Abstract
By using the way of weight coefficients, the technique of real analysis, and Hermite-Hadamard’s inequality, a more accurate Hardy-Mulholland-type inequality with multi-parameters and a best possible constant factor is given. The equivalent forms, the reverses, the operator expressions and some particular cases are considered.
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1 Introduction
Assuming that \(p > 1\), \(\frac{1}{p} + \frac{1}{q} = 1\), \(a_{m},b_{n} \ge 0\), \(a = \{ a_{m}\}_{m = 1}^{\infty} \in l^{p}\), \(b = \{ b_{n}\}_{n = 1}^{\infty} \in l^{q}\), \(\Vert a \Vert _{p} = (\sum_{m = 1}^{\infty} a_{m}^{p} )^{\frac{1}{p}} > 0\), and \(\Vert b \Vert _{q} > 0\), we have the following Hardy-Hilbert’s inequality with the best possible constant \(\frac{\pi}{\sin (\pi /p)}\) (cf. [1], Theorem 315):
A more accurate inequality of (1) is given as follows (cf. [1], Th. 323 and [2]):
where the constant factor \(\frac{\pi}{\sin (\pi /p)}\) is still the best possible.
Also we have the following Mulholland’s inequality similar to (1) with the same best possible constant factor \(\frac{\pi}{\sin (\pi /p)}\) (cf. [3] or [1], Th. 343, replacing \(\frac{a_{m}}{m}\), \(\frac{b_{n}}{n}\) by \(a_{m}\), \(b_{n}\)):
Inequalities (1)-(3) are important in analysis and its applications (cf. [1, 2, 4–18]).
Suppose that \(\mu_{i},\upsilon_{j} > 0\) (\(i,j \in \mathbb{N} = \{ 1,2, \ldots \}\)),
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 \(b_{n}\)): If \(a_{m},b_{n} \ge 0\), \(0 < \sum_{m = 1}^{\infty} \frac{a_{m}^{p}}{m^{p - 1}} < \infty\), \(0 < \sum_{n = 1}^{\infty} \frac{b_{n}^{q}}{n^{q - 1}} < \infty\), then
For \(\mu_{i} = \upsilon_{j} = 1\) (\(i,j \in \mathbb{N}\)), inequality (5) reduces to (1).
In 2015, Yang [19] gave an extension of (5) as follows: If \(0 < \lambda_{1},\lambda_{2} \le 1\), \(\lambda_{1} + \lambda_{2} = \lambda\), \(\{ \mu_{m}\}_{m = 1}^{\infty}\) and \(\{ \upsilon_{n}\}_{n = 1}^{\infty}\) are positive and decreasing, with \(U_{\infty} = V_{\infty} = \infty\), then we have the following inequality with the best possible constant factor \(\pi /\sin (\frac{\pi \lambda_{1}}{\lambda} )\):
In this paper, by using the way of weight coefficients, the technique of real analysis, and Hermite-Hadamard’s inequality, a new Hardy-Mulholland-type inequality with a best possible constant factor is given as follows: If \(\mu_{1} = \upsilon_{1} = 1\), \(\{ \mu_{m}\}_{m = 1}^{\infty}\) and \(\{ \upsilon_{n}\}_{n = 1}^{\infty}\) are positive and decreasing, with \(U_{\infty} = V_{\infty} = \infty\), we have the following inequality:
which is an extension of (3). Moreover, the more accurate inequality of (7) and its extension with multi-parameters and the best possible constant factors are obtained. The equivalent forms, the reverses, the operator expressions and some particular cases are considered.
2 Some lemmas and an example
In the following, we agree that \(p \ne 0,1\), \(\frac{1}{p} + \frac{1}{q} = 1\), \(- 1 < \gamma \le 0\), \(0 < \lambda_{1},\lambda_{2} < 1\), \(\lambda_{1} + \lambda_{2} = \lambda\), \(\mu_{i},\upsilon_{j} > 0\) (\(i,j \in \mathbb{N}\)), with \(\mu_{1} = \upsilon_{1} = 1\), \(U_{m}\) and \(V_{n}\) are defined by (4),
\(a_{m},b_{n} \ge 0\), \(\Vert a \Vert _{p,\Phi_{\lambda}}: = (\sum_{m = 2}^{\infty} \Phi_{\lambda} (m)a_{m}^{p})^{\frac{1}{p}}\) and \(\Vert b \Vert _{q,\Psi_{\lambda}}: = (\sum_{n = 2}^{\infty} \Psi_{\lambda} (n)b_{n}^{q})^{\frac{1}{q}}\), where
Lemma 1
If \(n \in \mathbb{N}\backslash \{ 1\}\), \(a \in (n - \frac{1}{2},n)\), \(f(x)\) is continuous in \((n - \frac{1}{2}, n + \frac{1}{2})\), and \(f'(x)\) is strictly increasing in the intervals \((n - \frac{1}{2},a)\), \((a,n)\) and \((n,n + \frac{1}{2})\), respectively, satisfying
then we have the following Hermite-Hadamard’s inequality (cf. [20]).
