Abstract
Let \(x=(x_{1},x_{2},\ldots,x_{n})\), and let \(K(u(x),v(y))\) satisfy \(u(rx)=ru(x)\), \(v(ry)=rv(y)\), \(K(ru,v)=r^{\lambda\lambda_{1}}K(u, r^{-\frac{\lambda_{1}}{\lambda_{2}}}v)\), and \(K(u,rv)=r^{\lambda\lambda_{2}}K(r^{-\frac{\lambda_{2}}{\lambda_{1}}}u, v)\). In this paper, we obtain a necessary and sufficient condition and the best constant factor for the Hilbert-type multiple integral inequality with kernel \(K(u(x),v(y))\) and discuss its applications in the theory of operators.
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1 Preliminary
Let \(n\ge1\), \(x=(x_{1},x_{2},\ldots, x_{n})\), \(\|x\|_{\rho}=(x_{1}^{\rho}+\cdots +x_{n}^{\rho})^{1/\rho}\), and \(\mathbf {R}_{+}^{n}=\{x=(x_{1},\ldots, x_{n}): x_{1}>0, \ldots, x_{n}>0\}\).
Define the function space
Definition 1
Let λ, \(\lambda_{1}\), and \(\lambda_{2}\) be constants, and let \(u(x)\), \(v(y)\) and \(K(u,v)\) satisfy: for all \(r>0\), \(u(rx)=ru(x)\), \(v(ry)=rv(y)\), and
Then we call \(K(u(x), v(y))\) a generalized homogeneous function with parameters \((\lambda,\lambda_{1},\lambda_{2})\). Obviously, \(K(u(x), v(y))\) is a homogeneous function of order \(\lambda\lambda_{1}\) when \(\lambda _{1}=\lambda_{2}\).
If \(p>1\) and \(\frac{1}{p}+\frac{1}{q}=1\), then we call the inequality
the Hilbert-type multiple integral inequality with \(f\in L^{p}_{u^{\alpha}(x)}(\mathbf {R}^{n}_{+})\) and \(g\in L^{q}_{v^{\beta}(y)}(\mathbf {R}^{n}_{+})\).
Define the integral operator T with kernel \(K(u(x),v(y))\) as follows:
If there exists a constant M such that
then T is called a bounded operator from \(L^{p}_{\omega_{1}}(\mathbf {R}^{n}_{+})\) to \(L^{p}_{\omega_{2}}(\mathbf {R}^{n}_{+})\). If T is a bounded operator from \(L^{p}_{\omega_{1}}(\mathbf {R}^{n}_{+})\) to itself, then we call T a bounded operator in \(L^{p}_{\omega_{1}}(\mathbf {R}^{n}_{+})\). The operator norm of T is defined as
By (1.2) inequality (1.1) can be rewritten as
It is not hard to prove that this inequality is equivalent to
In this paper, we discuss a necessary and sufficient condition and the best constant factor for the Hilbert-type multiple integral inequality with the integral kernel of the generalized homogeneous function \(K(u(x),v(y))\). Our research is of some theoretical and application value for the research of Hilbert-type inequalities. Further, these results are used to study the boundedness and norm of the operator. Related studies can be found in [1–16].
Lemma 1
Let\(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(n \ge1\), \(\lambda>0\), \(\lambda _{1}\lambda_{2}>0\), and let a nonnegative measurable function\(K(u(x), v(y))\)be a generalized homogeneous function with parameters\((\lambda, \lambda_{1}, \lambda_{2})\). Denote
Then
Proof
Since \(K(u(x), v(y))\) is a generalized homogeneous function with parameters \((\lambda, \lambda_{1}, \lambda_{2})\), we have
By the same method we can obtain \(\omega_{2}(y)=[v(y)]^{\lambda\lambda _{2}-\frac{\lambda_{2}}{\lambda_{1}}(\frac{\alpha+n}{p}-n)}W_{2}\). □
Lemma 2
([17])
Let\(p_{i}>0\), \(a_{i}>0\), \(\alpha_{i}>0\) (\(i=1,2,\ldots,n\)), and let\(\psi(u)\)be measurable. Then
whereΓis the gamma function.
