0 引言

令f:I⊆R→R是一个凸函数且a, b∈I, a < b, 则有

$ f\left( {\frac{{a + b}}{2}} \right) \le \frac{1}{{b - a}}\int_a^b {f\left( x \right){\rm{d}}x} \le \frac{{f\left( a \right) + f\left( b \right)}}{2}, $

(1)

这就是著名的Hermite-Hadamard不等式.

近年来, 众多研究者根据函数具备不同的凸性对Hermite-Hadamard型不等式进行了推广和改进^[1-7].随着分数次积分的广泛应用, 不少学者开始研究涉及分数次积分的Hermite-Hadamard型不等式, 并且得到了越来越多关于分数次积分的结果^[8-11].

KIRMACI^[12]证明了以下与不等式(1) 左端有关联的一些结果:

引理1  令f:I^*⊂R→R是区间I^*上的一个可微映射(I^*是I的内部), 若f′∈L[a, b], a, b∈I^*, a < b, 则

$ \begin{array}{l} \frac{1}{{b - a}}\int_a^b {f\left( x \right){\rm{d}}x} - f\left( {\frac{{a + b}}{2}} \right) = \\ \;\;\;\;\;\;\;\left( {b - a} \right)\left[ {\int_0^{1/2} {tf'\left( {ta + \left( {1 - t} \right)b} \right){\rm{d}}t} } \right. + \\ \;\;\;\;\;\;\;\left. {\int_{1/2}^0 {\left( {t - 1} \right)f'\left( {ta + \left( {1 - t} \right)b} \right){\rm{d}}t} } \right]. \end{array} $

(2)

根据引理1, KIRMACI证明了以下有关凸函数的3个定理.

定理1  令f:I^*⊂R→R是区间I^*上的可微映射, 若|f′|是区间[a, b]上的凸函数, a, b∈I^*, a < b, 则

$ \begin{array}{l} \left| {\frac{1}{{b - a}}\int_a^b {f\left( x \right){\rm{d}}x} - f\left( {\frac{{a + b}}{2}} \right)} \right| \le \\ \;\;\;\;\;\;\;\frac{{\left( {b - a} \right)}}{8}\left( {\left| {f'\left( a \right)} \right| + \left| {f'\left( b \right)} \right|} \right). \end{array} $

(3)

定理2  令f:I^*⊂R→R是I^*上的可微映射, 若映射${{\left| {{f}'} \right|}^{\frac{p}{p-1}}}$是[a, b]上的凸函数, a, b∈I^*, a < b, 且p>1，则

$ \begin{array}{l} \left| {\frac{1}{{b - a}}\int_a^b {f\left( x \right){\rm{d}}x} - f\left( {\frac{{a + b}}{2}} \right)} \right| \le \frac{{b - a}}{{16}}{\left( {\frac{4}{{p + 1}}} \right)^{1/p}} \times \\ \left[ {{{\left( {{{\left| {f'\left( a \right)} \right|}^{p/\left( {p - 1} \right)}} + 3{{\left| {f'\left( b \right)} \right|}^{p/\left( {p - 1} \right)}}} \right)}^{\left( {p - 1} \right)/p}} + } \right.\\ \left. {{{\left( {3{{\left| {f'\left( a \right)} \right|}^{p/\left( {p - 1} \right)}} + {{\left| {f'\left( b \right)} \right|}^{p/\left( {p - 1} \right)}}} \right)}^{\left( {p - 1} \right)/p}}} \right]. \end{array} $

(4)

定理3  令f:I^*⊂R→R是区间I^*上的可微映射.若映射${{\left| {{f}'} \right|}^{\frac{p}{p-1}}}$是[a, b]上的凸函数, a, b∈I^*, a < b, 且p>1, 则

$ \begin{array}{l} \left| {\frac{1}{{b - a}}\int_a^b {f\left( x \right){\rm{d}}x} - f\left( {\frac{{a + b}}{2}} \right)} \right| \le \\ \;\;\;\;\;\frac{{b - a}}{4}{\left( {\frac{4}{{p + 1}}} \right)^{1/p}}\left( {\left| {f'\left( a \right)} \right| + \left| {f'\left( b \right)} \right|} \right). \end{array} $

(5)

KIRMACI等^[13]还证明了对于凹函数，有

定理4  令f:I^*⊂R→R是区间I^*上的可微映射, 且p≥1.如果|f′|^p是[a, b]上的凹函数, a, b∈I^*, a < b, 并且|f′|是一个线性映射, 则

$ \begin{array}{l} \left| {\frac{1}{{b - a}}\int_a^b {f\left( x \right){\rm{d}}x} - f\left( {\frac{{a + b}}{2}} \right)} \right| \le \\ \;\;\;\;\;\;\left( {\frac{{b - a}}{8}} \right)\left( {f'\left( {a + b} \right)} \right). \end{array} $

(6)

为了得到新的结果, 特引入下面3个定义.

定义1^[14]  若f:[0, ∞)→R, 对于x, y∈[0, ∞), λ∈[0, 1]，并且对于某一固定的s∈(0, 1], 若不等式

$ f\left( {\lambda x + \left( {1 - \lambda } \right)y} \right) \le {\lambda ^s}f\left( x \right) + {\left( {1 - \lambda } \right)^s}f\left( y \right) $

成立, 则称函数f为第2种意义下的s-凸函数，记为f∈K_s².若(-f)是s-凸的, 则称f是s-凹函数.

易知, 当s=1时, s-凸性即为通常意义下的凸函数.

DRAGOMIR等^[15]证明了对于第2种意义下s-凸函数的Hadamard不等式的一个变式:

定理5  假设f:[0, ∞]→[0, ∞]是一个第2种意义下的s-凸函数, 其中s∈(0, 1) 且a, b∈[0, ∞), a < b.若f∈L¹([a, b]), 则有

$ {2^{s - 1}}f\left( {\frac{{a + b}}{2}} \right) \le \frac{1}{{b - a}}\int_a^b {f\left( x \right){\rm{d}}x} \le \frac{{f\left( a \right) + f\left( b \right)}}{{s + 1}}, $

(7)

常数$k=\frac{1}{s+1}$是式(7) 第2个不等式的最佳常系数.