Proof
In view of \(f'(n - 0) \le f'(n + 0) = \lim_{x \to n^{ +}} f'(x)\) is finite, we set the linear function \(g(x)\) as follows:
Since \(f'(x)\) is strictly increasing in \([n - \frac{1}{2},a)\) and \((a,n)\), then for \(x \in [n - \frac{1}{2},a)\),
for \(x \in (a,n)\), \(f'(x) < \lim_{x \to n^{ -}} f'(x) = f'(n - 0)\). Hence,
Since \(f(x)-g(x)\) is continuous in \((n - \frac{1}{2},n]\) with \(f(n)-g(n)=0\), it follows that
In the same way, since \(f'(x)\) is strictly increasing in \((n,n + \frac{1}{2})\), then for \(x \in (n,n + \frac{1}{2})\), \(f'(x) > f'(n + 0) \ge f'(n - 0)\). Hence,
Since \(f(x)-g(x)\) is continuous in \([n,n + \frac{1}{2})\) with \(f(n)-g(n)=0\), it follows that
Therefore, we have \(f(x) - g(x) > 0\), \(x \in (n - \frac{1}{2},n + \frac{1}{2})\backslash \{ n\}\). Then we find
namely, (9) follows. The lemma is proved. □
Note
With the assumptions of Lemma 1, if (i) \(a \in (n,n + \frac{1}{2})\), \(f'(x)\) is strictly increasing in the intervals \((n - \frac{1}{2},n)\), \((n,a)\) and \((a,n + \frac{1}{2})\), respectively, or (ii) \(a = n\), \(f'(x)\) is strictly increasing in the intervals \((n - \frac{1}{2},n)\) and \((n,n + \frac{1}{2})\), respectively, then in the same way, we still can obtain (9).
Example 1
\(\{ \mu_{m}\}_{m = 1}^{\infty}\) and \(\{ \upsilon_{n}\}_{n = 1}^{\infty}\) are decreasing, we set functions \(\mu (t): = \mu_{m}\), \(t \in (m - 1,m]\) (\(m \in \mathbb{N}\)), \(\upsilon (t): = \upsilon_{n}\), \(t \in (n - 1,n]\) (\(n \in \mathbb{N}\)), and
Then it follows that \(U(m) = U_{m}\), \(V(n) = V_{n}\), \(U(\infty ) = U_{\infty}\), \(V(\infty ) = V_{\infty}\) and
For \(0 < \lambda \le 1\), \(- 1 < \gamma \le 0\), we set
We find
namely, \(K_{\gamma} (\lambda_{1}) \in \mathbb{R}_{ +}\). In the following, we express \(K_{\gamma} (\lambda_{1})\) in other forms.
(i) For \(\gamma = 0\), we obtain
(ii) for \(- 1 < \gamma < 0\), \(0 < \frac{1 + \gamma}{1 - \gamma} < 1\), by the Lebesgue term by term integration theorem (cf. [21]), we find
(iii) for \(\lambda_{1} = \lambda_{2} = \frac{\lambda}{2}\), \(- 1 < \gamma < 0\), we find
For fixed \(m \in \mathbb{N}\backslash \{ 1\}\), we define the function \(f(y)\) as follows:
Then \(f(y)\) is continuous in \((n - \frac{1}{2},n + \frac{1}{2})\) (\(n \in \mathbb{N}\backslash \{ 1\} \)). There exists a unified number \(y_{0} > \frac{3}{2}\) satisfying \(V(y_{0}) = \frac{\alpha}{\beta} U_{m}\).
