在本节中,我们发现了一个Hermite–Hadamard不等式\(tgs\)-通过Katugampola分数积分的凸函数。
定理3.1
让
\(阿尔法>0)和
\(\rho>0\).让
\(\chi:[h_{1}^{\rho},h_{2}^{\ rho}]\子集\mathbb{R}\rightarrow\mathbb{R}\)是一个非负函数
\(0\leqh{1}<h{2}\)和
\(X中的chi^{p}_{c} (h{1}^{\rho},h{2}^{).如果χ也是一个
\(tgs\)-上的凸函数
\([h{1}^{\rho},h{2}^{\ rho}]\),那么以下不等式成立:
$$开始{对齐}和2\chi\biggl(\frac{h{1}^{\rho}+h{2}^{\tho}}{2}\biggr)alpha}{h{1}+}\chi\bigl(h{2}^{\rho}\bigr)+{}^{\ rho}I^{\alpha}_{h_{2}-}\chi\bigl(h{1}^{\rho}\biger)\bigr]\\&\quad\leq\frac{\alpha。\结束{对齐}$$
(5)
证明
让\(在[0,1]\中).考虑\(x,y\在[h{1},h{2}]\中),\(h{1}\geq0\),由定义\(x^{\rho}=r^{\rro}h{1}^{\ρo}+(1-r^{\ρo})h{2}^{\ rho}\),\(y^{\rho}=r^{\rro}h{2}^{\ρo}+(1-r^{\ρo})h{1}^{\ rho}\).自χ是一个\(tgs\)-上的凸函数\([h{1}^{\rho},h{2}^{\ rho}]\),我们有
$$\chi\biggl(\frac{x^{\rho}+y^{\rro}}{2}\biggr)\leq\frac}\chi(x^{\ rho})+\chi$$
然后我们有
$$4\chi\bigl(\frac{h{1}^{\rho}+h{2}^{\rho}}{2}\biggr)\leq\chi\bigl(r^{\rho}h{1}^{\rho}+\bigl(1-r^{\rho}\bigr)h{2}^{\rho}\bigr)+\chi\bigl(r^{\rho}\bigr)h{1}^{\rho}\bigr)$$
(6)
将的两边相乘(6)由\(r^{\alpha\rho-1}\),\(阿尔法>0)然后将所得不等式与第页结束\([0,1]\),我们获得
$$开始{aligned}\frac{4}{\alpha\rho}\chi\biggl(\frac}h{1}^{\rho{+h{2}^{\ rho}}{2}\biggr)\leq{}&\int^{1}_{0}r^{\alpha\rho-1}\chi\bigl^{1}_{0}r^{\alpha\rho-1}\chi\bigl}-h{1}^{\rho}}\biggr)^{\alpha-1}\chi\bigl(g^{\rro}\bigr)\frac{g^{\rho-1}}{h{1}^{\hro}-h{2}^{-h{1}^{\rho}}{h{2}^{\ rho}-h{1}^{\rro}}\biggr)varGamma(\alpha)}{(h{2}^{\rho}-h{1}^{\ rho})^{\alpha}}\bigl[{}^{\\rho}I^{\alpha}_{h{1}+}\chi\bigl_{2}-}\chi\bigl(h{1}^{\rho}\bigr)\bigr]。\结束{对齐}$$
(7)
这建立了第一个不等式。为了证明第二个不等式(5),我们首先观察到,对于\(tgs\)-凸函数χ,我们有
$$\chi\bigl(r^{\rho}h_{1}^{\rro}+\bigle(1-r^{\ rho}\bigr)h_{2}^{\\rho}\ bigr$$
和
$$\chi\bigl(r^{\rho}h_{2}^{\hro}+\bigl$$
通过添加这些不等式,我们得到
$$\chi\bigl(r^{\rho}h_{1}^{\rro}+\bigr(1-r^{\\rho}\biger)h_{2}^{\ rho}\ biger)+\chi\bigl}\bigr)\bigl(\chi\bigle(h{1}^{\rho}\biger)+\chi\bill(h{2}^{\ rho}\ biger)\bigr$$
(8)
将的两边相乘(8)由\(r^{\alpha\rho-1}\),\(阿尔法>0)然后将所得不等式与第页结束\([0,1]\),我们获得