下面给出α(∈R⁺)阶左侧和右侧分数次积分的定义.

定义2  令f∈L¹[a, b], α(∈R⁺)阶Riemann-Liouville积分J_a+^αf和J_b-^αf(a≥0) 分别定义为

$ J_{a + }^\alpha f\left( x \right) = \frac{1}{{\Gamma \left( \alpha \right)}}\int_a^x {{{\left( {x - t} \right)}^{\alpha - 1}}f\left( t \right){\rm{d}}t} ,x > a, $

$ J_{b - }^\alpha f\left( x \right) = \frac{1}{{\Gamma \left( \alpha \right)}}\int_x^b {{{\left( {t - x} \right)}^{\alpha - 1}}f\left( t \right){\rm{d}}t} ,x < b, $

其中$\Gamma \left( \alpha \right)=\int_{0}^{\infty }{{{\text{e}}^{-u}}{{u}^{a-1}}}$du, 这里J_a+⁰f(x)=J_b-⁰f(x)=f(x).

可见, 当α=1时, 分数次积分即为经典积分.

为了简化结果的表示形式，引入不完全Beta函数的定义^[16-18]:

定义3  对于a, b>0且0 < x < 1, 不完全Beta函数B_x(a, b)定义为

$ {B_x}\left( {a,b} \right) = \int_0^x {{t^{a - 1}}{{\left( {1 - t} \right)}^{b - 1}}{\rm{d}}t} . $

本文利用Riemann-Liouville分数次积分, 对具有s-凸性的函数, 建立了与Hermite-Hadamard不等式左端相关的不等式.将Hermite-Hadamard积分不等式推广到分数次积分, 文中结果推广了已有文献中的结论, 如不等式(3)~(6) 是本文结果的特殊情况.

1 Hermite-Hadamard型分数次积分不等式的主要结果及证明

引理2  令f:[a, b]→R是区间(a, b)上的可微映射, a < b.如果f′∈L[a, b], 则以下关于分数次积分的等式成立:

$ \begin{array}{l} \frac{{\Gamma \left( {\alpha + 1} \right)}}{{{{\left( {b - a} \right)}^\alpha }}}\left[ {{{\left( { - 1} \right)}^\alpha }J_{\frac{{a + b}}{2} - }^\alpha f\left( b \right) - J_{\frac{{a + b}}{2} + }^\alpha f\left( a \right)} \right] + \\ \;\;\;\;\;\;\;\;\;\left[ { - 1 + {{\left( { - 1} \right)}^\alpha }} \right]\frac{1}{{{2^\alpha }}}f\left( {\frac{{a + b}}{2}} \right) = \\ \;\;\;\;\;\;\;\;\;\left( {b - a} \right)\left[ {\int_0^{1/2} {{t^\alpha }f'\left( {ta + \left( {1 - t} \right)b} \right){\rm{d}}t} + } \right.\\ \;\;\;\;\;\;\;\;\;\left. {\int_{1/2}^1 {{{\left( {t - 1} \right)}^\alpha }f'\left( {ta + \left( {1 - t} \right)b} \right){\rm{d}}t} } \right]. \end{array} $

(8)

证明  由分部积分, 有

$ \begin{array}{l} \int_0^{1/2} {{t^\alpha }f'\left( {ta + \left( {1 - t} \right)b} \right){\rm{d}}t} = \\ {t^\alpha }\frac{{f\left( {ta + \left( {1 - t} \right)b} \right)}}{{a - b}}\left| {_0^{1/2}} \right. - \int_0^{1/2} {\alpha {t^{\alpha - 1}}\frac{{f\left( {ta + \left( {1 - t} \right)b} \right)}}{{a - b}}{\rm{d}}t} = \\ - \frac{1}{{{2^\alpha }\left( {b - a} \right)}}f\left( {\frac{{a + b}}{2}} \right) + \frac{\alpha }{{b - a}}\int_b^{\frac{{a + b}}{2}} {{{\left( {\frac{{x - b}}{{a - b}}} \right)}^{\alpha - 1}}\frac{{f\left( x \right)}}{{a - b}}{\rm{d}}x} = \\ - \frac{1}{{{2^\alpha }\left( {b - a} \right)}}f\left( {\frac{{a + b}}{2}} \right) + {\left( { - 1} \right)^\alpha }\frac{{\Gamma \left( {\alpha + 1} \right)}}{{{{\left( {b - a} \right)}^{\alpha + 1}}}}J_{\frac{{a + b}}{2} - }^\alpha f\left( b \right) \end{array} $

(9)

和

$ \begin{array}{l} \int_{1/2}^1 {{{\left( {t - 1} \right)}^\alpha }f'\left( {ta + \left( {1 - t} \right)b} \right){\rm{d}}t} = \\ {\left( {t - 1} \right)^\alpha }\frac{{f\left( {ta + \left( {1 - t} \right)b} \right)}}{{a - b}}\left| {_{1/2}^1} \right. - \\ \int_{1/2}^1 {\alpha {{\left( {t - 1} \right)}^{\alpha - 1}}\frac{{f\left( {ta + \left( {1 - t} \right)b} \right)}}{{a - b}}{\rm{d}}t} = \\ {\left( { - 1} \right)^\alpha }\frac{1}{{{2^\alpha }\left( {b - a} \right)}}f\left( {\frac{{a + b}}{2}} \right) + \\ \frac{\alpha }{{b - a}}\int_{\frac{{a + b}}{2}}^a {{{\left( {\frac{{x - a}}{{a - b}}} \right)}^{\alpha - 1}}\frac{{f\left( x \right)}}{{a - b}}{\rm{d}}x} = \\ {\left( { - 1} \right)^\alpha }\frac{1}{{{2^\alpha }\left( {b - a} \right)}}f\left( {\frac{{a + b}}{2}} \right) - \frac{{\Gamma \left( {\alpha + 1} \right)}}{{{{\left( {b - a} \right)}^{\alpha + 1}}}}J_{\frac{{a + b}}{2} + }^\alpha f\left( a \right), \end{array} $

(10)

其中, 用到变量代换x=ta+(1-t)b, t∈[0, 1].