2 Main results
Theorem 1
Let\(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(n\ge1\), \(\rho>0\), \(\lambda>0\), \(\lambda _{1}\lambda_{2}>0\), let there exist positive constants\(C_{1}\)and\(C_{2}\)such that\(C_{1}\|x\|_{\rho}\le u(x)\le C_{2}\|x\|_{\rho}\), \(C_{1}\|y\|_{\rho}\le v(y)\le C_{2}\|y\|_{\rho}\), let a nonnegative measurable function\(K(u(x),v(y))\)be a generalized homogeneous function with parameters\((\lambda, \lambda_{1}, \lambda_{2})\), and let the convergent integrals\(W_{1}\)and\(W_{2}\)be defined as in Lemma 1. Then we have:
- (i)
There exists a constantMsuch that the Hilbert-type multiple integral inequality in (1.1) holds if and only if\(\frac{\lambda_{2}\alpha-n\lambda_{1}}{p} +\frac{\lambda_{1}\beta-n\lambda_{2}}{q}=\lambda\lambda_{1}\lambda_{2}\).
- (ii)
The best constant factor in (1.1) is\(\inf M=W_{1}^{\frac{1}{p}}W_{2}^{\frac{1}{q}}\).
Proof
Let \(\varOmega(a< b)=\{x=(x_{1},\ldots, x_{n}):a< \|x\|_{\rho}< b \}\).
(i) Suppose there exists a constant M such that (1.1) holds. Denote \(l=\frac{\lambda_{2}\alpha-n\lambda_{1}}{p}+\frac{\lambda_{1}\beta -n\lambda_{2}}{q}-\lambda\lambda_{1}\lambda_{2}\). First, we let \(\lambda_{1}>0\), \(\lambda_{2}>0\). For \(l>0\) and \(\varepsilon>0\) sufficiently small, we set
Thus we have
In view of \(\lambda_{1}>0\), \(\lambda_{2}>0\), \(C_{1}\|x\|_{\rho}\le u(x)\le C_{2}\|x\| _{\rho}\), \(C_{1}\|y\|_{\rho}\le v(y)\le C_{2}\|y\|_{\rho}\), the two integrals in (2.1) are all convergent.
Also, since \(-\frac{\lambda_{1}}{\lambda_{2}}<0\) and \((C_{2}\|x\|_{\rho})^{-\frac{\lambda_{1}}{\lambda_{2}}}\le u^{-\frac{\lambda_{1}}{\lambda _{2}}}(x)\le(C_{1}\|x\|_{\rho})^{-\frac{\lambda_{1}}{\lambda_{2}}}\), we have
Combining this with (1.1) and (2.1), we get
Since \(l>0\) and ε is sufficiently small, \(-n+\lambda _{1}\varepsilon-\frac{l}{\lambda_{2}}<-n\), and additionally \(C_{1}\|x\|_{\rho}\le u(x)\le C_{2}\|x\|_{\rho}\), then \(\int_{\varOmega(0<1)}[u(x)]^{-n+\lambda_{1}\varepsilon-\frac{l}{\lambda _{2}}}\,dx=+\infty\). So (2.2) is a contradiction to \(l>0\).
If \(l<0\), let \(\varepsilon>0\) be sufficient small. Then we set
Similarly, we can get
Since \(C_{1}\|x\|_{\rho}\le u(x)\le C_{2}\|x\|_{\rho}\), \(C_{1}\|y\|_{\rho}\le v(y)\le C_{2}\|y\|_{\rho}\), \(l<0\), \(\lambda_{1}>0\), \(\lambda_{2}>0\), and \(\varepsilon>0\) is sufficient small, the right-hand side of (2.3) converges; also, \(\int_{\varOmega(1<+\infty)}[v(y)]^{-n-\lambda_{2}\varepsilon-\frac {l}{\lambda_{1}}}\,dy\) diverges, and thus (2.3) is a contradiction to \(l<0\).
