Abstract
By the use of weight coefficients and techniques of real analysis, we establish a new Hardy–Mulholland-type inequality with a mixed kernel and a best possible constant factor in terms of the hypergeometric function. Equivalent forms, an operator expression with the norm and reverses are also considered.
1 Introduction
If \(p>1,\) \(\frac{1}{p}+\frac{1}{q}=1,\) \(a_{m}, b_{n}\ge 0,\) \(a=\{a_{m}\}_{m=1}^{\infty }\in l^{p},\) \(b=\{b_{n}\}_{n=1}^{\infty }\in l^{q},\)
then we have the following Hardy–Hilbert inequality with the best possible constant factor \(\frac{\pi }{\sin (\pi /p)}\):
We also have the following Mulholland inequality
with the same best possible constant factor \(\frac{\pi }{\sin (\pi /p)}\) (cf. [5]). The inequalities (1) and (2) are important in Mathematical Analysis and its various applications (cf. [5, 15, 35,36,37,38]).
If \(\mu _{i},\upsilon _{j}>0\,(i,j\in {\mathbf {N}}=\{1,2,\ldots \}),\)
then we have the following Hardy–Hilbert-type inequality (cf. Theorem 321 of [5], replacing \(\mu _{m}^{1/q}a_{m}\) and \(\upsilon _{n}^{1/p}b_{n}\) with \(a_{m}\) and \(b_{n}\)) :
For \(\mu _{i}=\upsilon _{j}=1\,(i,j\in {\mathbf {N}}),\) inequality (4) reduces to (1).
In 1998, by introducing an independent parameter \(\lambda \in (0,1]\), Yang [34] proved an extension of the integral analogous of (1) with the kernel \(\frac{1}{(x+y)^{\lambda }}\) for \(p=q=2\). Recently, Yang [36] presented the following extensions of (1) and (2): If \(\lambda _{1},\lambda _{2}\in {\mathbf {R}},\lambda _{1}+\lambda _{2}=\lambda ,k_{\lambda }(x,y)\) is a finite non-negative homogeneous function of degree \( -\lambda ,\) with
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 \(a_{m},b_{n}\ge 0,\)
\(b=\{b_{n}\}_{n=1}^{\infty }\in l_{q,\psi },\) \(\Vert a\Vert _{p,\phi },\Vert b\Vert _{q,\psi }>0,\) we have
where the constant factor \(k(\lambda _{1})\) is still the best possible.
Clearly, for \(\lambda =1,\) \(\lambda _{1}=\frac{1}{q},\lambda _{2}=\frac{1}{p} ,\) \(k_{1}(x,y)=\frac{1}{x+y},\) the inequality (5) reduces to (1). Some other new results including multidimensional Hilbert-type inequalities, Hardy–Mulholland-type inequalities and Hardy–Hilbert-type inequalities are established in [1,2,3,4, 6,7,8,9,10, 12, 14, 16,17,31, 33, 39,40,48].
In this paper, by the use of weight coefficients and techniques of real analysis, we prove a new Hardy–Mulholland-type inequality with the following mixed kernel
and a best possible constant factor expressed in terms of the hypergeometric function. This inequality constitutes an extension of (2). Equivalent forms, operator expressions with the norm and reverses are considered as well.