用(b-a)分别乘以式(9) 和(10) 两边, 分别得到

$ \begin{array}{l} \left( {b - a} \right)\int_0^{1/2} {{t^\alpha }f'\left( {ta + \left( {1 - t} \right)b} \right){\rm{d}}t} = \\ - \frac{1}{{{2^\alpha }}}f\left( {\frac{{a + b}}{2}} \right) + {\left( { - 1} \right)^\alpha }\frac{{\Gamma \left( {\alpha + 1} \right)}}{{{{\left( {b - a} \right)}^\alpha }}}J_{\frac{{a + b}}{2} - }^\alpha f\left( b \right) \end{array} $

(11)

和

$ \begin{array}{l} \left( {b - a} \right)\int_{1/2}^1 {{{\left( {t - 1} \right)}^\alpha }f'\left( {ta + \left( {1 - t} \right)b} \right){\rm{d}}t} = \\ {\left( { - 1} \right)^\alpha }\frac{1}{{{2^\alpha }}}f\left( {\frac{{a + b}}{2}} \right) - \frac{{\Gamma \left( {\alpha + 1} \right)}}{{{{\left( {b - a} \right)}^\alpha }}}J_{\frac{{a + b}}{2} + }^\alpha f\left( a \right). \end{array} $

(12)

由式(11) 和(12)，引理2得证.

由引理2, 可以得到关于第2种意义下s-凸函数的分数次积分不等式.

定理6  令f:[a, b]⊂[0, ∞]→R是区间(a, b)上的可微映射, a < b, 且使得f′∈L[a, b].如果对某一个固定的s∈(0, 1], |f′|在区间[a, b]上是第2种意义下s-凸的, 则下面的分数次积分不等式成立:

$ \begin{array}{l} \left| {\frac{{\Gamma \left( {\alpha + 1} \right)}}{{{{\left( {b - a} \right)}^\alpha }}}\left[ {{{\left( { - 1} \right)}^\alpha }J_{\frac{{a + b}}{2} - }^\alpha f\left( b \right) - J_{\frac{{a + b}}{2} + }^\alpha f\left( a \right)} \right] + \left[ { - 1 + } \right.} \right.\\ \left. {\left. {{{\left( { - 1} \right)}^\alpha }} \right]\frac{1}{{{2^\alpha }}}f\left( {\frac{{a + b}}{2}} \right)} \right| \le \left( {b - a} \right)\left[ {\frac{1}{{{2^{\alpha + s + 1}}\left( {\alpha + s + 1} \right)}} + } \right.\\ \left. {{B_{\frac{1}{2}}}\left( {\alpha + 1,s + 1} \right)} \right]\left( {\left| {f'\left( a \right)} \right| + \left| {f'\left( b \right)} \right|} \right). \end{array} $

(13)

其中${{B}_{\frac{1}{2}}}$(α+1, s+1) 是不完全Beta函数.

证明  由引理2以及|f′|是第2种意义下的s-凸函数, 可得

$ \begin{array}{l} \left| {\frac{{\Gamma \left( {\alpha + 1} \right)}}{{{{\left( {b - a} \right)}^\alpha }}}\left[ {{{\left( { - 1} \right)}^\alpha }J_{\frac{{a + b}}{2} - }^\alpha f\left( b \right) - J_{\frac{{a + b}}{2} + }^\alpha f\left( a \right)} \right] + \left[ { - 1 + } \right.} \right.\\ \left. {\left. {{{\left( { - 1} \right)}^\alpha }} \right]\frac{1}{{{2^\alpha }}}f\left( {\frac{{a + b}}{2}} \right)} \right| \le \\ \left( {b - a} \right)\left[ {\int_0^{1/2} {\left| {{t^\alpha }} \right|\left| {f'\left( {ta + \left( {1 - t} \right)b} \right)} \right|{\rm{d}}t} + } \right.\\ \left. {\int_{1/2}^1 {\left| {{{\left( {t - 1} \right)}^\alpha }} \right|\left| {f'\left( {ta + \left( {1 - t} \right)b} \right)} \right|{\rm{d}}t} } \right] \le \\ \left( {b - a} \right)\left[ {\int_0^{1/2} {{t^\alpha }\left[ {{t^s}\left| {f'\left( a \right)} \right| + {{\left( {1 - t} \right)}^s}\left| {f'\left( b \right)} \right|} \right]{\rm{d}}t} + } \right.\\ \left. {\int_{1/2}^1 {{{\left( {1 - t} \right)}^\alpha }\left[ {{t^s}\left| {f'\left( a \right)} \right| + {{\left( {1 - t} \right)}^s}\left| {f'\left( b \right)} \right|} \right]{\rm{d}}t} } \right] = \\ \left( {b - a} \right)\left[ {\frac{1}{{{2^{\alpha + s + 1}}\left( {\alpha + s + 1} \right)}} + {B_{\frac{1}{2}}}\left( {\alpha + 1,s + 1} \right)} \right] \times \\ \left( {\left| {f'\left( a \right)} \right| + \left| {f'\left( b \right)} \right|} \right), \end{array} $

其中用到以下计算结果：

$ \int_0^{1/2} {{t^{\alpha + s}}{\rm{d}}t} = \int_{1/2}^1 {{{\left( {1 - t} \right)}^{\alpha + s}}{\rm{d}}t} = \frac{1}{{{2^{\alpha + s + 1}}\left( {\alpha + s + 1} \right)}} $

和

$ \begin{array}{*{20}{c}} {\int_0^{1/2} {{t^\alpha }{{\left( {1 - t} \right)}^s}{\rm{d}}t} = \int_{1/2}^1 {{{\left( {1 - t} \right)}^\alpha }{t^s}{\rm{d}}t} = }\\ {{B_{\frac{1}{2}}}\left( {\alpha + 1,s + 1} \right).} \end{array} $

结论得证.