(i) If \(y_{0} \in (n - \frac{1}{2},n + \frac{1}{2})\), we find
For \(y_{0} \ne n\), we obtain for \(y \ne n\) that
for \(y_{0} \ne n\), we obtain for \(y = n\) that
Since \(0 < \lambda \le 1\), \(- 1 < \gamma \le 0\), \((1 - \gamma )\upsilon_{n} \ge (1 + \gamma )\upsilon_{n + 1}\), in view of the above results, we find \(f'(n - 0) \le f'(n + 0)\) (\(n \ne y_{0}\)), and \(f'(y)\) (<0) is strictly increasing in \((n - \frac{1}{2},y_{0})\), \((y_{0},n)\) and \((n,n + \frac{1}{2})\) for \(y_{0} < n\) or in \((n - \frac{1}{2},n)\), \((n,y_{0})\) and \((y_{0}, n + \frac{1}{2})\) for \(y_{0} > n\).
We obtain
Since for \(y_{0} = n\), \(V'(y_{0} - 0) = \upsilon_{n}\), \(V'(y_{0} + 0) = \upsilon_{n + 1}\) and for \(y_{0} \ne n\), \(V'(y_{0} - 0) = V'(y_{0})\), then we have \(\lambda (1 - \gamma )V'(y_{0} - 0) \ge \lambda (1 + \gamma )V'(y_{0} + 0)\), namely, \(f'(y_{0} - 0) \le f'(y_{0} + 0)\).
(ii) If \(y_{0} \notin (n - \frac{1}{2},n + \frac{1}{2})\), then it follows that \(f'(y) = \frac{V'(y)}{V(y)}\frac{d}{dy}k_{\lambda} (\ln \alpha U_{m},\ln \beta V(y)) < 0\), \(y \in (n - \frac{1}{2},n + \frac{1}{2})\backslash \{ n\}\). We still can find that
and \(f'(y)\) (<0) is strictly increasing in \((n - \frac{1}{2},n)\) and \((n,n + \frac{1}{2})\).
Therefore, \(f(y)\) satisfies the conditions of Lemma 1 with Note. So does \(g(y) = \frac{f(y)}{V(y)\ln^{1 - \lambda_{2}}\beta V(y)}\). Hence, by (9), we have
Definition 1
Define the following weight coefficients:
Lemma 2
If \(\{ \mu_{m}\}_{m = 1}^{\infty}\) and \(\{ \upsilon_{n}\}_{n = 1}^{\infty}\) are decreasing and \(U_{\infty} = V_{\infty} = \infty\), then for \(m,n \in \mathbb{N}\backslash \{ 1\}\), we have the following inequalities:
where \(K_{\gamma} (\lambda_{1})\) is determined by (12).
Proof
For \(y \in (n - \frac{1}{2},n + \frac{1}{2})\backslash \{ n\}\), \(\upsilon_{n + 1} \le V'(y)\), by (16), we find
Setting \(t = \frac{\ln \beta V(y)}{\ln \alpha U_{m}}\) in the above, since \(\beta V(\frac{3}{2}) = \beta (1 + \frac{\upsilon_{2}}{2}) \ge 1\) and \(\frac{V'(y)}{V(y)}\,dy = (\ln \alpha U_{m})\,dt\), we find
Hence, we obtain (18). In the same way, we obtain (19). □
Note
For example, \(\mu_{n} = \upsilon_{n} = \frac{1}{n^{\sigma}}\) (\(0 \le \sigma \le 1\)) satisfies the conditions of Lemma 2.
Lemma 3
With regard to the assumptions of Lemma 2, (i) for \(m,n \in \mathbb{N}\backslash \{ 1\}\), we have
where
(ii) for any \({c}>0\), we have
Proof
In view of \(\beta \le 1\) and \(\beta \ge \frac{1}{1 + \upsilon_{2}/2} > \frac{1}{1 + \upsilon_{2}}\), it follows that \(1 \le \frac{1 - \beta}{\beta \upsilon_{2}} + 1 < 2\). Since, by Examples 1, \(g(y)\) is strictly decreasing in \([n,n + 1)\), then for \(m \in \mathbb{N}\backslash \{ 1\}\), we find
Setting \(t = \frac{\ln \beta V(y)}{\ln \alpha U_{m}}\), we have \(\ln \beta V(\frac{1 - \beta}{\beta \upsilon_{2}} + 1) = \ln \beta (1 + \frac{1 - \beta}{\beta \upsilon_{2}}\upsilon_{2}) = 0\) and
where
In view of the integral mid-value theorem, for fixed \(m \in \mathbb{N}\backslash \{ 1\}\), there exists \(\theta (m) \in (\frac{1 - \beta}{\beta \upsilon_{2}},1)\) such that
Hence, we find
namely, \(\theta (\lambda_{2},m) = O(\frac{1}{\ln^{\lambda_{2}}\alpha U_{m}})\). Then we obtain (20) and (22). In the same way, we obtain (21) and (23).
