Abstract
We establish necessary and sufficient conditions for the one-dimensional differential 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.
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1 Introduction
Let \(I=(a,b)\), \(-\infty\leq a< b\leq\infty\), \(1\leq p, q\leq\infty\), \(\frac{1}{p}+\frac{1}{p'}=1\), and \(\frac{1}{q}+\frac{1}{q'}=1\). Let \(u\geq0\) and \(\rho>0\) be weight functions such that \(u\in L_{q}^{\mathrm {loc}}\) and \(\rho^{-1}\equiv\frac{1}{\rho}\in L_{p'}^{\mathrm {loc}}\), where \(L_{p}\equiv L_{p}(I)\) stands for the space of measurable functions f on I with finite norm
Let \(\operatorname {AC}(I)\) be the set of all functions locally absolutely continuous on I. Let be the set of functions from \(\operatorname {AC}(I)\) with compact supports on I.
We consider the following Hardy inequality in the differential form
In [1] it is shown that if
for some \(c\in I\), then inequality (1) does not hold. In the case
or
then inequality (1) is satisfied for all functions \(f\in \operatorname {AC}(I)\) such that \(f(a)=0\) or \(f(b)=0\), respectively. For example, in case (3), it is equivalent to the weighted integral Hardy inequality (see [1])
which has been studied for all values of the parameters \(0< p,q\leq \infty\) (see [2, 3], and [4]).
In [1] it is also shown that in the last case
inequality (1) is satisfied for all functions \(f\in \operatorname {AC}(I)\) such that \(f(a)=0\) and \(f(b)=0\), that is, it is an overdetermined case. This case is studied in [1, 2], and [4].
In the present work, for \(1\leq p\leq q\leq\infty\), we establish a criterion for the validity of inequality (1) with an estimate of the type
for the least constant C in (1), where \(B(u,\rho)\) is some functional depending on u and ρ. Moreover, we present a calculation formula for the least value of \(C_{1}\) and its two-sided estimate.
We suppose that only condition (2) does not hold. In case (3) or (4), our criterion coincides with the well-known Muckenhoupt result. However, our upper estimate in (7) is worse than the known one (see [2–4], and Remark 3.2 further). In case (6), our criterion is given in terms different from those in [1, 2], and [4]. The terms in [4] are close to ours, but the comparison analysis shows that our results and an estimate of type (7) are better than in [4] (see Remark 3.3).
At the end of the paper, we find a criterion for the compactness of the set in \(L_{q}(I)\).
2 Auxiliary statements
Lemma 2.1
Let \(1< q<\infty\) and \(\varphi(\lambda)=\frac{\lambda^{q}}{\lambda ^{q}-1}-\frac{1}{\lambda-1}\), \(\lambda>1\). There exists a point \(\lambda _{1}\equiv\lambda_{1}(q)\) such that
and
In addition, \(\frac{\lambda^{q}}{\lambda^{q}-1}<\frac{1}{\lambda-1}\) for \(1<\lambda<\lambda_{1}\) and \(\frac{\lambda^{q}}{\lambda^{q}-1}>\frac{1}{\lambda-1}\) for \(\lambda>\lambda_{1}\).
Proof
Since
the sign of the function φ is defined by the value of the function \(d(\lambda)=\lambda^{q+1}-2\lambda^{q}+1\). Moreover, \(\varphi (\lambda)=0\), \(\lambda>1\), if and only if \(d(\lambda)=0\).
For \(q=2\), we have \(d(\lambda)=\lambda^{3}-2\lambda^{2}+1=(\lambda -1)(\lambda^{2}-\lambda-1)\). This means that \(d(\lambda)=0\) for \(\lambda =\frac{1+\sqrt{5}}{2}\).
Let \(q\neq2\). Let \(\lambda=1+\varepsilon\), \(\varepsilon>0\). Using the Lagrange finite-increment formula, we have
This gives that \(\varphi(\lambda)=\varphi(1+\varepsilon)\geq0\) for \(\varepsilon\geq q-1\), that is, \(\varphi(\lambda)>0\) for \(\lambda>q\).