推论1  令f:[a, b]⊂[0, ∞]→R是区间(a, b)上的一个可微映射, a < b, 并且使得f′∈L[a, b].如果|f′|在[a, b]上是凸的, 则以下分数次积分不等式成立:

$ \begin{array}{l} \left| {\frac{{\Gamma \left( {\alpha + 1} \right)}}{{{{\left( {b - a} \right)}^\alpha }}}\left[ {{{\left( { - 1} \right)}^\alpha }J_{\frac{{a + b}}{2} - }^\alpha f\left( b \right) - J_{\frac{{a + b}}{2} + }^\alpha f\left( a \right)} \right] + \left[ { - 1 + } \right.} \right.\\ \left. {\left. {\;\;\;\;\;\;\;\;{{\left( { - 1} \right)}^\alpha }} \right]\frac{1}{{{2^\alpha }}}f\left( {\frac{{a + b}}{2}} \right)} \right| \le \\ \;\;\;\;\;\;\;\;\frac{{\left( {b - a} \right)}}{{{2^{\alpha + 1}}\left( {\alpha + 1} \right)}}\left( {\left| {f'\left( a \right)} \right| + \left| {f'\left( b \right)} \right|} \right). \end{array} $

(14)

证明  在式(13) 中, 取s=1, 由于

$ \begin{array}{l} {B_{\frac{1}{2}}}\left( {\alpha + 1,2} \right) = \int_0^{1/2} {{t^\alpha }\left( {1 - t} \right){\rm{d}}t} = \\ \;\;\;\;\;\int_{1/2}^1 {{{\left( {1 - t} \right)}^\alpha }{t^s}{\rm{d}}t} = \frac{1}{{{2^{\alpha + 2}}}}\frac{{\alpha + 3}}{{\left( {\alpha + 1} \right)\left( {\alpha + 2} \right)}}, \end{array} $

于是, 结论得证.

注1  在推论1中, 如果令α=1, 则由式(14) 可得到定理1中的不等式(3).

推论2  令f:[a, b]⊂[0, ∞]→R, 在区间(a, b)上是一个可微映射, a < b, 并且使得f′∈L[a, b].如果对某一个固定的s∈(0, 1], |f′|在[a, b]上是第2种意义下s-凸的, 则以下不等式成立:

$ \begin{array}{l} \left| {\frac{1}{{b - a}}\int_a^b {f\left( x \right){\rm{d}}x} - f\left( {\frac{{a + b}}{2}} \right)} \right| \le \\ \left( {b - a} \right)\left[ {\frac{{{2^{s + 1}} - 1}}{{{2^{s + 1}}\left( {s + 1} \right)\left( {s + 2} \right)}}} \right]\left( {\left| {f'\left( a \right)} \right| + \left| {f'\left( b \right)} \right|} \right). \end{array} $

(15)

证明  在式(13) 中, 取α=1, 由于

$ \begin{array}{l} {B_{\frac{1}{2}}}\left( {2,s + 1} \right) = \int_0^{1/2} {t{{\left( {1 - t} \right)}^s}{\rm{d}}t} = \\ \;\;\;\;\;\int_{1/2}^1 {\left( {1 - t} \right){t^s}{\rm{d}}t} = \frac{1}{{{2^{s + 2}}}}\frac{{{2^{s + 2}} - s - 3}}{{\left( {s + 1} \right)\left( {s + 2} \right)}}. \end{array} $

则结论成立.

注2  在推论2中, 令s=1, 则式(15) 也能得到定理1中的不等式(3).

定理7  令f:[a, b]⊂[0, ∞]→R, 在区间(a, b)上是一个可微映射, a < b, 并且使得f′∈L[a, b].如果对某一固定的s∈(0, 1], |f′|^p/(p-1)在区间[a, b]上是第2种意义下s-凸的, p>1, 则以下分数次积分不等式成立:

$ \begin{array}{l} \left| {\frac{{\Gamma \left( {\alpha + 1} \right)}}{{{{\left( {b - a} \right)}^\alpha }}}\left[ {{{\left( { - 1} \right)}^\alpha }J_{\frac{{a + b}}{2} - }^\alpha f\left( b \right) - J_{\frac{{a + b}}{2} + }^\alpha f\left( a \right)} \right]} \right. + \left[ { - 1 + } \right.\\ \left. {\left. {{{\left( { - 1} \right)}^\alpha }} \right]\frac{1}{{{2^\alpha }}}f\left( {\frac{{a + b}}{2}} \right)} \right| \le \frac{{b - a}}{{{2^{\alpha + s + 1}}\left( {s + 1} \right)}}{\left( {\frac{{{2^s}\left( {s + 1} \right)}}{{\alpha p + 1}}} \right)^{\frac{1}{p}}} \times \\ \left[ {{{\left( {{{\left| {f'\left( a \right)} \right|}^{p/\left( {p - 1} \right)}} + \left( {{2^{s + 1}} - 1} \right){{\left| {f'\left( b \right)} \right|}^{p/\left( {p - 1} \right)}}} \right)}^{\left( {p - 1} \right)/p}} + } \right.\\ \left. {{{\left( {\left( {{2^{s + 1}} - 1} \right){{\left| {f'\left( a \right)} \right|}^{p/\left( {p - 1} \right)}} + {{\left| {f'\left( b \right)} \right|}^{p/\left( {p - 1} \right)}}} \right)}^{\left( {p - 1} \right)/p}}} \right]. \end{array} $

(16)

证明  由引理2以及Hölder不等式, 得到

$ \begin{array}{l} \left| {\frac{{\Gamma \left( {\alpha + 1} \right)}}{{{{\left( {b - a} \right)}^\alpha }}}\left[ {{{\left( { - 1} \right)}^\alpha }J_{\frac{{a + b}}{2} - }^\alpha f\left( b \right) - J_{\frac{{a + b}}{2} + }^\alpha f\left( a \right)} \right]} \right. + \\ \left. {\left[ { - 1 + {{\left( { - 1} \right)}^\alpha }} \right]\frac{1}{{{2^\alpha }}}f\left( {\frac{{a + b}}{2}} \right)} \right| \le \\ \left( {b - a} \right)\left[ {\int_0^{1/2} {\left| {{t^\alpha }} \right|\left| {f'\left( {ta + \left( {1 - t} \right)b} \right)} \right|{\rm{d}}t} + } \right.\\ \left. {\int_{1/2}^1 {\left| {{{\left( {t - 1} \right)}^\alpha }} \right|\left| {f'\left( {ta + \left( {1 - t} \right)b} \right)} \right|{\rm{d}}t} } \right] \le \\ \left( {b - a} \right)\left[ {{{\left( {\int_0^{1/2} {{t^{\alpha p}}{\rm{d}}t} } \right)}^{\frac{1}{p}}}{{\left( {\int_0^{1/2} {{{\left| {f'\left( {ta + \left( {1 - t} \right)b} \right)} \right|}^q}{\rm{d}}t} } \right)}^{\frac{1}{q}}}} \right] + \\ \left( {b - a} \right)\left[ {{{\left( {\int_{1/2}^1 {\left| {{{\left( {t - 1} \right)}^{\alpha p}}} \right|{\rm{d}}t} } \right)}^{\frac{1}{p}}}\left( {\int_{1/2}^1 {\left| {f'\left( {ta + } \right.} \right.} } \right.} \right.\\ \left. {{{\left. {{{\left. {\left. {\left( {1 - t} \right)b} \right)} \right|}^q}{\rm{d}}t} \right)}^{\frac{1}{q}}}} \right], \end{array} $