2 An example and some lemmas
In the sequel, we consider that
with \(\mu _{1}=\upsilon _{1}=1,\ U_{m}\) and \(V_{n}\) are defined in (3)\(,a_{m},b_{n}\ge 0,\)
where
We introduce the following hypergeometric function (cf. [32]):
where \(\mathrm{{Re}}(\gamma )>\mathrm{{Re}}(\beta )>0,|\arg (1-z)|<\pi ;(1-zt)^{-\alpha }=1,\) for \(z=0.\)
Example 2.1
For \(-\alpha<\lambda _{1},\lambda _{2}\le 1\,(-2\alpha <\lambda =\lambda _{1}+\lambda _{2}\le 2),\) we set
-
(1)
Since \(\lambda +\alpha >-\alpha ,\) there exists a constant \(L_{\alpha }=\max \{2^{\alpha },1\}>0,\) such that
$$\begin{aligned} \frac{1}{(t+1)^{\lambda +\alpha }}\le (t+1)^{\alpha }\le L_{\alpha }\quad (t\in (0,1)). \end{aligned}$$For \(\lambda _{1},\lambda _{2}>-\alpha ,\) we get that
$$\begin{aligned} 0< & {} k_{\alpha }(\lambda _{1}):=\int _{0}^{\infty }k_{\lambda }(1,t)t^{\lambda _{2}-1}\,{\mathrm{d}}t=\int _{0}^{\infty }k_{\lambda }(t,1)t^{\lambda _{1}-1}\,{\mathrm{d}}t \\= & {} \int _{0}^{\infty }\frac{(\min \{t,1\})^{\alpha }}{(t+1)^{\lambda +\alpha } }t^{\lambda _{1}-1}\,{\mathrm{d}}t=\int _{0}^{1}\frac{t^{\lambda _{1}+\alpha -1}+t^{\lambda _{2}+\alpha -1}}{(t+1)^{\lambda +\alpha }}\,{\mathrm{d}}t \\\le & {} L_{\alpha }\int _{0}^{1}(t^{\lambda _{1}+\alpha -1}+t^{\lambda _{2}+\alpha -1})\,{\mathrm{d}}t=L_{\alpha }\left( \frac{1}{\lambda _{1}+\alpha }+\frac{1}{ \lambda _{2}+\alpha }\right) <\infty , \end{aligned}$$(7)and thus by (6) and (7), it follows that
$$\begin{aligned} k_{\alpha }(\lambda _{1})=\sum _{i=1}^{2}\frac{1}{\lambda _{i}+\alpha } F(\lambda +\alpha ,\lambda _{i}+\alpha ,\lambda _{i}+\alpha +1,-1)\in {\mathbf {R}}_{+}. \end{aligned}$$(8)-
(i)
For \(\alpha =0,\) we obtain \(k_{\lambda }(x,y)=\frac{1}{(x+y)^{\lambda }}\, (0<\lambda \le 2)\) and
$$\begin{aligned} k_{0}(\lambda _{1})=\int _{0}^{\infty }\frac{t^{\lambda _{1}-1}\,{\mathrm{d}}t}{ (t+1)^{\lambda }}=B(\lambda _{1},\lambda _{2})=\sum _{i=1}^{2}\frac{1}{ \lambda _{i}}F(\lambda ,\lambda _{i},\lambda _{i}+1,-1); \end{aligned}$$(9) -
(ii)
for \(-\alpha<\lambda +\alpha \le 1\,(<2+\alpha ),\) we derive that
$$\begin{aligned} k_{\alpha }(\lambda _{1})= & {} \int _{0}^{1}\frac{t^{\lambda _{1}+\alpha -1}+t^{\lambda _{2}+\alpha -1}}{(t+1)^{\lambda +\alpha }}\,{\mathrm{d}}t \\= & {} \int _{0}^{1}\sum _{k=0}^{\infty }\left( _{_{{}}\,\,\,\,\,\,\,\,k}^{-\lambda -\alpha }\right) t^{k}(t^{\lambda _{1}+\alpha -1}+t^{\lambda _{2}+\alpha -1})\,{\mathrm{d}}t \\= & {} \int _{0}^{1}\sum _{k=0}^{\infty }(-1)^{k}\left( _{_{{}}\,\,\,\,\,\,\,\,k}^{\lambda +\alpha +k-1}\right) t^{k}(t^{\lambda _{1}+\alpha -1}+t^{\lambda _{2}+\alpha -1})\,{\mathrm{d}}t\\= & {} \int _{0}^{1}\sum _{k=0}^{\infty }\left( _{_{{}}\,\,\,\,\,\,\,\,2k}^{\lambda +\alpha +2k-1}\right) \left( 1-\frac{\lambda +\alpha +2k}{2k+1}t\right) t^{2k}(t^{\lambda _{1}+\alpha -1}+t^{\lambda _{2}+\alpha -1})\,{\mathrm{d}}t. \end{aligned}$$
Since
$$\begin{aligned} 1-\frac{\lambda +\alpha +2k}{2k+1}t\ge \frac{1-(\lambda +\alpha )+2k(1-t)}{2k+1}\ge 0, \end{aligned}$$in view of the Lebesgue term by term integration theorem (cf. [13]), we obtain that
$$\begin{aligned} k_{\alpha }(\lambda _{1})= & {} \sum _{k=0}^{\infty }\int _{0}^{1}\left( _{_{{}}\,\,\,\,\,\,\,\,2k}^{\lambda +\alpha +2k-1}\right) \left( 1-\frac{\lambda +\alpha +2k }{2k+1}t\right) t^{2k}(t^{\lambda _{1}-1}+t^{\lambda _{2}-1})\,{\mathrm{d}}t \\= & {} \sum _{k=0}^{\infty }\left( _{_{{}}\,\,\,\,\,\,\,\,k}^{-\lambda -\alpha }\right) \int _{0}^{1}t^{k}(t^{\lambda _{1}+\alpha -1}+t^{\lambda _{2}+\alpha -1})\,{\mathrm{d}}t \\= & {} \sum _{k=0}^{\infty }\left( _{_{{}}\,\,\,\,\,\,\,\,k}^{-\lambda -\alpha }\right) \left( \frac{1}{k+\lambda _{1}+\alpha }+\frac{1}{k+\lambda _{2}+\alpha }\right) . \end{aligned}$$(10) -
(i)
-
(2)
Suppose that \(\alpha \le 0\,(\alpha >-1).\) We have
$$\begin{aligned} \lambda +\alpha >-\alpha \ge 0,\quad 0<\lambda _{i}+\alpha \le \lambda _{i}\,(i=1,2). \end{aligned}$$
For \( \lambda _{2}\le 1\,(\lambda _{2}+\alpha \le \lambda _{2}\le 1)\) and fixed \( x>0, \) we deduce that
is strictly decreasing with respect to \(y>0\). Similarly, for \(\lambda _{1}\le 1\) and fixed \(y>0,\)
is strictly decreasing with respect to \(x>0\).
