to replace all measures with Lebesgue measure .

Now let's get into it. For a measurable function in this place we should have . If we put this function inside this inequality, we see Note we have used change-of-variable twice and the inequality once.

A constructive approach

I have no idea how people came up with this solution. Take where . Hölder's inequality gives us Hence

Note we have used the fact that and . Fubini's theorem gives us the final answer: It remains to find the minimum of . This is an elementary calculus problem. By taking its derivative, we see when it attains its minimum . Hence we get which is exactly what we want. Note the constant cannot be replaced with a smaller one. We simply proved the case when . For the general case, one simply needs to take absolute value.

Integration by parts

This approach makes use of properties of space. Still we assume that but we also assume , that is, is continuous and has compact support. Hence is differentiable in this situation. Integration by parts gives Note since has compact support, there are some such that only if and hence . Next it is natural to take a look at . Note we have hence . A substitution gives us which is equivalent to say Hölder's inequality gives us Together with the identity above we get which is exactly what we want since and all we need to do is divide on both sides. So what's next? Note is dense in . For any , we can take a sequence of functions such that with respect to -norm. Taking and , we need to show that pointwise, so that we can use Fatou's lemma. For , there exists some such that . Thus Hence pointwise, which also implies that pointwise. For we have note the third inequality follows since we have already proved it for . By Fatou's lemma, we have


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