Let us find an extremum of the function d for \(\lambda>1\). We have that \(d'(\lambda)=(q+1)\lambda^{q}-2q\lambda^{q-1}=\lambda ^{q-1}((q+1)\lambda-2q)\). This gives that \(d'(\lambda)=0\) for \(\lambda =\frac{2q}{q+1}\), \(d'(\lambda)>0\) for \(\lambda>\frac{2q}{q+1}\), and \(d'(\lambda)<0\) for \(1<\lambda<\frac{2q}{q+1}\). Therefore, the function \(d(\lambda)\) decreases for \(1<\lambda\leq\frac{2q}{q+1}\) and increases for \(\lambda>\frac{2q}{q+1}\). Moreover, it has a minimum at \(\frac {2q}{q+1}\). Since \(d(2)=1>0\), \(d(\lambda)>0\) for \(\lambda\geq2\). Hence, \(\varphi(\lambda)>0\) for \(\lambda>2\).
Thus,
Since \(d(1)=0\) and d has a minimum at \(\frac{2q}{q+1}\), it follows that \(d (\frac{2q}{q+1} )<0\) for \(\lambda>1\) and \(d(\lambda )<0\) for \(1<\lambda\leq\frac{2q}{q+1}\). Therefore, \(\varphi (\frac {2q}{q+1} )<0\), and
In view of the continuity of φ for \(\lambda>1\), from (10) and (11) there follows the existence of a point that satisfies (8) and (9).
The last statement of Lemma 2.1 follows from the intersection of graphs of two decreasing and concave upward functions \(\frac{\lambda ^{q}}{\lambda^{q}-1}\) and \(\frac{1}{\lambda-1}\) at the point \(\lambda =\lambda_{1}\). The proof of Lemma 2.1 is complete. □
Lemma 2.2
Let \(1< q<\infty\) and \(f(\lambda)=\frac{\lambda(\lambda^{q}-1)^{\frac {1}{q}}}{\lambda-1}\), \(\lambda>1\). Then
and, for \(q\neq2\), we have the estimate
where \(\gamma_{0}=\frac{2q}{q+1} (\frac{2q}{q+1} )^{\frac {1}{q}} (\frac{2q}{q-1} )^{\frac{1}{q'}}\), \(\gamma_{1}=\frac {2q}{q+1}q^{\frac{1}{q}} (\frac{2q}{q-1} )^{\frac{1}{q'}}\), and \(\gamma_{2}=qq^{\frac{1}{q}} (q' )^{\frac{1}{q'}}\).
Proof
By Lemma 2.1 for \(q=2\) we have \(\lambda_{1}(2)=\frac{1+\sqrt {5}}{2}\), so that \(f(\lambda_{1}(2))= (\frac{3\sqrt{5}+7}{\sqrt {5}-1} )^{\frac{1}{2}}\).
Let \(q\neq2\). The function f is continuous when \(\lambda>1\), and \(\lim_{\lambda\rightarrow1+}f(\lambda)=\infty\) and \(\lim_{\lambda \rightarrow\infty}f(\lambda)=\infty\). Therefore, it has a minimum. Since \(f'(\lambda)=\frac{(\lambda^{q}-1)^{\frac{1}{q}}}{\lambda-1} (\frac {\lambda^{q}}{\lambda^{q}-1}-\frac{1}{\lambda-1} )\), by Lemma 2.1 we have that \(f'(\lambda_{1})=0\), \(f'(\lambda)<0\) for \(1<\lambda<\lambda _{1}\), and \(f'(\lambda)>0\) for \(\lambda>\lambda_{1}\), that is, the function f decreases for \(1<\lambda<\lambda_{1}\), increases for \(\lambda>\lambda _{1}\), and has a minimum at \(\lambda=\lambda_{1}\). Thus, \(\inf_{\lambda >1}f(\lambda)=f(\lambda_{1})\). Again by Lemma 2.1 we have that \(\frac {\lambda_{1}^{q}}{\lambda_{1}^{q}-1}=\frac{1}{\lambda_{1}-1}\). Substituting this equality into the expression of \(f(\lambda_{1})\), we get (12).
The function \(g(x)=\frac{\lambda^{2}}{(\lambda-1)^{\frac{1}{q'}}}\) has a minimum at the point \(\lambda=\frac{2q}{q+1}\). Therefore,
By Lemma 2.1 we have that \(\frac{2q}{q+1}<\lambda_{1}<\min\{q,2\}\). Hence,
It is easy to see that \(f(2)<4\). Since \(\lambda_{1}< q\), we have \(\frac {q^{q}}{q^{q}-1}>\frac{1}{q-1}\) or \(q(q-1)^{\frac{1}{q}}>(q^{q}-1)^{\frac {1}{q}}\). Therefore,
Let us estimate \(f (\frac{2q}{q+1} )\):
From (14), (15), (16), and (17), taking into account that \(f(2)<4\), we have (13). The proof of Lemma 2.2 is complete. □
3 Main results
Let \(a< c< b\), \(d=d(a,c,b)=\min\{c-a, b-c\}\). Assume that
and
Theorem 3.1
Let \(1\leq p\leq q<\infty\). Inequality (1) holds if and only if \(A_{p,q}<\infty\). Moreover, for the least constant C in (1), we have the estimate
In turn, for \(f(\lambda_{1})\), by Lemma 2.2 we have \(f(\lambda _{1}(2))= (\frac{3\sqrt{5}+7}{\sqrt{5}-1} )^{\frac{1}{2}}\) and the estimate \(\gamma_{0}< f(\lambda_{1})<\min\{\gamma_{1},\gamma_{2},4\}\) for \(q\neq2\).