$$\frac{\rho^{\alpha-1}\varGamma(\alpha)}{(h_{2}^{\rho}-h_{1}^{\ rho})^{\alpha}}\bigl[{}^{\\rho}I^{\阿尔法}_{h_{1{+}\chi\bigl_{2}-}\chi\bigl(h{1}^{\rho}\bigr)\bigr]\leq 2\int_{0}^{1} 第页^{\alpha\rho+\rho-1}\bigl(1-r^{\rho}\bigr$$
(9)
自
$$\int美元^{1}_{0}\bigl(r^{\alpha\rho+\rho-1}-r^{\阿尔法\rho+2\rho-1}\bigr)\,dt=\frac{1}{\rho(\alpha+1)(\阿尔法+2)}$$
(9)成为
$$\frac{\rho^{\alpha-1}\varGamma(\alpha)}{(h_{2}^{\rho}-h_{1}^{\ rho})^{\alpha}}\bigl[{}^{\\rho}I^{\阿尔法}_{h_{1{+}\chi\bigl_{2}-}\chi\bigl(h{1}^{\rho}\bigr)\bigr]\leq\frac{2(\chi(h_1}^{\ rho})+\chi$$
(10)
因此(7)和(10)给予(5). □
备注3.1
(1) 通过出租\(\rho\右箭头1\)英寸(5)定理的3.1我们得到了中定理3.1的不等式3.1[18].
(2) 通过让\(\rho\右箭头1\)和\(阿尔法=1)英寸(5)定理的3.1我们得到了中定理2.1的不等式2.2[18].
定理3.2
让
\(阿尔法>0)和
\(\rho>0\).让
\(\chi:[h_{1}^{\rho},h_{2}^{\ rho}]\子集\mathbb{R}\rightarrow\mathbb{R}\)是上的可微非负映射
\(((h{1}^{\rho},h{2}^{)\)具有
\(0\leqh{1}<h{2}\).如果
\(|\chi'|\)是
\(tgs\)-凸的
\([h{1}^{\rho},h{2}^{\ rho}]\),那么下面的不等式成立:
$$\开始{aligned}&\biggl\vert\frac{\chi(h{1}^{\rho})+\chi{2}-\frac{\rho^{\alpha}\varGamma(\alpha+1)}{2(h{2}^{\rho}-h{1}^{\rho})_{2}-}\chi\bigl(h{1}^{\rho}\bigr)\bigr]\biggr\vert\\&\quad\leq\frac{h{2}^{\ rho}-h{1}^{\rro}}{(\alpha+2)(\alfa+3)}\bigl[\bigl\vert\chi'\bigl(h{1\rho}\ bigr rho}\bigr)\bigr\vert\bigr]。\结束{对齐}$$
(11)
证明
发件人(7)一个人可以拥有
$$开始{对齐}和\frac{\rho^{\alpha-1}\varGamma(\alpha_{2}-}\chi\bigl(h{1}^{\rho}\bigr)\bigr]\\&\quad=\int^{1}_{0}r^{\alpha\rho-1}\chi\bigl(r^{\rho}h{1}^{\rho}+\bigl(1-r^{\rho}\bigr)h{2}^{\rho}\bigr)\,dr+\int^{1}_{0}r^{\alpha\rho-1}\chi\bigl$$
(12)
通过按部分积分,我们得到
$$开始{对齐}和\frac{chi(h{1}^{\rho})+\chi(h{2}^{\rho}}+}\chi\bigl(h{2}^{\rho}\bigr)+^{\rro}I^{\alpha}_{h_{2}-}\chi\bigl(h{1}^{\rho}\bigr)\bigr]\\&\quad=\frac{h{2}^{\ rho}-h{1}^{\rro}}{\alpha}\int_{0}^{1} 对^{\rho(\alpha+1)-1}\bigl[\chi'\bigl(r^{\rho}h{2}^{\rro}+\bigr \,dr.\end{对齐}$$
(13)
通过使用三角形不等式和\(tgs\)-的凸性\(|\chi'|\),我们获得
