Abstract
By applying the weight functions, the technique of real analysis and Hermite-Hadamard’s inequality, a half-discrete Hardy-Hilbert-type inequality related to the kernel of exponential function with the best possible constant factor expressed by the gamma function is given. The more accurate equivalent forms, the operator expressions with the norm, the reverses, and some particular cases are considered.
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1 Introduction
Suppose that \(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(f(x),g(y)\geq 0\), \(f\in L^{p}(\mathbf{R}_{+})\), \(g\in L^{q}(\mathbf{R}_{+})\), \(\Vert f\Vert _{p} =(\int_{0}^{\infty }f^{p}(x)\,dx)^{\frac{1}{p}}>0\), \(\Vert g\Vert _{q}>0\), and we have the following Hardy-Hilbert integral inequality (cf. [1]):
where the constant factor \(\frac{\pi }{\sin (\pi /p)}\) is the best possible. If \(a_{m},b_{n}\geq 0\), \(a=\{a_{m}\}_{m=1}^{\infty }\in l^{p}\), \(b=\{b_{n}\}_{n=1}^{\infty }\in l^{q}\), \(\Vert a\Vert _{p}=(\sum_{m=1}^{\infty }a_{m}^{p})^{\frac{1}{p}}>0\), \(\Vert b\Vert _{q}>0\), then we have the following discrete analogy of (1) with the same best possible constant \(\frac{\pi }{\sin (\pi /p)}\) (cf. [1]):
Inequalities (1) and (2) are important in analysis and its applications (cf. [1–5]).
If \(\mu _{i},\upsilon _{j}>0\) (\(i,j\in \mathbf{N}=\{1,2,\ldots \}\)),
then we have the following Hardy-Hilbert-type inequality (cf. [1], Theorem 321, replacing \(\mu _{m}^{1/q}a_{m}\) and \(\upsilon _{n}^{1/p}b_{n}\) by \(a_{m}\) and \(b_{n}\)):
For \(\mu _{i}=\upsilon _{j}=1\) (\(i,j\in \mathbf{N}\)), inequality (4) reduces to (2).
Note
The authors of [1] did not prove that (4) is valid with the best possible constant factor.
In 1998, by introducing an independent parameter \(\lambda \in (0,1]\), Yang [6] gave an extension of (1) with the kernel \(\frac{1}{(x+y)^{\lambda }}\) for \(p=q=2\). Following [6], Yang [5] gave some extensions of (1) and (2) as follows:
If \(\lambda _{1},\lambda _{2}\in \mathbf{R}\), \(\lambda _{1}+\lambda _{2}=\lambda \), \(k_{\lambda }(x,y)\) is a non-negative homogeneous function of degree −λ, with \(k(\lambda _{1})=\int_{0}^{\infty }k_{\lambda }(t,1)t^{\lambda _{1}-1}\,dt\in \mathbf{R}_{+}\), \(\phi (x)=x^{p(1-\lambda _{1})-1}\), \(\psi (x)=x^{q(1-\lambda _{2})-1}\), \(f(x),g(y)\geq 0\),
\(g\in L_{q,\psi }(\mathbf{R}_{+})\), \(\Vert f\Vert _{p,\phi }, \Vert g\Vert _{q,\psi }>0\), then we have
where the constant factor \(k(\lambda _{1})\) is the best possible. Moreover, if \(k_{\lambda }(x,y)\) keeps a finite value and \(k_{\lambda }(x,y)x^{\lambda _{1}-1}\) (\(k_{\lambda }(x,y)y^{\lambda _{2}-1}\)) is decreasing with respect to \(x>0\) (\(y>0\)), then, for \(a_{m,}b_{n}\geq 0\),
\(b=\{b_{n}\}_{n=1}^{\infty }\in l_{q,\psi }\), \(\Vert a\Vert _{p,\phi }, \Vert b\Vert _{q,\psi }>0\), we have
where the constant factor \(k(\lambda _{1})\) is still the best possible.
In 2015, by adding some conditions, Yang [7] gave an extension of (4) as follows:
where the constant \(B(\lambda _{1},\lambda _{2})\) is still the best possible.
Some other results including multidimensional Hilbert-type inequalities are provided by [8–30].
