A new approach to Hardy-type inequalities

We introduce a new method which can be used to establish sharp Hardy-type inequalities on the positive halfline. As an illustration, we present a new proof of a classical result due to Bliss.


Introduction.
A classical Hardy inequality states that for any nonnegative function f on the positive halfline (0, ∞), we have the sharp estimate for any number k > 1 (see e.g. Hardy, Littlewood, and Pólya [4]). Our motivation comes from the extension established by Hardy and Littlewood [3]: where ≥ k > 1 and α = /k − 1. Unfortunately, as Hardy and Littlewood observed, the constant (k/(k − 1)) k above is no longer optimal when > k. However, they guessed what the best value is, and their conjecture was confirmed a few years later by Bliss [1]. Here is the precise statement. (1.3) Equipped with the above statement, one easily proves the following extension involving a power weight on the right hand side. Theorem 1.2. Suppose that 1 < k < are fixed constants, and let α, β be positive numbers satisfying β < k − 1 and (1 + α)(1 + β) −1 = /k. Then for any nonnegative function f on (0, ∞), we have

The inequality is sharp.
To see how this theorem follows from the previous one, simply apply (1.2) to the function t → t β/(k−1−β) f (t (k−1)/(k−1−β) ) and change the variables under the integrals.
There is a vast literature concerning various versions and applications of these results. It is impossible to review it here, and we refer the interested reader to the works [4][5][6][7][8][9][10][11] for an overview, history, and much more.
The main purpose of this paper is to introduce a novel method which can be used to prove general Hardy-type inequalities on the positive halfline. Roughly speaking, the method will allow to deduce a given estimate from the existence of a certain special function, possessing appropriate domination and monotonicity properties. In a sense, the technique is closely related to the so-called Bellman function method, a powerful tool used widely in harmonic analysis and probability theory: see e.g. [2,[12][13][14] and the references therein. The method is described in the next section. In Section 3, we illustrate the technique by providing a novel proof of (1.2).

A method.
Let f be a nonnegative function on (0, 1], and let k > 1 be fixed. Define Hardy inequalities 167 for t ∈ (0, 1]. A straightforward application of Hölder's inequality shows that the pair (X t , Y t ) takes values in the set for all f . For instance, choose V (x, y, t) = t α x and G(x, y) = Cy /k : then (2.2) becomes the "localized" Hardy inequality The key idea in the proof of (2. by (2.4). Moreover, using the monotonicity property (iii), we may write for any t ∈ (0, 1). Consequently, applying (2.3), we obtain for any t. Letting t → 0 yields the claim, by virtue of Fatou's lemma.
A very interesting feature of the method is that the implication of the above lemma can be reversed. If the estimate (2.2) holds true, one can write an abstract (but non-explicit) formula for a special function B.
For each (x, y, t), the class is nonempty: it contains, for instance, the function where the supremum is taken over all f ∈ M(x, y, t). Let us verify that B satisfies the required properties (2.3), (2.4), and (2.5). The first condition is evident since V is nonnegative. The second property follows from (2.2): for any f ∈ M(x, y, 1) we have y). Thus, taking the supremum over all f as above, we get (2.4). So, it remains to establish (2.5), for which the reasoning is slightly more complicated. Pick any function f : (0, 1] → [0, ∞) and fix 0 ≤ t < u ≤ 1. Take an arbitraryf ∈ M(X t (f ), Y t (f ), t), and let us "splice" f andf into one function f =fχ (0,t] + fχ (t,1] . Observe that for any r ∈ [t, 1], where in the third equality we used the assumption Vol. 104 (2015) Hardy inequalities 169 Taking the supremum over allf , we get the desired property (iii).
Suppose we have fixed V , G and we want to prove (2.2) with the use of the above method. How to find an appropriate B and how do we check that it has the desired properties? To address these issues, let us make a few comments on the properties (2.3), (2.4), and (2.5). The first two of them are just appropriate dominations for the function B: it can be neither too small nor too large. The most mysterious condition is the third one, and at the first glance it seems to be difficult to verify. Thus, it would be desirable to rephrase it in a more convenient form. We can offer a partial result in this direction. Namely, in many situations one can restrict oneself to showing (2.2) for continuous f only; furthermore, the functions V and G are often smooth, and one can expect the desired B to have this regularity as well. Note that for any t ∈ (0, 1). So, a direct differentiation of the function in (2.5) leads to the following, slightly stronger form of (iii), which we state separately.