其中，

$ \frac{1}{p} + \frac{1}{q} = 1. $

因为|f′|^q在[a, b]上是第2种意义下s-凸的, 则有

$ \begin{array}{l} \int_0^{1/2} {{{\left| {f'\left( {ta + \left( {1 - t} \right)b} \right)} \right|}^q}{\rm{d}}t} \le \\ \int_0^{1/2} {\left( {{t^s}{{\left| {f'\left( a \right)} \right|}^q} + {{\left( {1 - t} \right)}^s}{{\left| {f'\left( b \right)} \right|}^q}} \right){\rm{d}}t} = \\ \frac{1}{{{2^{s + 1}}\left( {s + 1} \right)}}\left( {{{\left| {f'\left( a \right)} \right|}^q} + \left( {{2^{s + 1}} - 1} \right){{\left| {f'\left( b \right)} \right|}^q}} \right) \end{array} $

和

$ \begin{array}{l} \int_{1/2}^1 {{{\left| {f'\left( {ta + \left( {1 - t} \right)b} \right)} \right|}^q}{\rm{d}}t} \le \\ \int_{1/2}^1 {\left( {{t^s}{{\left| {f'\left( a \right)} \right|}^q} + {{\left( {1 - t} \right)}^s}{{\left| {f'\left( b \right)} \right|}^q}} \right){\rm{d}}t} = \\ \frac{1}{{{2^{s + 1}}\left( {s + 1} \right)}}\left( {\left( {{2^{s + 1}} - 1} \right){{\left| {f'\left( a \right)} \right|}^q} + {{\left| {f'\left( b \right)} \right|}^q}} \right). \end{array} $

通过计算得到

$ \int_0^{1/2} {{t^{\alpha p}}{\rm{d}}t} = \int_{1/2}^1 {\left| {{{\left( {t - 1} \right)}^{\alpha p}}} \right|{\rm{d}}t} = \frac{1}{{{2^{\alpha p + 1}}\left( {\alpha p + 1} \right)}}. $

从而可得

$ \begin{array}{l} \left| {\frac{{\Gamma \left( {\alpha + 1} \right)}}{{{{\left( {b - a} \right)}^\alpha }}}\left[ {{{\left( { - 1} \right)}^\alpha }J_{\frac{{a + b}}{2} - }^\alpha f\left( b \right) - J_{\frac{{a + b}}{2} + }^\alpha f\left( a \right)} \right]} \right. + \\ \left. {\left[ { - 1 + {{\left( { - 1} \right)}^\alpha }} \right]\frac{1}{{{2^\alpha }}}f\left( {\frac{{a + b}}{2}} \right)} \right| \le \\ \frac{{b - a}}{{{2^{\alpha + s + 1}}\left( {s + 1} \right)}}{\left( {\frac{{{2^s}\left( {s + 1} \right)}}{{\alpha p + 1}}} \right)^{\frac{1}{p}}} \times \\ \left[ {{{\left( {{{\left| {f'\left( a \right)} \right|}^{p/\left( {p - 1} \right)}} + \left( {{2^{s + 1}} - 1} \right){{\left| {f'\left( b \right)} \right|}^{p/\left( {p - 1} \right)}}} \right)}^{\left( {p - 1} \right)/p}} + } \right.\\ \left. {{{\left( {\left( {{2^{s + 1}} - 1} \right){{\left| {f'\left( a \right)} \right|}^{p/\left( {p - 1} \right)}} + {{\left| {f'\left( b \right)} \right|}^{p/\left( {p - 1} \right)}}} \right)}^{\left( {p - 1} \right)/p}}} \right], \end{array} $

定理7得证.

推论3  令f:[a, b]⊂[0, ∞]→R, 在区间(a, b)上是一个可微映射, a < b, 并且使得f′∈L[a, b].如果|f′|^p/(p-1)在区间[a, b]上是凸的, p>1, 则以下分数次积分不等式成立:

$ \begin{array}{l} \left| {\frac{{\Gamma \left( {\alpha + 1} \right)}}{{{{\left( {b - a} \right)}^\alpha }}}\left[ {{{\left( { - 1} \right)}^\alpha }J_{\frac{{a + b}}{2} - }^\alpha f\left( b \right) - J_{\frac{{a + b}}{2} + }^\alpha f\left( a \right)} \right]} \right. + \left[ { - 1 + } \right.\\ \left. {\left. {{{\left( { - 1} \right)}^\alpha }} \right]\frac{1}{{{2^\alpha }}}f\left( {\frac{{a + b}}{2}} \right)} \right| \le \\ \frac{{b - a}}{{{2^{\alpha + 3}}}}{\left( {\frac{4}{{\alpha p + 1}}} \right)^{\frac{1}{p}}}\left[ {\left( {{{\left| {f'\left( a \right)} \right|}^{p/\left( {p - 1} \right)}} + } \right.} \right.\\ {\left. {3{{\left| {f'\left( b \right)} \right|}^{p/\left( {p - 1} \right)}}} \right)^{\left( {p - 1} \right)/p}} + \\ \left. {{{\left( {3{{\left| {f'\left( a \right)} \right|}^{p/\left( {p - 1} \right)}} + {{\left| {f'\left( b \right)} \right|}^{p/\left( {p - 1} \right)}}} \right)}^{\left( {p - 1} \right)/p}}} \right]. \end{array} $

(17)

证明  在式(16) 中, 取s=1, 即可得到式(7).