About the topic of half-discrete Hilbert-type inequalities with the non-homogeneous kernels, Hardy et al. provided a few results in Theorem 351 of [1]. But they did not prove that the constant factors are the best possible. However, Yang [31] gave a result with the kernel \(\frac{1}{(1+nx)^{\lambda }}\) by introducing a variable and proved that the constant factor is the best possible. In 2011 Yang [32] gave the following half-discrete Hardy-Hilbert inequality with the best possible constant factor \(B ( \lambda _{1},\lambda _{2} ) \):
where \(\lambda _{1}>0\), \(0<\lambda _{2}\leq 1\), \(\lambda _{1}+\lambda _{2}=\lambda \). Zhong et al. ([17, 33, 34]) investigated several half-discrete Hilbert-type inequalities with particular kernels. Applying weight functions, a half-discrete Hilbert-type inequality with a general homogeneous kernel of degree \(-\lambda \in \mathbf{R}\) and a best constant factor \(k ( \lambda _{1} ) \) are obtained as follows:
which is an extension of (8) (cf. [35]). At the same time, a half-discrete Hilbert-type inequality with a general non-homogeneous kernel and a best constant factor are given by Yang [36]. In 2012-2014, Yang et al. published three books [37, 38] and [39] concerned with building the theory of half-discrete Hilbert-type inequalities.
In this paper, by applying weight functions, the technique of real analysis, and Hermite-Hadamard’s inequality, a half-discrete Hardy-Hilbert-type inequality related to the kernel of exponential function with a best possible constant factor expressed by the gamma function is given, which is similar to (7) and an extension of (9) in the following particular kernel:
Furthermore, the more accurate equivalent forms, the operator expressions with the norm, the reverses, and some particular cases are considered.
2 An example and some lemmas
In the following, we agree that \(\nu _{n}>0\), \(0\leq\tau _{n}\leq \frac{\nu _{n}}{2}\) (\(n\in \mathbf{N}\)), \(V_{n}=\sum_{i=1}^{n}\nu _{i}\), \(\mu (t) \) is a positive continuous function in \(\mathbf{R}_{+}=(0,\infty )\),
\(\nu (t):=\nu _{n}\), \(t\in (n-\frac{1}{2},n+\frac{1}{2}]\) (\(n\in \mathbf{N}\)), and
\(p\neq 0,1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(\delta \in \{-1,1\}\), \(f(x),a_{n}\geq 0\) (\(x\in \mathbf{R}_{+}\), \(n\in \mathbf{N}\)), \(\Vert f\Vert _{p,\Phi _{\delta }}=(\int_{0}^{\infty }\Phi _{\delta }(x)f^{p}(x)\,dx)^{\frac{1}{p}}\), \(\Vert a\Vert _{q,\widehat{\Psi }}=(\sum_{n=1}^{\infty }\widehat{\Psi }(n)b_{n}^{q})^{\frac{1}{q}}\), where
Example 1
For \(\alpha >0\), \(0<\gamma \), \(\sigma \leq 1\), we set \(h(t)=\frac{1}{e^{\alpha t^{\gamma }}}\) (\(t\in \mathbf{R}_{+}\)).
(i) Setting \(u=\alpha t^{\gamma }\), we find
where
is called the gamma function (cf. [40]).