For \({c}>0\), we find
Hence, we obtain (20). In the same way, we obtain (21). □
Lemma 4
If \(- 1 < \gamma \le 0\), \(0 < \lambda_{1},\lambda_{2} < 1\), \(\lambda_{1} + \lambda_{2} \le 1\), \(K_{\gamma} (\lambda_{1})\) is determined by (12), then for \(0 < \delta < \min \{ \lambda_{1},\lambda_{2}\}\), we have
Proof
We find, for \(0 < \delta < \min \{ \lambda_{1},\lambda_{2}\}\),
In the same way, we find
and then we have (26). □
3 Main results
In the following, we also set
Theorem 1
-
(i)
For \({p} >1\), we have the following equivalent inequalities:
$$\begin{aligned}& I: = \sum_{n = 2}^{\infty} \sum _{m = 2}^{\infty} k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n})a_{m}b_{n} \le \Vert a \Vert _{p,\tilde{\Phi}_{\lambda}} \Vert b \Vert _{q,\tilde{\Psi}_{\lambda}}, \end{aligned}$$(28)$$\begin{aligned}& J: = \Biggl\{ \sum_{n = 2}^{\infty} \frac{\upsilon_{n + 1}\ln^{p\lambda_{2} - 1}\beta V_{n}}{(\varpi (\lambda_{1},n))^{p - 1}V_{n}} \Biggl( \sum_{m = 2}^{\infty} k_{\lambda} (\ln \alpha U_{m},\ln \beta V_{n})a_{m} \Biggr)^{p} \Biggr\} ^{\frac{1}{p}} \le \Vert a \Vert _{p,\tilde{\Phi}_{\lambda}}; \end{aligned}$$(29) -
(ii)
for \(0< p<1\) (or \(p<0\)), we have the equivalent reverse of (28) and (29).
Proof
(i) By Hölder’s inequality with weight (cf. [20]) and (17), we have
Then, by (16), we find
and then (29) follows.
By Hölder’s inequality (cf. [20]), we have
On the other hand, assuming that (28) is valid, we set
Then we find \(J^{p} = \Vert b \Vert _{q,\tilde{\Psi}_{\lambda}}^{q}\). If \(J = 0\), then (29) is trivially valid; if \(J = \infty\), then by (31), (29) takes the form of equality. Suppose that \(0 < J < \infty\). By (28), it follows that
and then (29) follows, which is equivalent to (28).
(ii) For \(0<{p}<1\) (or \({p}<0\)), by the reverse Hölder’s inequality with weight (cf. [20]) and (13), we obtain the reverse of (30) (or (30)), then we have the reverse of (31), and then the reverse of (29) follows. By Hölder’s inequality (cf. [20]), we have the reverse of (32), and then by the reverse of (29), the reverse of (28) follows.
On the other hand, assuming that the reverse of (28) is valid, we set \(b_{n}\) as (33). Then we find \(J^{p} = \Vert b \Vert _{q,\tilde{\Psi}_{\lambda}}^{q}\). If \(J = \infty\), then the reverse of (29) is trivially valid; if \(J = 0\), then by the reverse of (31), (29) takes the form of equality (=0). Suppose that \(0 < J < \infty\). By the reverse of (28), it follows that the reverses of (34) and (35) are valid, and then the reverse of (29) follows, which is equivalent to the reverse of (28). □
Theorem 2
If \({p}>1\), \(\{ \mu_{m}\}_{m = 1}^{\infty}\) and \(\{ \upsilon_{n}\}_{n = 1}^{\infty}\) are decreasing, \(U_{\infty} = V_{\infty} = \infty\), \(\Vert a \Vert _{p,\Phi_{\lambda}} \in \mathbb{R}_{ +}\) and \(\Vert b \Vert _{q,\Psi_{\lambda}}^{q} \in \mathbb{R}_{ +}\), then we have the following equivalent inequalities:
where the constant factor \(K_{\gamma} (\lambda_{1})\) is the best possible.
Proof
Using (18) and (19) in (28) and (29), we obtain the equivalent inequalities (36) and (37).