Proof
Necessity. Let inequality (1) hold with the least constant \(C>0\).
Suppose that \(1< p<\infty\). Let \(c\in I\), \(0< h< d\), and \(a<\alpha <c-h<c+h<\beta<b\). We introduce the following function:
It is obvious that . Substituting \(f_{c,h}\) into (1), we get
From the last inequality, taking into account that its left-hand side does not depend on α, β such that \(a<\alpha<\beta<b\), we have
for all \(c\in I\) and \(0< h< d\).
In the case \(p=1\), we construct f in the following way. Let numbers c and h be defined as before, \(\delta>0\), and \(a< x-\delta< x+\delta \leq c-h< c+h\leq y-\delta< y+\delta< b\). Assume that
Substituting \(f_{c,h}\) into (1), we get
Taking the limit in this inequality as \(\delta\rightarrow0\), we get
for almost all \(x:(a< x\leq c-h)\) and almost all \(y:(c+h\leq y< b)\).
For \(\alpha>1\), there exist points \(x:(a< x\leq c-h)\) and \(y:(c+h\leq y< b)\) such that
Then
This, together with \(\Vert u\Vert _{q,(x,y)}\geq \Vert u\Vert _{q,(c-h,c+h)}\), yields that
Since the left-hand side of this inequality does not depend on \(\alpha >1\), letting \(\alpha\rightarrow1\), we get (19) for \(p=1\). Thus, for all \(1\leq p\leq q<\infty\), we have that
Sufficiency. Let \(A_{p,q}<\infty\) be correct. Let f be a nontrivial function from . Without loss of generality, we assume that \(f\geq0\). Let \(\lambda>1\). For any integer k, we assume that \(T_{k}=\{t\in I:f(t)>\lambda^{k}\}\), \(\Delta T_{k}=T_{k}\setminus T_{k+1}\). Due to the boundedness of the function f, there exists an integer \(n=n(f)\) such that \(T_{n}\neq\emptyset\) and \(T_{n+1}=\emptyset\). It is obvious that \(I=\bigcup_{k\leq n}T_{k}=\bigcup_{k\leq n}\Delta T_{k}\). The set \(T_{k}\) is open. Therefore, there exists a family of mutually disjoint intervals \(\{J_{j}^{k}\}\), \(J_{j}^{k}=(c_{j}^{k},d_{j}^{k})\), such that \(T_{k}=\bigcup_{j}J_{j}^{k}\). For \(n-1\geq k>-\infty\), we assume that \(M_{j}^{k}=T_{k+1}\cap J_{j}^{k}\). For \(M_{j}^{k}\neq\emptyset\), we define \(\alpha _{j}^{k}=\inf M_{j}^{k}\) and \(\beta_{j}^{k}=\sup M_{j}^{k}\). Then
In view of the continuity of the function f, we get that \(f(\alpha _{j}^{k})=f(\beta_{j}^{k})=\lambda^{k+1}\) and \(f(c_{j}^{k})=f(d_{j}^{k})=\lambda^{k}\). Hence,
From (22) by Hölder’s inequality we have that
In view of \(f(t)<\lambda^{k+1}\) for \(t\in\Delta T_{k}\) and \(\lambda ^{qk}=(1-\lambda^{-q})\sum_{i\leq k}\lambda^{qi}\), we have that
(by (21))
Since \(A_{p,q}<\infty\), from (23) and (24), taking into account that \(\frac{q}{p}\geq1\), this gives
(by the second relation from (21))
Therefore,
Since the left-hand side of this inequality does not depend on \(\lambda >1\), by Lemma 2.2 we have
that is, inequality (1) holds with the estimate \(C\leq f(\lambda _{1}) A_{p,q}\) for the least constant C in (1). This fact, together with (20), gives (18). The proof of Theorem 3.1 is complete. □
Remark 3.1
Let us notice that in [5], for inequality (1), an estimate of the type (18) has been obtained in the case \(1< q=p<\infty\).