(ii) We obtain, for \(t>0\), \(\alpha >0\), \(0<\gamma \leq 1\), \(h(t)=\frac{1}{e^{\alpha t^{\gamma }}}>0\), \(h^{\prime }(t)=-\alpha \gamma t^{\gamma -1}\frac{1}{e^{\alpha t^{\gamma }}}<0\) and
(iii) If \(g(u)>0\), \(g^{\prime }(u)<0\), \(g^{\prime \prime }(u)>0\), then we find that, for \(y\in (n-\frac{1}{2},n+\frac{1}{2})\), \(g(V(y))>0\), \(\frac{d}{dy}g(V(y)) =g^{\prime }(V(y))\nu _{n}<0\), and
For \(g_{1}(u)>0\), \(g_{1}^{\prime }(u)<0\), \(g_{1}^{\prime \prime }(u)>0\), \(g_{2}(u)>0\), \(g_{2}^{\prime }(u)\leq 0\), \(g_{2}^{\prime \prime }(u)\geq 0\) (\(u>0\)), we obtain \(g_{1}(u)g_{2}(u)>0\), \((g_{1}(u)g_{2}(u))^{\prime }=g_{1}^{\prime }(u)g_{2}(u)+g_{1}(u)g_{2}^{\prime }(u)<0\), and
(iv) For \(\alpha >0\), \(0<\gamma \), \(\sigma \leq 1\), \(c>0\), we have \(h(cV(y))V^{\sigma -1}(y)>0\), \(\frac{d}{dy}(h(cV(y))V^{\sigma -1}(y))<0\), and
Then by Hermite-Hadamard’s inequality (cf. [41]), we have
Lemma 1
If \(g(t)\) (>0) is a strictly decreasing continuous function in \((\frac{1}{2},\infty )\), which is strictly convex satisfying \(\int_{\frac{1}{2}}^{\infty }g(t)\,dt\in \mathbf{R}_{+}\), then we have
Proof
By Hermite-Hadamard’s inequality and the decreasing property, we have
and, for \(n_{0}\in \mathbf{N}\), it follows that
Hence, choosing plus for the above two inequalities, we have (12). □
Lemma 2
If \(\alpha >0\), \(0<\gamma \), \(\sigma \leq 1\), define the following weight coefficients:
Then we have the following inequalities:
where \(k(\sigma )\) is indicated by (10).
Proof
Since \(V_{n}-\tau _{n}\geq \int_{\frac{1}{2}}^{n+\frac{1}{2}}\nu (t)\,dt-\frac{\nu _{n}}{2}=\int_{\frac{1}{2}}^{n}\nu (t)\,dt=V(n)\), and, for \(t\in (n-\frac{1}{2},n+\frac{1}{2})\), \(\nu _{n}=V^{\prime }(t)\), by (11) (for \(c=U^{\delta }(x)\)) and (12), we have
Setting \(u=U^{\delta }(x)V(t)\), by (10), we find
Hence, (16) follows.
Setting \(u=(V_{n}-\tau _{n})U^{\delta }(x)\) in (15), we find \(du=\delta (V_{n}-\tau _{n})U^{\delta -1}(x)\mu (x)\,dx\) and
If \(\delta =1\), then
if \(\delta =-1\), then
Hence, by (10), we have (17). □
Remark 1
(i) We do not need the condition of \(\sigma \leq 1\) in obtaining (17). (ii) If \(U(\infty )=\infty \), then we have
For example, we set \(\mu (t)=\frac{1}{(1+t)^{a}}\) (\(t>0\); \(0\leq a\leq 1\)), then for \(x\geq 0\), we find
\(U(0)=0\), and \(U(\infty )=\int_{0}^{\infty }\frac{dt}{(1+t)^{a}}=\infty \).
Lemma 3
If \(\alpha >0\), \(0<\gamma \), \(\sigma \leq 1\), there exists \(n_{0}\in \mathbf{N}\), such that \(\{\nu _{n}\}_{n=n_{0}}^{\infty }\) is decreasing and \(V(\infty )=\infty \), then: (i) for \(x\in \mathbf{R}_{+}\), we have
where
(ii) for any \(b>0\), we have
Proof
Since \(V_{n}-\tau _{n}\leq V_{n}\leq V_{n+1}-\frac{\nu _{n+1}}{2}=V(n+1)\), and \(\nu _{n}\geq V^{\prime }(t)\) (\(t\in (n,n+1)\); \(n\geq n_{0}\)), by (13), we find
Setting \(u=U^{\delta }(x)V(t)\), in view of \(V(\infty )=\infty \), by (10), we find
We find
and then (19) follows.
For \(b>0\), we find
Since \(\frac{V^{-b}(n_{0}+1)-\nu _{1}^{-b}}{b}\rightarrow \mathrm{Constant}\) (\(b\rightarrow 0^{+}\)), we have (20). □
Note
For example, \(\nu _{n}=\frac{1}{n^{a}}\) (\(n\in \mathbf{N}\); \(0\leq a\leq 1\)) satisfies the conditions of \(\{\nu _{n}\}_{n=1}^{\infty }\) in Lemma 3 (for \(n_{0}=1\)).