For \(\varepsilon \in (0,\min \{ p\lambda_{1},p(1 - \lambda_{2})\} )\), we set \(\tilde{\lambda}_{1} = \lambda_{1} - \frac{\varepsilon}{p}\) (\(\in (0,1)\)), \(\tilde{\lambda}_{2} = \lambda_{2} + \frac{\varepsilon}{p}\) (\(\in (0,1)\)), and
Then, by (24), (25) and (21), we have
If there exists a positive constant \(K \le K_{\gamma} (\lambda_{1})\) such that (36) is valid when replacing \(K_{\gamma} (\lambda_{1})\) by K, then, in particular, we have \(\varepsilon \tilde{I} < \varepsilon K \Vert \tilde{a} \Vert _{p,\Phi_{\lambda}} \Vert \tilde{b} \Vert _{q,\Psi_{\lambda}}\), namely,
In view of (26), it follows that \(K_{\gamma} (\lambda_{1}) \le K(\varepsilon \to 0^{ +} )\). Hence, \(K = K_{\gamma} (\lambda_{1})\) is the best possible constant factor of (36).
Similarly to (32), we still can find the following inequality:
Hence, we can prove that the constant factor \(K_{\gamma} (\lambda_{1})\) in (37) is the best possible. Otherwise, we would reach the contradiction by (39) that the constant factor in (36) is not the best possible. □
Remark 1
(i) For \(\alpha = \beta = 1\) in (36) and (37), setting
we have the following equivalent Mulholland-type inequalities:
(40) is an extension of (7) and the following inequality (for \(\lambda = 1\), \(\lambda_{1} = \frac{1}{q}\), \(\lambda _{2} = \frac{1}{p}\), \(\gamma = 0\)):
(ii) For \(\lambda = 1\), \(\lambda_{1} = \frac{1}{q}\), \(\lambda _{2} = \frac{1}{p}\) in (36) and (37), we have the following equivalent inequalities:
where
(iii) For \(\gamma = 0\), (43) reduces to the following more accurate Hardy-Mulholland-type inequality (7):
In particular, for \(\mu_{i} = \upsilon_{j} = 1\) (\(i,j \in \mathbb{N}\)), (46) reduces to the following more accurate Mulholland’s inequality (\(\frac{2}{3} \le \alpha,\beta \le 1\)):
For \(p > 1\), \(\Psi_{\lambda}^{1 - p}(n) = \frac{\upsilon_{n + 1}}{V_{n}}(\ln \beta V_{n})^{p\lambda_{2} - 1}\), we define the following normed spaces:
Assuming that \(a = \{ a_{m}\}_{m = 2}^{\infty} \in l_{p,\Phi_{\lambda}}\), setting
we can rewrite (37) as follows:
namely, \(c \in l_{p,\Psi_{\lambda}^{1 - p}}\).
Definition 2
Define a Hardy-Mulholland-type operator \(T:l_{p,\Phi_{\lambda}} \to l_{p,\Psi_{\lambda}^{1 - p}}\) 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 (36) and (37) as follows:
Define the norm of operator T as follows:
Then, by (50), we find \(\Vert T \Vert \le K_{\gamma} (\lambda_{1})\). Since the constant factor in (50) is the best possible, we have
4 Some reverses
In the following, we also set
For \(0 < p < 1\) or \(p < 0\), we still use the formal symbols \(\Vert a \Vert _{p,\Phi_{\lambda}}\), \(\Vert b \Vert _{q,\Psi_{\lambda}}\), \(\Vert a \Vert _{p,\tilde{\Omega}_{\lambda}}\) and \(\Vert b \Vert _{q,\tilde{\Upsilon}_{\lambda}}\) et al.
Theorem 3
If \(0 < p < 1\), \(\{ \mu_{m}\}_{m = 1}^{\infty}\) and \(\{ \upsilon_{n}\}_{n = 1}^{\infty}\) are decreasing, \(U_{\infty} = V_{\infty} = \infty\), \(\Vert a \Vert _{p,\Phi_{\lambda}} \in \mathbb{R}_{ +}\) and \(\Vert b \Vert _{q,\Psi_{\lambda}}^{q} \in \mathbb{R}_{ +}\), then we have the following equivalent inequalities with the best possible constant factor \(K_{\gamma} (\lambda_{1})\):
Proof
Using (20) and (19) in the reverses of (28) and (29), since
and
we obtain equivalent inequalities (53) and (54).