Lemma 2.2
If \(0\le -\alpha <\lambda _{1},\lambda _{2}\le 1,k_{\alpha }(\lambda _{1})\) is defined as in (7), and we define the following weight coefficients:
then the following inequalities hold true:
Proof
In view of (3), we set
and
Then it follows that
For \( x\in (m-1,m),\)
for \(y\in (n-1,n),\)
Since V(y) is strictly increasing in \((n-1,n](n\in {\mathbf {N}})\), and \(-\alpha <\lambda _{2}\le 1,\) \(\lambda _{1}>-\alpha \), in view of Example 2.1(2), we derive that
Setting \(t=\frac{\ln V(y)}{\ln U_{m}},\) we obtain that
and
Hence, we obtain (13). Similarly, for \(-\alpha <\lambda _{1}\le 1,\lambda _{2}>-\alpha ,\) we have (14). \(\square \)
Lemma 2.3
If \(0\le -\alpha <\lambda _{1},\lambda _{2}\le 1,k_{\alpha }(\lambda _{1})\) is defined in (7), \(U_{\infty }=V_{\infty }=\infty ,\) there exist \(m_{0},n_{0}\in {\mathbf {N}},\) such that \( \{\mu _{m}\}_{m=m_{0}}^{\infty }\) and \(\{\upsilon _{n}\}_{n=n_{0}}^{\infty }\) are decreasing, then:
-
(i)
for \(m,n\in {\mathbf {N}}\backslash \{1\},\) we have
$$\begin{aligned} k_{\alpha }(\lambda _{1})(1-\theta (\lambda _{2},m))< & {} \omega (\lambda _{2},m)\quad (-\alpha <\lambda _{2}\le 1,\lambda _{1}>-\alpha ), \end{aligned}$$(16)$$\begin{aligned} k_{\alpha }(\lambda _{1})(1-\vartheta (\lambda _{1},n))< & {} \varpi (\lambda _{1},n)\quad (-\alpha <\lambda _{1}\le 1,\lambda _{2}>-\alpha ), \end{aligned}$$(17)where
$$\begin{aligned} \theta (\lambda _{2},m):= & {} \frac{1}{k_{\alpha }(\lambda _{1})}\int _{0}^{ \frac{\ln V_{n_{0}+1}}{\ln U_{m}}}\frac{(\min \{t,1\})^{\alpha }}{ (t+1)^{\lambda +\alpha }}t^{\lambda _{2}-1}\,{\mathrm{d}}t \\= & {} O\left( \frac{1}{\ln ^{\alpha +\lambda _{2}}U_{m}}\right) \in (0,1), \end{aligned}$$(18)$$\begin{aligned} \vartheta (\lambda _{1},n):= & {} \frac{1}{k_{\alpha }(\lambda _{1})}\int _{0}^{ \frac{\ln U_{m_{0}+1}}{\ln V_{n}}}\frac{(\min \{t,1\})^{\alpha }}{ (t+1)^{\lambda +\alpha }}t^{\lambda _{1}-1}\,{\mathrm{d}}t \\= & {} O\left( \frac{1}{\ln ^{\alpha +\lambda _{1}}V_{n}}\right) \in (0,1); \end{aligned}$$(19) -
(ii)
for any \(a>0,\) we have
$$\begin{aligned} \sum _{m=2}^{\infty }\frac{\mu _{m}}{U_{m}\ln ^{1+a}U_{m}}= & {} \frac{1}{a} \left[ \frac{1}{\ln ^{a}U_{m_{0}+1}}+aO_{1}(1)\right] , \end{aligned}$$(20)$$\begin{aligned} \sum _{n=2}^{\infty }\frac{\upsilon _{n}}{V_{n}\ln ^{1+a}V_{n}}= & {} \frac{1}{a} \left[ \frac{1}{\ln ^{a}V_{n_{0}+1}}+aO_{2}(1)\right] . \end{aligned}$$(21)
Proof
Since \(\upsilon _{n}\ge \upsilon _{n+1}\,\,(n\ge n_{0}),-\alpha <\lambda _{2}\le 1,\lambda _{1}>-\alpha \) and \(V(\infty )=\infty ,\) by Example 2.1(2), we have
For \(U_{m}\ge V_{n_{0}+1}\,(m\ge 2),\) by Example 2.1(1), we obtain that
namely, (16) and (18) follow. Similarly, for \(-\alpha <\lambda _{1}\le 1,\lambda _{2}>-\alpha , \) we obtain (17) and (19).