3 Main results and operator expressions
Theorem 1
If \(\alpha >0\), \(0<\gamma \), \(\sigma \leq 1\), then for \(p>1\), \(0<\Vert f\Vert _{p,\Phi _{\delta }}\), \(\Vert a\Vert _{q,\widehat{\Psi }}<\infty \), we have the following equivalent inequalities:
Proof
By Hölder’s inequality with weight (cf. [41]), we have
In view of (17) and the Lebesgue term by term integration theorem (cf. [42]), we find
By Hölder’s inequality (cf. [41]), we have
Then by (22), we have (21). On the other hand, assuming that (21) is valid, we set
Then we find \(J_{1}^{p}=\Vert a\Vert _{q,\widehat{\Psi }}^{q}\). If \(J_{1}=0\), then (22) is trivially valid; if \(J_{1}=\infty \), then (22) remains impossible. Suppose that \(0< J_{1}<\infty \). By (21), we have
and then (22) follows, which is equivalent to (21).
Still by Hölder’s inequality with weight (cf. [41]), we have
Then by (16) and the Lebesgue term by term integration theorem (cf. [42]), it follows that
By Hölder’s inequality (cf. [41]), we have
Then by (23), we have (21). On the other hand, assuming that (23) is valid, we set
Then we find \(J_{2}^{q}=\Vert f\Vert _{p,\Phi _{\delta }}^{p}\). If \(J_{2}=0\), then (23) is trivially valid; if \(J_{2}=\infty \), then (23) keeps impossible. Suppose that \(0< J_{2}<\infty \). By (21), we have
and then (23) follows, which is equivalent to (21).
Therefore, (21), (22), and (23) are equivalent. □
Theorem 2
As regards the assumptions of Theorem 1, if there exists \(n_{0}\in \mathbf{N}\), such that \(\{\nu _{n}\}_{n=n_{0}}^{\infty }\) is decreasing and \(U(\infty )=V(\infty )=\infty \), then the constant factor \(k(\sigma )=\frac{\Gamma (\sigma /\gamma )}{\gamma \alpha ^{\sigma /\gamma }}\) in (21), (22), and (23) is the best possible.
Proof
For \(\varepsilon \in (0,q\sigma )\), we set \(\widetilde{\sigma }=\sigma -\frac{\varepsilon }{q}\) (\(\in (0,1)\)), and \(\widetilde{f}=\widetilde{f}(x)\), \(x\in \mathbf{R}_{+}\), \(\widetilde{a}=\{\widetilde{a}_{n}\}_{n=1}^{\infty }\),
Then for \(\delta =\pm 1\), since \(U(\infty )=\infty \), we find
By (20), (32), and (19), we obtain
If there exists a positive constant \(K\leq k(\sigma )\), such that (21) is valid when replacing \(k(\sigma )\) to K, then in particular, by Lebesgue term by term integration theorem, we have \(\varepsilon \widetilde{I}<\varepsilon K\Vert \widetilde{f}\Vert _{p,\Phi _{\delta }}\Vert \widetilde{a}\Vert _{q,\Psi }\), namely,
It follows that \(k(\sigma )\leq K\) (\(\varepsilon \rightarrow 0^{+}\)). Hence, \(K=k(\sigma )\) is the best possible constant factor of (21).
The constant factor \(k(\sigma )\) in (22) ((23)) is still the best possible. Otherwise, we would reach a contradiction by (26) ((29)) that the constant factor in (21) is not the best possible. □
For \(p>1\), we find \(\widehat{\Psi }^{1-p}(n)=\frac{\nu _{n}}{(V_{n}-\tau _{n})^{1-p\sigma }}\) (\(n\in \mathbf{N}\)), \(\Phi _{\delta }^{1-q}(x)=\frac{\mu (x)}{U^{1-q\delta \sigma }(x)}\) (\(x\in \mathbf{R}_{+}\)), and we define the following real normed spaces:
Assuming that \(f\in L_{p,\Phi _{\delta }}(\mathbf{R}_{+})\), setting
we can rewrite (22) as \(\Vert c\Vert _{p,\widehat{\Psi }^{1-p}}< k(\sigma )\Vert f\Vert _{p,\Phi _{\delta }}<\infty \), namely, \(c\in l_{p,\widehat{\Psi }^{1-p}}\).