In conclusion, when \(\lambda_{1}>0\), \(\lambda_{2}>0\), then we have \(l=0\), that is, \(\frac{\lambda_{2}\alpha-n\lambda_{1}}{p}+\frac{\lambda_{1}\beta -n\lambda_{2}}{q}=\lambda\lambda_{1}\lambda_{2}\).
Again, suppose \(\lambda_{1}<0\), \(\lambda_{2}<0\). If \(l>0\), then taking \(\varepsilon>0\) sufficiently small, we set
We thus have
Meanwhile, using \(C_{1}\|x\|_{\rho}\le u(x)\le C_{2}\|x\|_{\rho}\), \((C_{2}\|x\| _{\rho})^{-\frac{\lambda_{1}}{\lambda_{2}}}\le u^{-\frac{\lambda_{1}}{\lambda _{2}}}\le(C_{1}\|x\|_{\rho})^{-\frac{\lambda_{1}}{\lambda_{2}}}\), we have
Combining this with (1.1) and (2.4), we obtain
Since the two integrals of the right-hand side of (2.5) converge, but the integral
diverges, (2.5) is a contradiction to \(l>0\).
If \(l<0\) and \(\varepsilon>0\) is sufficiently small, then we set
Similarly, we can get
We now easily get that both integrals on the right-hand side of (2.6) converge, but
diverges, and thus (2.6) is a contradiction to \(l<0\).
To sum up, when \(\lambda_{1}<0\), \(\lambda_{2}<0\), we also have \(l=0\), that is, \(\frac{\lambda_{2}\alpha-n\lambda_{1}}{p}+\frac{\lambda_{1}\beta-n\lambda _{2}}{q}=\lambda\lambda_{1}\lambda_{2}\).
On the contrary, if \(\frac{\lambda_{2}\alpha-n\lambda_{1}}{p}+\frac{\lambda _{1}\beta-n\lambda_{2}}{q}=\lambda\lambda_{1}\lambda_{2}\), then let \(a=\frac {\alpha}{pq}+\frac{n}{pq}\), \(b=\frac{\beta}{pq}+\frac{n}{pq}\). By the Hölder inequality and Lemma 1 we have
Taking arbitrary \(M\ge W_{1}^{\frac{1}{p}}W_{2}^{\frac{1}{q}}\), inequality (1.1) holds.
(ii) Suppose inequality (1.1) holds. If \(\inf M \neq W_{1}^{\frac {1}{p}}W_{2}^{\frac{1}{q}}\), then there exists a constant \(M_{0}< W_{1}^{\frac {1}{p}}W_{2}^{\frac{1}{q}}\) such that
for all \(f\in L^{p}_{u^{\alpha}(x)}(R_{+}^{n})\) and \(g\in L^{q}_{v^{\beta}(y)}(R_{+}^{n})\).
Let \(\varepsilon>0\) and \(\delta>0\) be sufficient small. We take
Then we have
Since \(\frac{\lambda_{2}\alpha-n\lambda_{1}}{p}+\frac{\lambda_{1}\beta -n\lambda_{2}}{q}=\lambda\lambda_{1}\lambda_{2}\) and \(v^{-\frac{\lambda _{2}}{\lambda_{1}}}(y)\le(C_{1}\|y\|_{\rho})^{-\frac{\lambda_{2}}{\lambda_{1}}}\), we have
Combining this with (2.7) and (2.8), we obtain
Thus
We also take
Similarly, we can get
Since \(C_{1}\|x\|_{\rho}\le u(x)\le C_{2}\|x\|_{\rho}\), \(\int_{\varOmega(\delta <1)}[u(x)]^{-n}\,dx\) is a usual integral, but \(\int_{\varOmega(1<+\infty )}[u(x)]^{-n}\,dx\) diverges, and thus