Lemma 2.3. Suppose that B is continuous, of class C 1 in the interior of its domain, and satisfies
for any (x, y, t) ∈ D × (0, 1) and d ≥ 0. Then (iii) holds true.
This lemma also indicates how to search for B: it is enough to construct a smooth function which satisfies (2.3), (2.4), and the above differential inequality (2.6). Though the latter still looks complicated, in many cases it can be handled due to some additional, "structural" properties of B: see below.
Before we proceed, let us mention that the above method admits many extensions. We will give just one example, concerning bounds of the form  Let D be the set in which (X · (f ), Y · (f )) might take its values (it depends on Φ and β), fix V : D × (0, 1] → [0, ∞), G : D → [0, ∞) and suppose that we want to show (2.2) for all nonnegative f on (0, 1]. Then, as one easily verifies, Lemmas 2.1 and 2.2 remain valid, but Lemma 2.3 needs to be slightly modified. With the above X · and Y · , the stronger version of (iii) is for (x, y, t) ∈ D × (0, 1]. This is easily checked by the differentiation in (2.5).

A new proof of Bliss' result.
For the sake of clarity, we have decided to split this section into three parts.

Some special functions.
We start with the following technical fact.
Proof. Let s > 1 be fixed, and denote by F (u) the difference of the left-hand side and the right-hand side of (3.1). Observe that F (u) equals After some manipulations we check that F (u) has the same sign as u). Now, we see that G(0) < 0, G(u) is positive for u sufficiently close to 1, and G (u) = k(s − u k−1 ) > 0. So, there is u 0 such that G is negative on (0, u 0 ) and positive on (u 0 , 1), and hence, F is decreasing on (0, u 0 ) and increasing on (u 0 , 1). But we easily check that F (0) > 0 and F (1) = 0; thus F has a unique zero inside (0, 1). The claimed regularity of the function u follows at once from the well-known facts on implicit functions.
Proof. Let u be the function introduced in the previous lemma, and put Vol. 104 (2015)

Hardy inequalities 171
We compute that (for brevity, we have decided to omit the argument s and write u instead of u(s), ϕ instead of ϕ(s), etc.) Now differentiate (3.1) with respect to s, the argument of u, to get Plug this into (3.4): then the right-hand side becomes 1, so we get which, in turn, implies . (3.6) Inserting this identity into (3.2) transforms it into (3.5); this shows that ϕ [given by (3.3)] enjoys the differential equation (3.2), as desired. The monotonicity of ϕ follows immediately from (3.5). It remains to show that this ϕ satisfies lim s↓1 ϕ(s) = k/ . To do this, observe that lim s→1 u(s) = 1, by the very definition of u, and consequently, using (3.1), by de l'Hospital rule. This completes the proof.
The final property of ϕ is the following boundedness condition. Proof. By (3.6), for any s > 1 we have so we will be done if we manage to establish the identity lim s→∞ ϕ(s)s − /k = C k, ,α . To do this, note first that by (3.1), (3.3), and (3.5), or, which is the same, .
A little calculation shows that the right-hand side is equal to C 1/(α+1) k, ,α . Therefore, if s 0 > 1 is a fixed number and s > s 0 is arbitrary, we may write (1.2). Fix , k and let α = /k − 1. By an easy dilation, it is enough to prove that

Proof of
for any continuous nonnegative function f on [0, 1]. This bound is of the form (2.2), with V (x, y, t) = t α x and G(x, y) = C k, ,α y /k . Thus we can apply the method of Section 2. As we will see, the right choice for B is given by where ϕ is the function of Lemma 3.2. Clearly, this function is nonnegative and, by (3.7), we have B(x, y, 1) = x ϕ(x −k y) ≤ C k, ,α y /k = G(x, y). So, the where s = y/x k . If we divide both sides by x t α , the bound becomes where ξ = d/x ≥ 0. By the previous subsection, we know that ϕ (s) > 0 for all s. So, the left-hand side, considered as a function of ξ, attains its minimum for ξ satisfying k k ϕ(s) − sϕ (s) + kϕ (s)ξ k−1 = 0, or . But for this ξ both sides of (3.9) are equal: indeed, which is zero, due to (3.2). This proves that B belongs to the class B(V, G). The next property comes from the observation that for any λ > 0, we have f ∈ M(x, y, 1) if and only if λf ∈ M(λx, λ k y, 1). So, by the definition of B, we get B(λx, λ k y, 1) = λ B(x, y, 1). Hence, taking λ = 1/x, we get B(x, y, t)

On the search for
We conclude the paper with the following comment.

Remark 3.4.
There is a natural question about other Hardy-type inequalities which can be successfully treated with the above method. As an example (a partial answer to this question), one can show that the estimate of Theorem 1.2 can be proved with the use of a certain special function B. However, we have decided not to include any details here, since, as we have noted in the introductory section, the result follows at once from Theorem 1.1.