Definition 1
Define a half-discrete Hardy-Hilbert-type operator \(T_{1}:L_{p,\Phi _{\delta }}(\mathbf{R}_{+})\rightarrow l_{p,\widehat{\Psi }^{1-p}}\) as follows: For any \(f\in L_{p,\Phi _{\delta }}(\mathbf{R}_{+})\), there exists a unique representation \(T_{1}f=c\in l_{p,\widehat{\Psi }^{1-p}}\). Define the formal inner product of \(T_{1}f\) and \(a=\{a_{n}\}_{n=1}^{\infty }\in l_{q,\widehat{\Psi }}\) as follows:
Then we can rewrite (21) and (22) as follows:
Define the norm of operator \(T_{1}\) as follows:
Then by (36), it follows that \(\Vert T_{1}\Vert \leq k(\sigma )\). Since, by Theorem 2, the constant factor in (36) is the best possible, we have
Assuming that \(a=\{a_{n}\}_{n=1}^{\infty }\in l_{q,\widehat{\Psi }}\), setting
we can rewrite (23) as \(\Vert h\Vert _{q,\Phi _{\delta }^{1-q}}< k(\sigma )\Vert a\Vert _{q,\widehat{\Psi }}<\infty \), namely, \(h\in L_{q,\Phi _{\delta }^{1-q}}(\mathbf{R}_{+})\).
Definition 2
Define a half-discrete Hardy-Hilbert-type operator \(T_{2}:l_{q,\widehat{\Psi }}\rightarrow L_{q,\Phi _{\delta }^{1-q}}(\mathbf{R}_{+})\) as follows: For any \(a=\{a_{n}\}_{n=1}^{\infty }\in l_{q,\widehat{\Psi }}\), there exists a unique representation \(T_{2}a=h\in L_{q,\Phi _{\delta }^{1-q}}(\mathbf{R}_{+})\). Define the formal inner product of \(T_{2}a\) and \(f\in L_{p,\Phi _{\delta }}(\mathbf{R}_{+})\) as follows:
Then we can rewrite (21) and (23) as follows:
Define the norm of operator \(T_{2}\) as follows:
Then by (40), we find \(\Vert T_{2}\Vert \leq k(\sigma )\). Since, by Theorem 2, the constant factor in (40) is the best possible, we have
4 Some equivalent reverses
In the following, we also set
For \(0< p<1\) or \(p<0\), we still use the formal symbols \(\Vert f\Vert _{p,\Phi _{\delta }}\), \(\Vert f\Vert _{p,\widetilde{\Phi }_{\delta }}\), and \(\Vert a\Vert _{q,\widehat{\Psi }}\).
Theorem 3
As regards the assumptions of Theorem 2, for \(p<0\), \(0<\Vert f\Vert _{p,\Phi _{\delta }}\), \(\Vert a\Vert _{q,\widehat{\Psi }}<\infty \), we have the following equivalent inequalities with the best possible constant factor \(k(\sigma )=\frac{\Gamma (\sigma /\gamma )}{\gamma \alpha ^{\sigma /\gamma }}\):
Proof
By the reverse Hölder inequality with weight (cf. [41]), since \(p<0\), in the similar way to obtaining (24) and (25), we have
Then by (18) and the Lebesgue term by term integration theorem, it follows that
By the reverse Hölder inequality (cf. [41]), we have
Then by (43), we have (42). On the other hand, assuming that (42) is valid, we set \(a_{n}\) as in Theorem 1. Then we find \(J_{1}^{p}=\Vert a\Vert _{q,\widehat{\Psi }}^{q}\). If \(J_{1}=\infty \), then (43) is trivially valid; if \(J_{1}=0\), then (43) keeps impossible. Suppose that \(0< J_{1}<\infty \). By (42), it follows that
and then (43) follows, which is equivalent to (42).