注3  在推论3中, 如果令α=1, 则由式(17) 可得到定理2中的不等式(4).

推论4  令f:[a, b]⊂[0, ∞]→R, 在区间(a, b)中是一个可微映射, a < b, 并且使得f′∈L[a, b].如果对某一个固定的s∈(0, 1], |f′|^p/(p-1)在区间[a, b]上是第2种意义下s-凸的, p>1, 则以下不等式成立:

$ \begin{array}{l} \left| {\frac{1}{{b - a}}\int_a^b {f\left( x \right){\rm{d}}x - f\left( {\frac{{a + b}}{2}} \right)} } \right| \le \\ \frac{{b - a}}{{{2^{s + 2}}\left( {s + 1} \right)}}{\left( {\frac{{{2^s}\left( {s + 1} \right)}}{{p + 1}}} \right)^{\frac{1}{p}}} \times \\ \left[ {{{\left( {{{\left| {f'\left( a \right)} \right|}^{p/\left( {p - 1} \right)}} + \left( {{2^{s + 1}} - 1} \right){{\left| {f'\left( b \right)} \right|}^{p/\left( {p - 1} \right)}}} \right)}^{\left( {p - 1} \right)/p}} + } \right.\\ \left. {{{\left( {\left( {{2^{s + 1}} - 1} \right){{\left| {f'\left( a \right)} \right|}^{p/\left( {p - 1} \right)}} + {{\left| {f'\left( b \right)} \right|}^{p/\left( {p - 1} \right)}}} \right)}^{\left( {p - 1} \right)/p}}} \right]. \end{array} $

(18)

证明  在式(16) 中, 取α=1, 即可得到式(18).

注4  在推论4中, 如果令s=1, 则由式(18) 也可得到定理2中的不等式(4).

定理8  令f:[a, b]⊂[0, ∞]→R, 在区间(a, b)上是一个可微映射, a < b, 并且使得f′∈L[a, b].如果对某一固定的s∈(0, 1], |f′|^p/(p-1)在区间[a, b]上是第2种意义下s-凸的, p>1, 则以下分数次积分不等式成立:

$ \begin{array}{l} \left| {\frac{{\Gamma \left( {\alpha + 1} \right)}}{{{{\left( {b - a} \right)}^\alpha }}}\left[ {{{\left( { - 1} \right)}^\alpha }J_{\frac{{a + b}}{2} - }^\alpha f\left( b \right) - J_{\frac{{a + b}}{2} + }^\alpha f\left( a \right)} \right]} \right. + \left[ { - 1 + } \right.\\ \left. {\left. {{{\left( { - 1} \right)}^\alpha }} \right]\frac{1}{{{2^\alpha }}}f\left( {\frac{{a + b}}{2}} \right)} \right| \le \\ \frac{{b - a}}{{{2^\alpha }\left( {s + 1} \right)}}{\left( {\frac{{{2^s}\left( {s + 1} \right)}}{{\alpha p + 1}}} \right)^{\frac{1}{p}}}\left( {\left| {f'\left( a \right)} \right| + \left| {f'\left( b \right)} \right|} \right). \end{array} $

(19)

证明  令a₁=|f′(a)|^p/(p-1), b₁=(2^s+1-1)|f′(b)|^p/(p-1), a₂=(2^s+1-1)|f′(a)|^p/(p-1), b₂=|f′(b)|^p/(p-1)，这里对p>1, 有0 < (p-1)/p < 1.因为对于0 < r < 1, a₁, a₂, …, a_n≥0和b₁, b₂, …, b_n≥0, 有

$ \sum\limits_{i = 1}^n {{{\left( {{a_i} + {b_i}} \right)}^r}} \le \sum\limits_{i = 1}^n {a_i^r} + \sum\limits_{i = 1}^n {b_i^r} . $

由定理7, 可得

$ \begin{array}{l} \left| {\frac{{\Gamma \left( {\alpha + 1} \right)}}{{{{\left( {b - a} \right)}^\alpha }}}\left[ {{{\left( { - 1} \right)}^\alpha }J_{\frac{{a + b}}{2} - }^\alpha f\left( b \right) - J_{\frac{{a + b}}{2} + }^\alpha f\left( a \right)} \right]} \right. + \\ \left. {\left[ { - 1 + {{\left( { - 1} \right)}^\alpha }} \right]\frac{1}{{{2^\alpha }}}f\left( {\frac{{a + b}}{2}} \right)} \right| \le \\ \frac{{b - a}}{{{2^{\alpha + s + 1}}\left( {s + 1} \right)}}{\left( {\frac{{{2^s}\left( {s + 1} \right)}}{{\alpha p + 1}}} \right)^{\frac{1}{p}}} \times \\ \left[ {{{\left( {{{\left| {f'\left( a \right)} \right|}^{p/\left( {p - 1} \right)}} + \left( {{2^{s + 1}} - 1} \right){{\left| {f'\left( b \right)} \right|}^{p/\left( {p - 1} \right)}}} \right)}^{\left( {p - 1} \right)/p}} + } \right.\\ \left. {{{\left( {\left( {{2^{s + 1}} - 1} \right){{\left| {f'\left( a \right)} \right|}^{p/\left( {p - 1} \right)}} + {{\left| {f'\left( b \right)} \right|}^{p/\left( {p - 1} \right)}}} \right)}^{\left( {p - 1} \right)/p}}} \right] \le \\ \frac{{b - a}}{{{2^{\alpha + s + 1}}\left( {s + 1} \right)}}{\left( {\frac{{{2^s}\left( {s + 1} \right)}}{{\alpha p + 1}}} \right)^{\frac{1}{p}}} \times \\ \left[ {1 + {{\left( {{2^{s + 1}} - 1} \right)}^{\left( {p - 1} \right)/p}}} \right]\left( {\left| {f'\left( a \right)} \right| + \left| {f'\left( b \right)} \right|} \right) \le \\ \frac{{b - a}}{{{2^\alpha }\left( {s + 1} \right)}}{\left( {\frac{{{2^s}\left( {s + 1} \right)}}{{\alpha p + 1}}} \right)^{\frac{1}{p}}}\left( {\left| {f'\left( a \right)} \right| + \left| {f'\left( b \right)} \right|} \right), \end{array} $

定理8得证.