In the same way, we have
Letting \(\varepsilon\to0^{+}\) in (2.11), we get
Letting \(\delta\to0^{+}\), we obtain
This is a contradiction, and hence \(\inf M=W_{1}^{\frac{1}{p}}W_{2}^{\frac{1}{q}}\). □
Theorem 2
Let\(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(n\ge1\), \(\lambda>0\), \(\lambda _{1}\lambda_{2}>0\), \(\gamma=(1-p)\beta\), and let there exist constants\(C_{1}\)and\(C_{2}\)such that\(C_{1}\|x\|_{\rho}\le u(x)\le C_{2}\|x\|_{\rho}\)and\(C_{1}\|y\|_{\rho}\le v(y)\le C_{2}\|y\|_{\rho}\). Let a nonnegative measurable function\(K(u(x), v(y))\)be a generalized homogeneous function for parameters\((\lambda, \lambda_{1}, \lambda_{2})\). Let the operatorTbe defined by (1.2), and let\(W_{1}\)and\(W_{2}\)defined by Lemma 1be also convergent. Then
- (i)
Tis a bounded operator from\(L^{p}_{u^{\alpha}(x)}(R^{n}_{+})\)to\(L^{p}_{v^{\gamma}(y)}(R^{n}_{+})\)if and only if\(\frac{1}{p}[\lambda_{2}(\alpha +n)-\lambda_{1}(\gamma+n)]=n\lambda_{2}+\lambda\lambda_{1}\lambda_{2}\).
- (ii)
IfTis a bounded operator from\(L^{p}_{u^{\alpha}(x)}(R^{n}_{+})\)to\(L^{p}_{v^{\gamma}(y)}(R^{n}_{+})\), then the operator norm ofTis\(\|T\|=W_{1}^{\frac{1}{p}}W_{2}^{\frac{1}{q}}\).
Proof
Since \(\frac{1}{p}+\frac{1}{q}=1\), \(\beta=\frac{\gamma}{1-p}\), \(\frac {\lambda_{2}\alpha-n\lambda_{1}}{p}+\frac{\lambda_{1}\beta-n\lambda _{2}}{q}=\lambda\lambda_{1}\lambda_{2}\) leads to \(\frac{1}{p} [\lambda_{2}(\alpha+n)-\lambda_{1}(\gamma+n)]=n\lambda_{2}+\lambda\lambda _{1}\lambda_{2}\), and since equality (1.1) is equivalent to (1.3), Theorem 2 holds by Theorem 1. □
3 Applications
Theorem 3
Let\(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(n\ge1\), \(\rho>0\), \(\lambda >0\), \(\lambda_{1}>0\), \(\lambda_{2}>0\), \(a_{i}>0\), \(b_{i}>0\), \(\alpha< n(p-1)\), \(\beta< n(q-1)\), \(u(x)=(\sum_{i=1}^{n} a_{i}x_{i}^{\rho})^{1/\rho}\), and\(v(y)=(\sum_{i=1}^{n} b_{i}y_{i}^{\rho})^{1/\rho}\). Then:
- (i)
There exists a constantMsuch that
$$\begin{aligned} \int_{R_{+}^{n}} \int_{R_{+}^{n}}\frac{1}{(u^{\lambda _{1}}(x)+v^{\lambda_{2}}(y))^{\lambda}}f(x)g(y)\,dx\,dy\le M \Vert f \Vert _{p,u^{\alpha}(x)} \Vert g \Vert _{q,v^{\beta}(y)} \end{aligned}$$(3.1)if and only if\(\frac{n\lambda_{1}-\lambda_{2}\alpha}{p}+\frac{n\lambda _{2}-\lambda_{1}\beta}{q}=\lambda\lambda_{1}\lambda_{2}\), where\(f\in L^{p}_{u^{\alpha}(x)}(R_{+}^{n})\)and\(g\in L^{q}_{v^{\beta}(y)}(R_{+}^{n})\).