For \(a>0,\) we deduce that
Hence we obtain (20). Similarly, we derive (21). \(\square \)
Lemma 2.4
If \(0\le -\alpha <\lambda _{1},\lambda _{2}\le 1,k_{\alpha }(\lambda _{1})\) is defined as in (7), then for
we have
Proof
For \(0<\delta <\min \{\alpha +\lambda _{1},\alpha +\lambda _{2}\},\) we get that
Similarly, we obtain
Hence, we derive (22). \(\square \)
3 Main results and operator expressions
We also set the following functions:
Theorem 3.1
If \(0\le -\alpha <\lambda _{1},\lambda _{2}\le 1,\) then:
-
(i)
for \(p>1,\) we have the following equivalent inequalities:
$$\begin{aligned} I:= & {} \sum _{n=2}^{\infty }\sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}a_{m}b_{n}\le \Vert a\Vert _{p,{\widetilde{\Phi }}_{\lambda }}\Vert b\Vert _{q,{\widetilde{\Psi }}_{\lambda }}, \end{aligned}$$(24)$$\begin{aligned} J:= & {} \left\{ \sum _{n=2}^{\infty }\frac{\upsilon _{n}\ln ^{p\lambda _{2}-1}V_{n}}{(\varpi (\lambda _{1},n))^{p-1}V_{n}}\left[ \sum _{m=2}^{\infty }\frac{(\min \{\ln U_{m},\ln V_{n}\})^{\alpha }}{(\ln U_{m}V_{n})^{\lambda +\alpha }}a_{m}\right] ^{p}\right\} ^{\frac{1}{p}} \\\le & {} \Vert a\Vert _{p,{\widetilde{\Phi }}_{\lambda }}; \end{aligned}$$(25) -
(ii)
for \(0<p<1\) (or \(p<0),\) we have the equivalent reverses of (24) and (25).
Proof
(i) By Hölder’s inequality with weight (cf. [11]) and (12), we have
Hence by (11), we deduce that
and then (25) follows.
By Hölder’s inequality (cf. [11]), we have
On the other hand, assuming that (24) holds true, we set
Then we obtain that \(J^{p}=\Vert b\Vert _{q,{\widetilde{\Psi }}_{\lambda }}^{q}.\) If \(J=0,\) then (25) is trivially valid; if \(J=\infty ,\) then by (27), (25) takes the form of equality (\(=\infty \)). Suppose that \(0<J<\infty .\) By (24), it follows that
and then (25) follows, which is equivalent to (24).
(ii) For \(0<p<1\) (or \(p<0),\) by the reverse Hölder inequality with weight (cf. [11]) and (12), we obtain the reverse of (26) (or (26)). Then by (11), we derive the reverse of (27), and thus the reverse of (25) follows. By Hölder’s inequality (cf. [11]), we obtain the reverse of (28) and then by the reverse of (25), we deduce the reverse of (24).
