Hardy's inequality is an inequality in mathematics, named after G. H. Hardy. It states that if is a sequence of non-negative real numbers, then for every real number p > 1 one has If the right-hand side is finite, equality holds if and only if a_n = 0 for all n. An integral version of Hardy's inequality states the following: if f is a measurable function with non-negative values, then If the right-hand side is finite, equality holds if and only if f(x) = 0 almost everywhere. Hardy's inequality was first published and proved (at least the discrete version with a worse constant) in 1920 in a note by Hardy. The original formulation was in an integral form slightly different from the above.

General one-dimensional version

The general weighted one dimensional version reads as follows:

Multidimensional versions

Multidimensional Hardy inequality around a point

In the multidimensional case, Hardy's inequality can be extended to L^{p}-spaces, taking the form where, and where the constant is known to be sharp; by density it extends then to the Sobolev space. Similarly, if p > n \ge 2, then one has for every

Multidimensional Hardy inequality near the boundary

If is an nonempty convex open set, then for every , and the constant cannot be improved.

Fractional Hardy inequality

If and, , there exists a constant C such that for every satisfying , one has

Proof of the inequality

Integral version

A change of variables gives which is less or equal than by Minkowski's integral inequality. Finally, by another change of variables, the last expression equals

Discrete version: from the continuous version

Assuming the right-hand side to be finite, we must have a_n\to 0 as n\to\infty. Hence, for any positive integer j , there are only finitely many terms bigger than 2^{-j}. This allows us to construct a decreasing sequence containing the same positive terms as the original sequence (but possibly no zero terms). Since for every n , it suffices to show the inequality for the new sequence. This follows directly from the integral form, defining f(x)=b_n if n-1<x<n and f(x)=0 otherwise. Indeed, one has and, for n-1<x<n, there holds (the last inequality is equivalent to, which is true as the new sequence is decreasing) and thus

Discrete version: Direct proof

Let p > 1 and let be positive real numbers. Set. First we prove the inequality Let and let \Delta_n be the difference between the n-th terms in the right-hand side and left-hand side of, that is,. We have: or According to Young's inequality we have: from which it follows that: By telescoping we have: proving. Applying Hölder's inequality to the right-hand side of we have: from which we immediately obtain: Letting we obtain Hardy's inequality.