Remark 3.2
If (3) or (4) is correct, then by Theorem 3.1 there follows the correctness of the corresponding integral Hardy inequality (see [1]). For example, if (3) holds, then
where the condition \(A_{p,q}<\infty\) coincides with the Muckenhoupt condition (see [3]), and inequality (1) is equivalent to the integral Hardy inequality (5). However, our upper estimate in (7) is worse than that in the known result (see e.g. [3], Thm. 5). For example, in case (3) with \(p=q=2\), from (12) we have \(A_{p,q}\leq C\leq (\frac{3\sqrt{5}+7}{\sqrt {5}-1} )^{\frac{1}{2}}A_{p,q}\approx3.33 A_{p,q}\), but from Theorem 5 of [3] it follows that \(A_{p,q}\leq C\leq2A_{p,q}\).
Theorem 3.2
Let \(1=p=q\). Inequality (1) holds if and only if \(A_{p,q}<\infty\). Moreover, \(A_{p,q}=C\), where C is the least constant in (1).
Proof
From (25) we have that \(\Vert uf\Vert _{q}\leq\lambda A_{p,q} \Vert \rho f'\Vert _{p}\). Taking \(\lambda\rightarrow1\), we get \(\Vert uf\Vert _{q}\leq A_{p,q} \Vert \rho f'\Vert _{p}\), that is, inequality (1) holds with the estimate \(A_{p,q}\leq C\), which, together with (20), gives \(A_{p,q}=C\). □
Theorem 3.3
Let \(1\leq p\leq q =\infty\). Inequality (1) holds if and only if \(A_{p,q}<\infty\). Moreover, \(A_{p,q}\leq C\leq4A_{p,q}\), where C is the least constant in (1).
Proof
Let \(1\leq p< q =\infty\). The necessity follows from Theorem 3.1. Let us prove the sufficiency. Let \(A_{p,q}<\infty\). For , we have
that is,
which, as before, means that
Let \(p=q=\infty\).
Sufficiency. From (22) we have
Using these relations instead of (23) and (24), we have
This gives that
and
Necessity. Substituting the function \(f_{c,h}\) into (1), we have
which means that
Therefore,
The proof of Theorem 3.3 is complete. □
Remark 3.3
The obtained results can be compared with the results of Theorem 8.2 of [4], where it is proved that, for \(1\leq p\leq q\leq\infty\), the validity of (1) is equivalent to the condition
Moreover, for the least constant C in (1), we have the estimates
and
It is easy to see that \(A_{p,q}< B_{p,q}\). Moreover, the estimates for the least constant C in (1) obtained in Theorems 3.1 and 3.3 are obviously better than in (27) and (28), respectively.
4 Compactness
Let \(I=(-\infty,+\infty)\) and .
Let
Theorem 4.1
Let \(1\leq p\leq q<\infty\). The set M is relatively compact in \(L_{q}(I)\) if and only if \(A_{p,q}<\infty\) and
Proof
Necessity. Let M be relatively compact in \(L_{q}(I)\). Then by Theorem 3.1 we have that \(A_{p,q}<\infty\). Let \(f_{c,h,\alpha ,\beta}\equiv f_{c,h}\) be the function introduced in the necessary part of Theorem 3.1.
We assume that
Since and \(\Vert \rho(g^{\pm}_{z,h,\alpha,\beta})'\Vert _{p}\leq1\), we have \(g^{\pm}_{z,h,\alpha,\beta}\in M\), and by the Frechet-Kolmogorov theorem [6], p.10 we get
Therefore,
Similarly, working with the function \(g^{-}_{z,h,\alpha,\beta}\), we get \(\lim_{z\rightarrow-\infty}A^{-}_{p,q}(z)=0\), which, together with (30), gives (29).
Sufficiency. Let \(A_{p,q}<\infty\) and (29) hold. Then, on the basis of Theorem 3.1, the set M is bounded in \(L_{q}(I)\). Therefore, by the Frechet-Kolmogorov theorem it suffices to show that
Let
Hence, by Theorem 3.1 we have
Then
This, together with (29), gives (31). The proof of Theorem 4.1 is complete. □
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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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Kalybay, A. One-dimensional differential Hardy inequality. J Inequal Appl 2017, 21 (2017). https://doi.org/10.1186/s13660-017-1293-3
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DOI: https://doi.org/10.1186/s13660-017-1293-3