For \(\varepsilon \in (0,\min \{ p\lambda_{1},p(1 - \lambda_{2})\})\), we set \(\tilde{\lambda}_{1}\), \(\tilde{\lambda}_{2}\), \(\tilde{a}_{m}\) and \(\tilde{b}_{n}\) as (38). Then, by (24), (25) and (19), we find
If there exists a positive constant \(K \ge K_{\gamma} (\lambda_{1})\) such that (53) is valid when replacing \(K_{\gamma} (\lambda_{1})\) by K, then, in particular, we have \(\varepsilon \tilde{I} > \varepsilon K \Vert \tilde{a} \Vert _{p,\Phi_{\lambda}} \Vert \tilde{b} \Vert _{q,\Psi_{\lambda}}\), namely,
It follows that \(K_{\gamma} (\lambda_{1}) \ge K\) (\(\varepsilon \to 0^{ +} \)). Hence, \(K = K_{\gamma} (\lambda_{1})\) is the best possible constant factor of (53).
The constant factor \(K_{\gamma} (\lambda_{1})\) in (54) is still the best possible. Otherwise, we would reach the contradiction by the reverse of (39) that the constant factor in (53) is not the best possible. □
Remark 2
For \(\alpha = \beta = 1\), set
It is evident that (53) and (54) are extensions of the following equivalent inequalities:
where the constant factor \(K_{\gamma} (\lambda_{1})\) is the best possible.
Theorem 4
If \(p < 0\), \(\{ \mu_{m}\}_{m = 1}^{\infty}\) and \(\{ \upsilon_{n}\}_{n = 1}^{\infty}\) are decreasing, \(U_{\infty} = V_{\infty} = \infty\), \(\Vert a \Vert _{p,\Phi_{\lambda}} \in \mathbb{R}_{ +}\) and \(\Vert b \Vert _{q,\Psi_{\lambda}}^{q} \in \mathbb{R}_{ +}\), then we have the following equivalent inequalities with the best possible constant factor \(K_{\gamma} (\lambda_{1})\):
Proof
Using (18) and (21) in the reverses of (28) and (29), since
and
we obtain equivalent inequalities (57) and (58).
For \(\varepsilon \in (0,\min \{ q\lambda_{2},q(1 - \lambda_{1})\})\), we set \(\tilde{\lambda}_{1} = \lambda_{1} + \frac{\varepsilon}{q}\) (\(\in (0,1)\)), \(\tilde{\lambda}_{2} = \lambda_{2} - \frac{\varepsilon}{q}\) (\(\in (0,1)\)), and
Then, by (24), (25) and (18), we have
If there exists a positive constant \(K \ge K_{\gamma} (\lambda_{1})\) such that (57) is valid when replacing \(K_{\gamma} (\lambda_{1})\) by K, then, in particular, we have \(\varepsilon \tilde{I} > \varepsilon K \Vert \tilde{a} \Vert _{p,\Phi_{\lambda}} \Vert \tilde{b} \Vert _{q,\tilde{\Upsilon}_{\lambda}}\), namely,
It follows that \(K_{\gamma} (\lambda_{1}) \ge K\) (\(\varepsilon \to 0^{ +} \)). Hence, \(K = K_{\gamma} (\lambda_{1})\) is the best possible constant factor of (57).
Similarly to the reverse of (32), we still can find that
Hence, the constant factor \(K_{\gamma} (\lambda_{1})\) in (58) is still the best possible. Otherwise, we would reach the contradiction by (59) that the constant factor in (57) is not the best possible. □
Remark 3
For \(\alpha = \beta = 1\), set
It is evident that (57) and (58) are extensions of the following equivalent inequalities:
where the constant factor \(K_{\gamma} (\lambda_{1})\) is the best possible.
5 Conclusions
In this paper, by using the way of weight coefficients, the technique of real analysis, and Hermite-Hadamard’s inequality, a more accurate Hardy-Mulholland-type inequality with multi-parameters and a best possible constant factor is given by Theorems 1, 2, and the equivalent forms are considered. The equivalent reverses with the best possible constant factor are obtained by Theorems 3, 4. Moreover, the operator expressions and some particular cases are considered. The method of weight coefficients is very important, which helps us to prove the main inequalities with the best possible constant factor. The lemmas and theorems provide an extensive account of this type of inequalities.
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Acknowledgements
This work is supported by the National Natural Science Foundation (No. 61370186, No. 61640222), and Appropriative Researching Fund for Professors and Doctors, Guangdong University of Education (No. 2015ARF25). We are grateful for this help.
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BY carried out the mathematical studies, participated in the sequence alignment and drafted the manuscript. 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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Yang, B., Chen, Q. On a more accurate Hardy-Mulholland-type inequality. J Inequal Appl 2017, 163 (2017). https://doi.org/10.1186/s13660-017-1442-8
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DOI: https://doi.org/10.1186/s13660-017-1442-8