- (ii)
If inequality (3.1) holds, then its best constant factor is
$$\inf M= \Biggl(\prod_{i=1}^{n} a_{i}^{-\frac{1}{\rho}} \Biggr)^{\frac{1}{q}} \Biggl(\prod _{i=1}^{n} b_{i}^{-\frac{1}{\rho}} \Biggr)^{\frac{1}{p}}\frac{\varGamma^{n}(\frac {1}{\rho})}{\rho^{n-1}\varGamma(\lambda)\varGamma(\frac{n}{\rho})} \varGamma \biggl( \frac{1}{\lambda_{1}} \biggl(\frac{n}{q}-\frac{\alpha}{p} \biggr) \biggr)\varGamma \biggl(\frac{1}{\lambda_{2}} \biggl(\frac{n}{p}- \frac{\beta}{q} \biggr) \biggr). $$
Proof
Set \(K(u(x), v(y))=\frac{1}{(u^{\lambda_{1}}(x)+v^{\lambda_{2}}(y))^{\lambda}}\). Then \(K(u(x),v(y))\) is a generalized homogeneous function for parameters \((\lambda, -\lambda_{1}, -\lambda_{2})\), and \(\frac{n\lambda_{1}-\lambda _{2}\alpha}{p}+\frac{n\lambda_{2}-\lambda_{1}\beta}{q}=\lambda\lambda_{1}\lambda _{2}\) is equivalent to \(\frac{(-\lambda_{2})\alpha-n(-\lambda_{1})}{ p}+\frac{(-\lambda_{1})\beta-n(-\lambda_{2})}{q}=\lambda(-\lambda _{1})(-\lambda_{2})\). Further, we have \(\lambda-\frac{1}{\lambda_{2}}(\frac {n}{p}-\frac{\beta}{q})=\frac{1}{\lambda_{1}}(\frac{n}{q}-\frac{\alpha}{p})\), and \(\frac{n}{p}-\frac{\beta}{q}>0\) and \(\frac{n}{q}-\frac{\alpha }{p}>0\) when \(\alpha< n(p-1)\) and \(\beta< n(q-1)\). By Lemma 1 we have
In the same way, we get
Thus
Hence Theorem 3 holds by Theorem 1. □
Corollary 1
Let\(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(n\ge1\), \(\rho>0\), \(\lambda>0\), \(\lambda_{1}>0\), \(\lambda_{2}>0\), \(u(x)=(\sum_{i=1}^{n} x_{i}^{\rho})^{\frac{1}{\rho }}\), and\(v(y)=(\sum_{i=1}^{n} y_{i}^{\rho})^{\frac{1}{\rho}}\). Then:
- (i)
The operatorTdefined by
$$T(f) (y)= \int_{R_{+}^{n}}\frac{1}{(u^{\lambda_{1}}(x)+v^{\lambda_{2}}(y))^{\lambda}}f(x)\,dx,\quad y\in R_{+}^{n}, $$is a bounded operator in\(L^{p}(R_{+}^{n})\)if and only if\(\frac{n\lambda _{1}}{p}+\frac{n\lambda_{2}}{q}=\lambda\lambda_{1}\lambda_{2}\).
- (ii)
WhenTis a bounded operator in\(L^{p}(R_{+}^{n})\), the operator norm ofTis
$$\Vert T \Vert =\frac{\varGamma^{n}(\frac{1}{\rho})}{\rho^{n-1}\lambda_{1}^{\frac {1}{q}}\lambda_{2}^{\frac{1}{p}}\varGamma(\lambda)\varGamma(\frac{n}{\rho })}\varGamma \biggl(\frac{n}{\lambda_{1}q} \biggr)\varGamma \biggl(\frac{n}{\lambda_{2}p} \biggr). $$
Proof
The corollary follows from Theorems 2 and 3. □
Theorem 4
Let\(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(n\ge1\), \(\rho>0\), \(\lambda >0\), \(\lambda_{1}>0\), \(\lambda_{2}>0\), \(\alpha< n(p-1)\), \(\beta< n(q-1)\), \(u(x)=(\sum_{i=1}^{n} x_{i}^{\rho})^{1/\rho}\), and\(v(y)=(\sum_{i=1}^{n} y_{i}^{\rho})^{1/\rho}\). Then
- (i)
There existsMsuch that
$$\begin{aligned} \int_{R_{+}^{n}} \int_{R_{+}^{n}}\frac{1}{(\max\{u^{\lambda _{1}}(x),v^{\lambda_{2}}(y)\})^{\lambda}}f(x)g(y)\,dx\,dy\le M \Vert f \Vert _{p,u^{\alpha}(x)} \Vert g \Vert _{q,v^{\beta}(y)} \end{aligned}$$(3.2)if and only if\(\frac{n\lambda_{1}-\lambda_{2}\alpha}{p}+\frac{n\lambda _{2}-\lambda_{1}\beta}{q}=\lambda\lambda_{1}\lambda_{2}\), where\(f\in L^{p}_{u^{\alpha}(x)}(R_{+}^{n})\)and\(g\in L^{q}_{v^{\beta}(y)}(R_{+}^{n})\).
