#### One-dimensional differential Hardy inequality

Kalybay Journal of Inequalities and Applications
One-dimensional differential Hardy
Aigerim Kalybay 0 1 2
0 University , Munaytpasov St. 5
1 Almaty , 050010 , Kazakhstan
2 Astana , 010008 , Kazakhstan
Hardy inequality to hold, including the overdetermined case. The solution is given in terms different from those of the known results. Moreover, the least constant for this inequality is estimated. © The Author(s) 2017. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
inequality; Hardy-type inequality; weight; differential operator
1 Introduction
space of measurable functions f on I with finite norm
◦
Let AC(I) be the set of all functions locally absolutely continuous on I. Let AC(I) be the
set of functions from AC(I) with compact supports on I.
We consider the following Hardy inequality in the differential form
f p =
ess supt∈I |f (t)|, p = ∞.
In [] it is shown that if
then inequality () is satisfied for all functions f ∈ AC(I) such that f (a) = or f (b) = ,
respectively. For example, in case (), it is equivalent to the weighted integral Hardy
inequality (see [])
≤ C
which has been studied for all values of the parameters < p, q ≤ ∞ (see [, ], and []).
In [] it is also shown that in the last case
inequality () is satisfied for all functions f ∈ AC(I) such that f (a) = and f (b) = , that is,
it is an overdetermined case. This case is studied in [, ], and [].
In the present work, for ≤ p ≤ q ≤ ∞, we establish a criterion for the validity of
inequality () with an estimate of the type
for the least constant C in (), where B(u, ρ) is some functional depending on u and ρ.
Moreover, we present a calculation formula for the least value of C and its two-sided
estimate.
We suppose that only condition () does not hold. In case () or (), our criterion
coincides with the well-known Muckenhoupt result. However, our upper estimate in () is
worse than the known one (see [–], and Remark . further). In case (), our criterion
is given in terms different from those in [, ], and []. The terms in [] are close to ours,
but the comparison analysis shows that our results and an estimate of type () are better
than in [] (see Remark .).
At the end of the paper, we find a criterion for the compactness of the set M = {uf : f ∈
◦
AC(I), ρf p ≤ } in Lq(I).
2 Auxiliary statements
Lemma . Let < q < ∞ and ϕ(λ) = λλq–q – λ– , λ > . There exists a point λ ≡ λ(q) such
that
= .
q+q < λ(q) < min{q, } for q = ,
Proof Since
ϕ(λ) = (λq – )(λ – )
This gives that ϕ(λ) = ϕ( + ε) ≥ for ε ≥ q – , that is, ϕ(λ) > for λ > q.
Let us find an extremum of the function d for λ > . We have that d (λ) = (q + )λq –
qλq– = λq–((q + )λ – q). This gives that d (λ) = for λ = q+q , d (λ) > for λ > q+q ,
and d (λ) < for < λ < q+q . Therefore, the function d(λ) decreases for < λ ≤ q+q and
increases for λ > q+q . Moreover, it has a minimum at q+q . Since d() = > , d(λ) > for
λ ≥ . Hence, ϕ(λ) > for λ > .
Thus,
ϕ(λ) < for < λ ≤ q+q .
where γ = q+q ( q+q ) q ( q–q ) q , γ = q+q q q ( q–q ) q , and γ = qq q (q ) q .
f (λ) = g(λ) > g q+q = q+q q+q
f (q) =
Let us estimate f ( q+q ):
f q+q = q–q
= q+q q q
is, the function f decreases for < λ < λ, increases for λ > λ, and has a minimum at
λq
λ = λ. Thus, infλ> f (λ) = f (λ). Again by Lemma . we have that λq– = λ– . Substituting
this equality into the expression of f (λ), we get ().
The function g(x) = λ has a minimum at the point λ = q+q . Therefore,
(λ–) q
From (), (), (), and (), taking into account that f () < , we have (). The proof of
Lemma . is complete.
3 Main results
Let a < c < b, d = d(a, c, b) = min{c – a, b – c}. Assume that
u q,(c–h,c+h)
, p = ∞,
Theorem . Let ≤ p ≤ q < ∞. Inequality () holds if and only if Ap,q < ∞. Moreover, for
the least constant C in (), we have the estimate
u q,(c–h,c+h) ≤ C
From the last inequality, taking into account that its left-hand side does not depend on α,
β such that a < α < β < b, we have
u q,(c–h,c+h) ≤ C
for all c ∈ I and < h < d.
In the case p = , we construct f in the following way. Let numbers c and h be defined as
before, δ > , and a < x – δ < x + δ ≤ c – h < c + h ≤ y – δ < y + δ < b. Assume that
Substituting fc,h into (), we get
for almost all x : (a < x ≤ c – h) and almost all y : (c + h ≤ y < b).