推论5  令f:[a, b]⊂[0, ∞]→R, 在区间(a, b)上是一个可微映射, a < b, 并且使得f′∈L[a, b].如果|f′|^p/(p-1)在区间[a, b]上是凸的, p>1, 则以下分数次积分不等式成立:

$ \begin{array}{l} \left| {\frac{{\Gamma \left( {\alpha + 1} \right)}}{{{{\left( {b - a} \right)}^\alpha }}}\left[ {{{\left( { - 1} \right)}^\alpha }J_{\frac{{a + b}}{2} - }^\alpha f\left( b \right) - J_{\frac{{a + b}}{2} + }^\alpha f\left( a \right)} \right]} \right. + \\ \left. {\left[ { - 1 + {{\left( { - 1} \right)}^\alpha }} \right]\frac{1}{{{2^\alpha }}}f\left( {\frac{{a + b}}{2}} \right)} \right| \le \\ \frac{{b - a}}{{{2^{\alpha + 1}}}}{\left( {\frac{4}{{\alpha p + 1}}} \right)^{\frac{1}{p}}}\left( {\left| {f'\left( a \right)} \right| + \left| {f'\left( b \right)} \right|} \right). \end{array} $

(20)

证明  在式(19) 中, 取s=1, 即可得到式(20).

注5  在推论5中, 取α=1, 则由式(20) 可得到定理3中的不等式(5).

推论6  令f:[a, b]⊂[0, ∞]→R, 在区间(a, b)上是一个可微映射, a < b, 并且使得f′∈L[a, b].如果对某一个固定的s∈(0, 1], |f′|^p/(p-1)在区间[a, b]上是第2种意义下s-凸的, p>1, 则以下分数次积分不等式成立:

$ \begin{array}{l} \left| {\frac{1}{{b + a}}\int_a^b {f\left( x \right){\rm{d}}x} - f\left( {\frac{{a + b}}{2}} \right)} \right| \le \\ \frac{{b - a}}{{2\left( {s + 1} \right)}}{\left( {\frac{{{2^s}\left( {s + 1} \right)}}{{p + 1}}} \right)^{\frac{1}{p}}}\left( {\left| {f'\left( a \right)} \right| + \left| {f'\left( b \right)} \right|} \right). \end{array} $

(21)

证明  在式(19) 中取α=1, 即可得到式(21).

注6  在推论6中, 如果令s=1, 则由式(21) 也可得到定理3中的不等式(5).

定理9  令f:[a, b]⊂[0, ∞]→R, 在区间(a, b)上是一个可微映射, a < b, 并且使得f′∈L[a, b].如果对某一固定的s∈(0, 1], |f′|^q在区间[a, b]上是第2种意义下s-凹的, p, q>1, 则以下分数次积分不等式成立:

$ \begin{array}{l} \left| {\frac{{\Gamma \left( {\alpha + 1} \right)}}{{{{\left( {b - a} \right)}^\alpha }}}\left[ {{{\left( { - 1} \right)}^\alpha }J_{\frac{{a + b}}{2} - }^\alpha f\left( b \right) - J_{\frac{{a + b}}{2} + }^\alpha f\left( a \right)} \right]} \right. + \\ \left. {\left[ { - 1 + {{\left( { - 1} \right)}^\alpha }} \right]\frac{1}{{{2^\alpha }}}f\left( {\frac{{a + b}}{2}} \right)} \right| \le \\ \frac{{b - a}}{{{2^{\alpha p + 1}}\left( {\alpha p + 1} \right)}}{2^{\frac{{s - 1}}{q}}}\left[ {\left| {f'\left( {\frac{{a + 3b}}{4}} \right)} \right| + \left| {f'\left( {\frac{{3a + b}}{4}} \right)} \right|} \right], \end{array} $

(22)

其中，$\frac{1}{p}+\frac{1}{q}=1$且α∈[0, 1].

证明  由引理2以及Hölder不等式, 可得

$ \begin{array}{l} \left| {\frac{{\Gamma \left( {\alpha + 1} \right)}}{{{{\left( {b - a} \right)}^\alpha }}}\left[ {{{\left( { - 1} \right)}^\alpha }J_{\frac{{a + b}}{2} - }^\alpha f\left( b \right) - J_{\frac{{a + b}}{2} + }^\alpha f\left( a \right)} \right]} \right. + \\ \left. {\left[ { - 1 + {{\left( { - 1} \right)}^\alpha }} \right]\frac{1}{{{2^\alpha }}}f\left( {\frac{{a + b}}{2}} \right)} \right| \le \\ \left( {b - a} \right)\left[ {\int_0^{1/2} {\left| {{t^\alpha }} \right|\left| {f'\left( {ta + \left( {1 - t} \right)b} \right)} \right|{\rm{d}}t + } } \right.\\ \left. {\int_{1/2}^1 {\left| {{{\left( {t - 1} \right)}^\alpha }} \right|\left| {f'\left( {ta + \left( {1 - t} \right)b} \right)} \right|{\rm{d}}t} } \right] \le \\ \left( {b - a} \right)\left[ {{{\left( {\int_0^{1/2} {{t^{\alpha p}}{\rm{d}}t} } \right)}^{\frac{1}{p}}}{{\left( {\int_0^{1/2} {{{\left| {f'\left( {ta + \left( {1 - t} \right)b} \right)} \right|}^q}{\rm{d}}t} } \right)}^{\frac{1}{q}}}} \right] + \\ \left( {b - a} \right)\left[ {{{\left( {\int_0^{1/2} {\left| {{{\left( {t - 1} \right)}^{\alpha p}}} \right|{\rm{d}}t} } \right)}^{\frac{1}{p}}} \times } \right.\\ \left. {{{\left( {\int_{1/2}^1 {{{\left| {f'\left( {ta + \left( {1 - t} \right)b} \right)} \right|}^q}{\rm{d}}t} } \right)}^{\frac{1}{q}}}} \right]. \end{array} $