Still by the reverse Hölder’s inequality with weight (cf. [41]), since \(0< q<1\), in the similar way to obtaining (27) and (28), we have
Then by (16) and the Lebesgue term by term integration theorem, it follows that
By the reverse Hölder inequality (cf. [41]), we have
Then by (44), we have (42). On the other hand, assuming that (44) is valid, we set \(f(x)\) as in Theorem 1. Then we find \(J_{2}^{q}=\Vert f\Vert _{p,\Phi _{\delta }}^{p}\). If \(J_{2}=\infty \), then (44) is trivially valid; if \(J_{2}=0\), then (44) remains impossible. Suppose that \(0< J_{2}<\infty \). By (42), it follows that
and then (44) follows, which is equivalent to (42).
Therefore, inequalities (42), (43), and (44) are equivalent.
For \(\varepsilon \in (0,q\sigma )\), we set \(\widetilde{\sigma }=\sigma -\frac{\varepsilon }{q}\) (\(\in (0,1)\)) and \(\widetilde{f}=\widetilde{f}(x)\), \(x\in \mathbf{R}_{+}\), \(\widetilde{a}=\{\widetilde{a}_{n}\}_{n=1}^{\infty }\),
By (20), (32), and (16), we obtain
If there exists a positive constant \(K\geq k(\sigma )\), such that (42) is valid when replacing \(k(\sigma )\) to K, then in particular, we have \(\varepsilon \widetilde{I}>\varepsilon K\Vert \widetilde{f}\Vert _{p,\Phi _{\delta }}\Vert \widetilde{a}\Vert _{q,\widehat{\Psi }}\), namely,
It follows that \(k(\sigma )\geq K\) (\(\varepsilon \rightarrow 0^{+}\)). Hence, \(K=k(\sigma )\) is the best possible constant factor of (42).
The constant factor \(k(\sigma )\) in (43) ((44)) is still the best possible. Otherwise, we would reach a contradiction by (45) ((46)) that the constant factor in (42) is not the best possible. □
Theorem 4
As regards the assumptions of Theorem 2, if \(0< p<1\), \(0<\Vert f\Vert _{p,\Phi _{\delta }}\), \(\Vert a\Vert _{q,\widehat{\Psi }}<\infty \), then we have the following equivalent inequalities with the best possible constant factor \(k(\sigma )=\frac{\Gamma (\sigma /\gamma )}{\gamma \alpha ^{\sigma /\gamma }} \):
Proof
By the reverse Hölder inequality with weight (cf. [41]), since \(0< p<1\), in a similar way to obtaining (24) and (25), we have
In view of (18) and the Lebesgue term by term integration theorem, we find
By the reverse Hölder inequality (cf. [41]), we have
Then by (48), we have (47). On the other hand, assuming that (47) is valid, we set \(a_{n}\) as in Theorem 1. Then we find \(J_{1}^{p}=\Vert a\Vert _{q,\widehat{\Psi }}^{q}\). If \(J_{1}=\infty \), then (48) is trivially valid; if \(J_{1}=0\), then (48) remains impossible. Suppose that \(0< J_{1}<\infty \). By (47), it follows that
and then (48) follows, which is equivalent to (47).
Still by the reverse Hölder inequality with weight (cf. [41]), since \(q<0\), we have
Then by (19) and the Lebesgue term by term integration theorem, it follows that
By the reverse Hölder inequality (cf. [41]), we have
Then by (49), we have (47). On the other hand, assuming that (47) is valid, we set \(f(x)\) as in Theorem 1. Then we find \(J^{q}=\Vert f\Vert _{p,\widetilde{\Phi }_{\delta }}^{p}\). If \(J=\infty \), then (49) is trivially valid; if \(J=0\), then (49) keeps impossible. Suppose that \(0< J<\infty \). By (47), it follows that
and then (49) follows, which is equivalent to (47).
Therefore, inequalities (47), (48), and (49) are equivalent.
For \(\varepsilon \in (0,p\sigma )\), we set \(\widetilde{\sigma }=\sigma +\frac{\varepsilon }{p}\) and \(\widetilde{f}=\widetilde{f}(x)\), \(x\in \mathbf{R}_{+}\), \(\widetilde{a}=\{\widetilde{a}_{n}\}_{n=1}^{\infty }\),
By (19), (20), and (32), we obtain
If there exists a positive constant \(K\geq k(\sigma )\), such that (42) is valid when replacing \(k(\sigma )\) to K, then, in particular, we have \(\varepsilon \widetilde{I}>\varepsilon K\Vert \widetilde{f}\Vert _{p,\widetilde{\Phi }_{\delta }}\Vert \widetilde{a}\Vert _{q,\widehat{\Psi }}\), namely,
It follows that \(k(\sigma )\geq K\) (\(\varepsilon \rightarrow 0^{+}\)). Hence, \(K=k(\sigma )\) is the best possible constant factor of (47).