For α > , there exist points x : (a < x ≤ c – h) and y : (c + h ≤ y < b) such that
ρ– ∞α,(a,c–h) ≤ ρ–(x) and
ρ– α∞,(c+h,b) ≤ ρ–(y).
∞,(a,c–h)
∞,(c+h,b)
Since the left-hand side of this inequality does not depend on α > , letting α → , we get
() for p = . Thus, for all ≤ p ≤ q < ∞, we have that
Ap,q ≤ C.
Tk+ ⊂
Tk ⊃
f (s) ds = –
f (s) ds.
From () by Hölder’s inequality we have that
k≤n–
i≤k
In view of f (t) < λk+ for t ∈
uf qq =
uf qq, Tk+ ≤
k≥i
uf qq ≤ λq λq – Ap,q
q
λpi ρ– –p q
p ,(cji,αji) + λpi ρ– –p
p ,(βji,dji) p
(by the second relation from ())
that is, inequality () holds with the estimate C ≤ f (λ)Ap,q for the least constant C in ().
This fact, together with (), gives (). The proof of Theorem . is complete.
Remark . Let us notice that in [], for inequality (), an estimate of the type () has
been obtained in the case < q = p < ∞.
uq(x) dx
Theorem . Let = p = q. Inequality () holds if and only if Ap,q < ∞. Moreover, Ap,q = C,
where C is the least constant in ().
Proof From () we have that uf q ≤ λAp,q ρf p. Taking λ → , we get uf q ≤
Ap,q ρf p, that is, inequality () holds with the estimate Ap,q ≤ C, which, together with
(), gives Ap,q = C.
Theorem . Let ≤ p ≤ q = ∞. Inequality () holds if and only if Ap,q < ∞. Moreover,
Ap,q ≤ C ≤ Ap,q, where C is the least constant in ().
Proof Let ≤ p < q = ∞. The necessity follows from Theorem .. Let us prove the
suffi◦
ciency. Let Ap,q < ∞. For ≤ f ∈ AC(I), we have
uf q = sup uf q, Tk+ ≤ λ sup λk u q, Tk+
k k
≤ λ sup λk u q,Tk+ ≤ λ sup λk u q,(αik,βik)
k k,i
λ λ
≤ λ – Ap,q sup ρf p, Tk ≤ λ – Ap,q ρf p,
k
which, as before, means that
C ≤ Ap,q.
Let p = q = ∞.
Sufficiency. From () we have
λ
uf q ≤ iλn>f λ – Ap,q ρf p = Ap,q ρf p,
λk ρ– –
,(cik,αik) ≤ λ – ρf p,(cik,αik),
λk ρ– –
,(βik,dik) ≤ λ – ρf p,(βik,dik).
Using these relations instead of () and (), we have
uf q ≤ λ sku,ip λk u q,(αik,βik) ≤ λAp,q sku,ip λk ρ– –,(cik,αik) + λk ρ– –,(βik,dik)
λ λ
≤ λ – Ap,q sup ρf p, Tk ≤ λ – Ap,q ρf p.
k
This gives that
C ≤ Ap,q.
uf q ≤ Ap,q ρf p
Necessity. Substituting the function fc,h into (), we have
u q,(c–h,c+h) ≤ C
which means that
Ap,q ≤ C.
The proof of Theorem . is complete.
u q,(c,d), min
< ∞.
Moreover, for the least constant C in (), we have the estimates
– p B ≤ C ≤ <inλ<f B, ≤ q < ∞,
– p B ≤ C ≤ B, q = ∞.
4 Compactness
Proof Necessity. Let M be relatively compact in Lq(I). Then by Theorem . we have that
Ap,q < ∞. Let fc,h,α,β ≡ fc,h be the function introduced in the necessary part of Theorem ..
We assume that
Kolmogorov theorem [], p. we get
≥ lim sup
N→∞ z,h,α,β
–
Similarly, working with the function gz,h,α,β , we get limz→–∞ Ap–,q(z) = , which, together
with (), gives ().
= .
u(t) =
u(t), t < N ,
t ≥ N ,
u(t) =
u(t), t > N .
t ≤ N ,
Hence, by Theorem . we have
–∞
–∞
≤ f (λ) sup Ap,q(x) for f ∈ M,
x<N
≤ f (λ) sup Ap,q(x) for f ∈ M.
x>N
This, together with (), gives (). The proof of Theorem . is complete.
Competing interests
The author declares that she has no competing interests.
Acknowledgements
The author would like to thank Professor Ryskul Oinarov and reviewers for their generous suggestions, which have
improved this paper. The paper was written under financial support by the Scientific Committee of the Ministry of
Education and Science of Kazakhstan, Grant No.5495/GF4, on priority area “Intellectual potential of the country.”
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