- (ii)
If inequality (3.2) holds, then its best constant factor is
$$\inf M=\frac{\varGamma^{n}(\frac{1}{\rho})}{\lambda_{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}\rho^{n-1}\varGamma(\frac{n}{\rho})} \biggl[ \biggl(\frac{1}{\lambda _{1}} \biggl( \frac{n}{q}-\frac{\alpha}{p} \biggr) \biggr)^{-1} + \biggl(\frac{1}{\lambda_{2}} \biggl(\frac{n}{p}-\frac{\beta}{q} \biggr) \biggr)^{-1} \biggr]. $$
Proof
Set \(K(u(x), v(y))=\frac{1}{(\max\{u^{\lambda_{1}}(x),v^{\lambda_{2}}(y)\} )^{\lambda}}\). Then \(K(u(x),v(y))\) is a generalized homogeneous function for parameters \((\lambda, -\lambda_{1}, -\lambda_{2})\). By Lemma 2 we get
Similarly, we obtain
Then we have
In summary, Theorem 4 holds by Theorem 1. □
Corollary 2
Let\(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(n\ge1\), \(\rho>0\), \(\lambda>0\), \(\lambda _{1}>0\), \(\lambda_{2}>0\), \(u(x)=(\sum_{i=1}^{n} x_{i}^{\rho})^{\frac{1}{\rho}}\), and\(v(y)=(\sum_{i=1}^{n} y_{i}^{\rho})^{\frac{1}{\rho}}\). Then
- (i)
The operatorTdefined by
$$T(f) (y)= \int_{R_{+}^{n}}\frac{1}{\max\{u^{\lambda_{1}}(x),v^{\lambda_{2}}(y)\} )^{\lambda}}f(x)\,dx, y\in R_{+}^{n}, $$is a bounded operator in\(L^{p}(R_{+}^{n})\)if and only if\(\frac{n\lambda _{1}}{p}+\frac{n\lambda_{2}}{q}=\lambda\lambda_{1}\lambda_{2}\).
- (ii)
WhenTis a bounded operator in\(L^{p}(R_{+}^{n})\), the operator norm ofTis
$$\Vert T \Vert =\frac{\varGamma^{n}(\frac{1}{\rho})}{\lambda_{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}\rho^{n-1}\varGamma(\frac{n}{\rho})} \biggl(\frac{\lambda_{1} q}{n}+ \frac{\lambda_{2} p}{n} \biggr). $$
Proof
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The authors thank the anonymous reviewers for their insightful and detailed comments on the paper.
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The first author was supported by the National Natural Science Foundation of China (No. 61300204). The second author was supported by the National Natural Science Foundation of China (No. 11401113) and the Characteristic Innovation Project (Natural Science) of Guangdong Province (No. 2017KTSCX133).
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YH and JL carried out the mathematical studies, participated in the sequence alignment, and drafted the manuscript. BY and QC participated in the design of the study and performed the numerical analysis. All authors read and approved the final manuscript.
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Hong, Y., Liao, J., Yang, B. et al. A class of Hilbert-type multiple integral inequalities with the kernel of generalized homogeneous function and its applications. J Inequal Appl 2020, 140 (2020). https://doi.org/10.1186/s13660-020-02401-0
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DOI: https://doi.org/10.1186/s13660-020-02401-0