On the other hand, assuming that the reverse of (24) holds true, we set \(b_{n}\) as in (29). Then we obtain that \(J^{p}=\Vert b\Vert _{q,{\widetilde{\Psi }} _{\lambda }}^{q}.\) If \(J=\infty ,\) then the reverse of (25) is trivially valid; if \(J=0,\) then by the reverse of (27), (25) takes the form of equality (\(=0\)). Suppose that \(0<J<\infty .\) By the reverse of (24), it follows that the reverses of (30) and (31) are valid and then the reverse of (25) follows, which is equivalent to the reverse of (24). \(\square \)
Theorem 3.2
If \(0\le -\alpha <\lambda _{1},\lambda _{2}\le 1,k_{\alpha }(\lambda _{1})\) is defined as in (8), there exist \(m_{0},n_{0}\in {\mathbf {N}},\) such that \(\{\mu _{m}\}_{m=m_{0}}^{\infty }\) and \(\{\upsilon _{n}\}_{n=n_{0}}^{\infty }\) are decreasing, \(U_{\infty }=V_{\infty }=\infty \), then for \(p>1\), \(\Vert a\Vert _{p,\Phi _{\lambda }}\in {\mathbf {R}}_{+}\) and \(\Vert b\Vert _{q,\Psi _{\lambda }}\in {\mathbf {R}}_{+},\) we have the following equivalent inequalities:
where the constant factor \(k_{\alpha }(\lambda _{1})\) is the best possible.
Proof
Applying (13) and (14) in (24) and (25), since
and
by simplification, we obtain the equivalent inequalities (32) and (33).
For \(\varepsilon \in (0,q(\alpha +\lambda _{2})),\) we set \({\widetilde{\lambda }}_{1}=\lambda _{1}+\frac{\varepsilon }{q}\,(>-\alpha ),{\widetilde{\lambda }}_{2}=\lambda _{2}-\frac{\varepsilon }{q}\,(\in (-\alpha ,1)),\) and \( {\widetilde{a}}=\{{\widetilde{a}}_{m}\}_{m=2}^{\infty },\) \({\widetilde{b}}=\{ {\widetilde{b}}_{n}\}_{n=2}^{\infty },\)
Then by (20), (21) and (19), we have
If there exists a positive constant \(K\le k_{\alpha }(\lambda _{1}),\) such that (32) is valid when replacing \(k_{\alpha }(\lambda _{1})\) by K, then in particular, we have \(\varepsilon {\widetilde{I}}<\varepsilon K\Vert {\widetilde{a}}\Vert _{p,\Phi _{\lambda }}\Vert {\widetilde{b}}\Vert _{q,\Psi _{\lambda }},\) namely
In view of (22), it follows that \(k_{\alpha }(\lambda _{1})\le K\,(\varepsilon \rightarrow 0^{+}).\) Hence, \(K=k_{\alpha }(\lambda _{1})\) is the best possible constant factor of (32).
Similarly to (28), we can still obtain that
Hence, we can prove that the constant factor \(k_{\alpha }(\lambda _{1})\) in (33) is the best possible. Otherwise, we would reach a contradiction by (36) that the constant factor in (32) is not the best possible.
For \(p>1,\)
we define the following normed spaces:
Assuming that \(a=\{a_{m}\}_{m=2}^{\infty }\in l_{p,\Phi _{\lambda }},\) setting
we can rewrite (33) as follows:
namely, \(c\in l_{p,\Psi _{\lambda }^{1-p}}.\)
Definition 3.3
Define a Hardy–Mulholland-type operator \(T:l_{p,\Phi _{\lambda }}\rightarrow l_{p,\Psi _{\lambda }^{1-p}}\) as follows: For any \( a=\{a_{m}\}_{m=2}^{\infty }\in l_{p,\Phi _{\lambda }},\) there exists a unique representation \(Ta=c\in l_{p,\Psi _{\lambda }^{1-p}}\). Define the formal inner product of Ta and \(b=\{b_{n}\}_{n=2}^{\infty }\in l_{q,\Psi _{\lambda }}\) as follows:
Then we can rewrite (32) and (33) as bellow:
Define the norm of operator T as follows:
Then by (39), we derive that \(\Vert T\Vert \le k_{\alpha }(\lambda _{1}).\) Since the constant factor in (39) 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 \(\Vert a\Vert _{p,\Phi _{\lambda }}\), \( \Vert b\Vert _{q,\Psi _{\lambda }},\Vert a\Vert _{p,{\widetilde{\Omega }}_{\lambda }}\) and \( \Vert b\Vert _{q,{\widetilde{\Upsilon }}_{\lambda }}\) as the formal symbols.