因为|f′|^q在区间[a, b]上是第2种意义下s-凹的, 利用不等式(7), 可得

$ \int_0^{1/2} {{{\left| {f'\left( {ta + \left( {1 - t} \right)b} \right)} \right|}^q}{\rm{d}}t} \le {2^{s - 1}}{\left| {f'\left( {\frac{{a + 3b}}{4}} \right)} \right|^q}, $

$ \int_{1/2}^1 {{{\left| {f'\left( {ta + \left( {1 - t} \right)b} \right)} \right|}^q}{\rm{d}}t} \le {2^{s - 1}}{\left| {f'\left( {\frac{{3a + b}}{4}} \right)} \right|^q}. $

通过计算, 可得

$ \int_0^{1/2} {{t^{\alpha p}}{\rm{d}}t} = \int_{1/2}^1 {\left| {{{\left( {t - 1} \right)}^{\alpha p}}} \right|{\rm{d}}t} = \frac{1}{{{2^{\alpha p + 1}}\left( {\alpha p + 1} \right)}}. $

由上述不等式, 有

$ \begin{array}{l} \left| {\frac{{\Gamma \left( {\alpha + 1} \right)}}{{{{\left( {b - a} \right)}^\alpha }}}\left[ {{{\left( { - 1} \right)}^\alpha }J_{\frac{{a + b}}{2} - }^\alpha f\left( b \right) - J_{\frac{{a + b}}{2} + }^\alpha f\left( a \right)} \right]} \right. + \\ \left. {\left[ { - 1 + {{\left( { - 1} \right)}^\alpha }} \right]\frac{1}{{{2^\alpha }}}f\left( {\frac{{a + b}}{2}} \right)} \right| \le \\ \frac{{b - a}}{{{2^{\alpha p + 1}}\left( {\alpha p + 1} \right)}}{2^{\frac{{s - 1}}{q}}}\left[ {\left| {f'\left( {\frac{{a + 3b}}{4}} \right)} \right| + \left| {f'\left( {\frac{{3a + b}}{4}} \right)} \right|} \right]. \end{array} $

推论7  在定理9的条件下, 如果|f′|是一个线性映射, 则

$ \begin{array}{l} \left| {\frac{{\Gamma \left( {\alpha + 1} \right)}}{{{{\left( {b - a} \right)}^\alpha }}}\left[ {{{\left( { - 1} \right)}^\alpha }J_{\frac{{a + b}}{2} - }^\alpha f\left( b \right) - J_{\frac{{a + b}}{2} + }^\alpha f\left( a \right)} \right]} \right. + \\ \left. {\left[ { - 1 + {{\left( { - 1} \right)}^\alpha }} \right]\frac{1}{{{2^\alpha }}}f\left( {\frac{{a + b}}{2}} \right)} \right| \le \\ \frac{{b - a}}{{{2^{\alpha p + 1}}\left( {\alpha p + 1} \right)}}{2^{\frac{{s - 1}}{q}}}\left| {f'\left( {a + b} \right)} \right|. \end{array} $

(23)

证明  根据定理9中的式(22), 由于|f′|是一个线性映射, 立即可得式(23)^[13].

注7  在推论7中, 取s=1, α=1, 由于p>1, 则由式(23) 可得到定理4中的式(6).

2 应用举例

文献[14]给出一个s凸函数的例子：令s∈(0, 1), a₁, b₁, c₁∈R，函数f:[0, ∞]→R定义为：

$ f\left( t \right) = \left\{ \begin{array}{l} {a_1},\;\;t = 0,\\ {b_1}{t^s} + {c_1},\;\;\;\;t > 0. \end{array} \right. $

如果b₁≥0, 且0≤c₁≤a₁，则f∈K_s².因此当取a₁=c₁=0，b₁=1时，对于f:[0, 1]→[0, 1](其中f(t)=t^s)，有f∈K_s².

注8  若在本文结论中取[a, b]=[0, 1]，即a=0, b=1, f(x)=x^λ-1(2>λ>1, x∈[0, 1])，结合不完全Beta函数定义可得：

$ \begin{array}{l} \frac{{\Gamma \left( {\alpha + 1} \right)}}{{{{\left( {b - a} \right)}^\alpha }}}{\left( { - 1} \right)^\alpha }J_{\frac{{a + b}}{2} - }^\alpha f\left( b \right) = {\left( { - 1} \right)^\alpha }\alpha \int_1^{\frac{1}{2}} {{{\left( {t - 1} \right)}^{\alpha - 1}}{t^{\lambda - 1}}{\rm{d}}t} = \\ \;\;\;\;\;\alpha \int_0^{\frac{1}{2}} {{x^{\alpha - 1}}{{\left( {1 - x} \right)}^{\lambda - 1}}{\rm{d}}x} = \alpha {B_{\frac{1}{2}}}\left( {\alpha ,\lambda } \right), \end{array} $

其中用到换元1-t=x，并且

$ \frac{{\Gamma \left( {\alpha + 1} \right)}}{{{{\left( {b - a} \right)}^\alpha }}}J_{\frac{{a + b}}{2} + }^\alpha f\left( a \right) = {\left( { - 1} \right)^{\alpha - 1}}\alpha \int_{\frac{1}{2}}^0 {{t^{\alpha + \lambda - 2}}{\rm{d}}t} = \frac{{{{\left( { - 1} \right)}^\alpha }\alpha }}{{\alpha + \lambda - 1}}{\left( {\frac{1}{2}} \right)^{\alpha + \lambda - 1}}. $

因此，有以下命题：

命题1  令2>λ>1, 由注8以及定理6可得不等式：

$ \begin{array}{l} \left| {\alpha {B_{\frac{1}{2}}}\left( {\alpha ,\lambda } \right) + \left[ {\frac{{{{\left( { - 1} \right)}^\alpha }\alpha }}{{\alpha + \lambda - 1}} - 1 + {{\left( { - 1} \right)}^\alpha }} \right]\frac{1}{{{2^{\alpha + \lambda - 1}}}}} \right| \le \\ \left[ {\frac{1}{{{2^{\alpha + s - 1}}\left( {\alpha + s - 1} \right)}} + {B_{\frac{1}{2}}}\left( {\alpha + 1,s + 1} \right)} \right]\left( {\lambda - 1} \right). \end{array} $

注9  按照命题1的方法，由文中其他结论可得到类似的不等式.