The constant factor \(k(\sigma )\) in (48) ((49)) is still the best possible. Otherwise, we would reach the contradiction by (50) ((51)) that the constant factor in (47) is not the best possible. □
5 Some corollaries and a remark
For \(\delta =1\) in Theorems 2-4, we have the following inequalities with the non-homogeneous kernel.
Corollary 1
As regards the assumptions of Theorem 2, (i) for \(p>1\), \(0<\Vert f\Vert _{p,\Phi _{1}}\), \(\Vert a\Vert _{q,\widehat{\Psi }}<\infty \), we have the following equivalent inequalities:
(ii) for \(p<0\), \(0<\Vert f\Vert _{p,\Phi _{1}}\), \(\Vert a\Vert _{q,\widehat{\Psi }}<\infty \), we have the following equivalent inequalities:
(iii) for \(0< p<1\), \(0<\Vert f\Vert _{p,\Phi _{1}}\), \(\Vert a\Vert _{q,\widehat{\Psi }}<\infty \), we have the following equivalent inequalities:
The above inequalities are with the best possible constant factor \(\frac{\Gamma (\sigma /\gamma )}{\gamma \alpha ^{\sigma /\gamma }}\).
For \(\delta =-1\) in Theorems 2-4, we have the following inequalities with the homogeneous kernel of degree 0:
Corollary 2
As regards the assumptions of Theorem 2, (i) for \(p>1\), \(0<\Vert f\Vert _{p,\Phi _{-1}}\), \(\Vert a\Vert _{q,\widehat{\Psi }}<\infty \), we have the following equivalent inequalities:
(ii) for \(p<0\), \(0<\Vert f\Vert _{p,\Phi _{-1}}\), \(\Vert a\Vert _{q,\widehat{\Psi }}<\infty \), we have the following equivalent inequalities:
(iii) for \(0< p<1\), \(0<\Vert f\Vert _{p,\Phi _{-1}}\), \(\Vert a\Vert _{q,\widehat{\Psi }}<\infty \), we have the following equivalent inequalities:
The above inequalities are with the best possible constant factor \(\frac{\Gamma (\sigma /\gamma )}{\gamma \alpha ^{\sigma /\gamma }}\).
Remark 2
(i) For \(\tau _{n}=0\) (\(n\in \mathbf{N}\)) in (21), setting \(\Psi (n):=\frac{V_{n}{}^{q(1-\sigma )-1}}{\nu _{n}^{q-1}}\) (\(n\in \mathbf{N}\)), we have the following inequality:
Hence, (21) is a more accurate inequality of (70) for \(0<\tau _{n}\leq \frac{\nu _{n}}{2}\).
(ii) For \(\mu (x)=\nu _{n}=1\) in (21), setting \(0\leq\tau \leq \frac{1}{2}\), we have the following inequality with the best possible constant factor \(\frac{\Gamma (\sigma /\gamma )}{\gamma \alpha ^{\sigma /\gamma }}\):
In particular, for \(\delta =1\), we have the following inequality with the non-homogeneous kernel:
for \(\delta =-1\), we have the following inequality with the homogeneous kernel of degree 0:
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Acknowledgements
This work is supported by the National Natural Science Foundation of China (No. 61370186) and the Appropriative Researching Fund for Professors and Doctors, Guangdong University of Education (No. 2015ARF25). We are grateful for their help.
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BY carried out the mathematical studies, participated in the sequence alignment and drafted the manuscript. JL participated in the design of the study and performed the numerical analysis. All authors read and approved the final manuscript.
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Liao, J., Yang, B. On a more accurate half-discrete Hardy-Hilbert-type inequality related to the kernel of exponential function. J Inequal Appl 2016, 162 (2016). https://doi.org/10.1186/s13660-016-1090-4
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DOI: https://doi.org/10.1186/s13660-016-1090-4