Theorem 4.1
With regard to the assumptions of Theorem 3.2, if \(0<p<1,\) \(\Vert a\Vert _{p,\Phi _{\lambda }}\in {\mathbf {R}}_{+}\) and \(\Vert b\Vert _{q,\Psi _{\lambda }}\in {\mathbf {R}}_{+},\) then we have the following equivalent inequalities with the best possible constant factor \(k_{\alpha }(\lambda _{1}):\)
Proof
Using (16) and (14) in the reverses of (24) and (25), since
and
by simplification, we obtain the equivalent inequalities (42) and (43).
For \(\varepsilon \in (0,p(\alpha +\lambda _{1})),\) we set
Then by (20), (21) and (14), we obtain that
If there exists a positive constant \(K\ge k_{\alpha }(\lambda _{1}),\) such that (42) is valid when replacing \(k_{\alpha }(\lambda _{1})\) by K, then in particular, we have \(\varepsilon {\widetilde{I}}>\varepsilon K\Vert {\widetilde{a}}\Vert _{p,\widetilde{\Omega }_{\lambda }}\Vert {\widetilde{b}}\Vert _{q,\Psi _{\lambda }},\) namely
In view of (22), it follows that \(k_{\alpha }(\lambda _{1})\ge K\,(\varepsilon \rightarrow 0^{+}).\) Hence, \(K=k_{\alpha }(\lambda _{1})\) is the best possible constant factor of (42). The constant factor \( k_{\alpha }(\lambda _{1})\) in (43) is still the best possible. Otherwise, we would reach a contradiction by the reverse of (36) that the constant factor in (42) is not the best possible.
Theorem 4.2
With regard to the assumptions of Theorem 3.2, if \(p<0,\ \Vert a\Vert _{p,\Phi _{\lambda }}\in {\mathbf {R}}_{+}\) and \(\Vert b\Vert _{q,\Psi _{\lambda }}\in {\mathbf {R}}_{+},\) then we have the following equivalent inequalities with the best possible constant factor \(B(\lambda _{1},\lambda _{2})\):
Proof
Using (13) and (17) in the reverses of (24) and (25), since
and
by simplification, we obtain equivalent inequalities (44) and (45).
For \(\varepsilon \in (0,q(\alpha +\lambda _{2})),\) we set
Then by (20), (21) and (12), we have
If there exists a positive constant \(K\ge k_{\alpha }(\lambda _{1}),\) such that (44) is satisfied when replacing \(k_{\alpha }(\lambda _{1})\) by K, then in particular, we have \(\varepsilon {\widetilde{I}}>\varepsilon K\Vert {\widetilde{a}}\Vert _{p,\Phi _{\lambda }}\Vert {\widetilde{b}}\Vert _{q,{\widetilde{\Upsilon }}_{\lambda }},\) namely
In view of (22), it follows that \(k_{\alpha }(\lambda _{1})\ge K(\varepsilon \rightarrow 0^{+}).\) Hence, \(K=k_{\alpha }(\lambda _{1})\) is the best possible constant factor of (44). Similarly to the reverse of (28), we still obtain that
Hence, the constant factor \(k_{\alpha }(\lambda _{1})\) in (45) is still the best possible. Otherwise, we would reach a contradiction by (46) that the constant factor in (44) is not the best possible. \(\square \)
Remark 4.3
For \(\alpha =0\) in (32), by (9), we have
For \(\lambda =1,\lambda _{1}=\frac{1}{q},\lambda _{2}=\frac{1}{p},\mu _{i}=\upsilon _{i}=1(i\in {\mathbf {N}})\) in (47), we deduce (2). Hence, (47) is an extension of (2); so is (32).
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Acknowledgements
We are thankful to the mathematicians who have read the manuscript of the paper, for their constructive comments that helped improve its presentation.
Funding
Open Access funding provided by University of Zurich. B. Yang: This work is supported by the National Natural Science Foundation (no. 61772140), and Science and Technology Planning Project Item of Guangzhou City (no. 201707010229). We are grateful for their support. A. M. Raigorodskii: The research was partially supported by the grant NSh-2540.2020.1 of the Russian President supporting leading scientific schools of Russia
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Communicated by Mario Krnic.
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Rassias, M.T., Yang, B. & Raigorodskii, A. A new Hardy–Mulholland-type inequality with a mixed kernel. Adv. Oper. Theory 6, 27 (2021). https://doi.org/10.1007/s43036-020-00123-0
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DOI: https://doi.org/10.1007/s43036-020-00123-0
Keywords
- Hardy–Mulholland-type inequality
- Weight coefficients
- Equivalent form
- Hypergeometric function
- Operator
- Reverse
Mathematics Subject Classification
- 26D15
- 47A07