$$\beign{aligned}&\biggl\vert\frac{\chi(h{1}^{\rho})+\chi(h{2}^{\rho})}{\alpha\rho}-\frac{\rho ^{\alpha-1}\varGamma(\alpha)}{(h{2}^{\rho}-h{1}^{\rho})^{\alpha}bigr)+{}^{\rho}I^{\alpha}_{h_{2}-}\chi\bigl(h{1}^{\rho}\bigr)\bigr]\biggr\vert\\&\quad\leq\frac{h{2}^{\ rho}-h{1}^{\rro}}{\alpha}\int_{0}^{1} 第页^{\rho(\alpha+1)-1}\bigl\vert\chi'\bigl \vert\,dr\\&\quad\leq\frac{h{2}^{\rho}-h{1}^{\ rho}}{\alpha}\int_{0}^{1} 第页^{\rho(\alpha+1)-1}\bigl[\chi'\bigl(r^{\rho}h{2}^{\rro}+\bigr(1-r^{\\rho}\biger)h_{1}^{\ rho}\ bigr)+\chi'\ bigl \,dr\\&\quad=\frac{2(h{2}^{\rho}-h{1}^{\ rho})}{\alpha}\int_{0}^{1} 第页^{\rho(\alpha+1)-1}r^{\rho}\bigl}^{\rho} . \结束{对齐}$$
(14)
将上述不等式的两边乘以\(\压裂{\alpha\rho}{2}\),我们得到了所需的不等式(11).□
推论3.3
考虑定理的类似假设3.2.
1如果
\(\rho=1\),然后
$$\begin{aligned}&\biggl\vert\frac{\chi(h{1})+\ch(h{2})}{2}-\frac{\varGamma(\alpha+1)}{2(h_{2} -小时_{1} )^{\alpha}}\bigl[J^{\alpha}_{h{1}+}\chi(h{2})+J^{\alpha}_{h_{2}-}\chi(h{1})\bigr]\biggr\vert\\&\quad\leq\frac{h_{2} -小时_{1} }{(\alpha+2)(\alfa+3)}\bigl[\bigl\vert\chi'(h{1})\bigr\vert+\bigl\ vert\chi`(h{2})\ bigr\ vert\bigr]。\结束{对齐}$$
(15)
2如果
\(\rho=\alpha=1\),然后
$$\开始{aligned}\biggl\vert\frac{\chi(h{1})+\chi{2}-\压裂{1}{h_{2} -小时_{1} }\int _{h{1}}^{h{2}}\chi(g)\,dg\biggr\vert\leq\frac{h_{2} -小时_{1} }{12}\bigl[\bigl\vert\chi'(h{1})\bigr\vert+\bigl\fort\chi'(h2})\ bigr\verst\bigr]。\结束{对齐}$$
(16)
为了获得更多结果,我们需要以下引理,在中也得到了证明[11].
引理3.1
([11])
让
\(阿尔法>0)和
\(\rho>0\).让
\(\chi:[h{1}^{\rho},h_{2}^{\rho}]\subet \mathbb){右}_{+}=[0,\infty)\rightarrow\mathbb{R}\)是上的可微映射
\(((h{1}^{\rho},h{2}^{)\)具有
\(0\leqh{1}<h{2}\).如果存在分数积分,则以下等式成立:
$$\开始{对齐}&\frac{\chi(h{1}^{\rho})+\chi{2}-\裂缝{\rho^{\alpha}\varGamma(\alpha+1)}{2_{2}-}\chi\bigl(h{1}^{\rho}\bigr)\bigr]\\&\quad=\frac{\rho}\chi'\bigl(r^{\rho}h{1}^{\rro}+\bigl-(1-r^{\ rho}\bigr)h{2}^{\ρ}\biger)\,dr.\end{aligned}$$
(17)
证明
通过使用中引理2的证明中的类似参数[15]。首先考虑
$$开始{对齐}&\int_{0}^{1}\bigl(1-r^{\rho}\bigr(r^{\rho}h{1}^{\rro}+(1-r^{\ rho})h{2}^{)}{\rho^{\rho}-h{2}^{\rho}}\int_{0}^{1}\bigl{2}^{\rho}}{h{2}^{\rho}-h{1}^{\\rho}}\biggr)^{\alpha-1}\cdot\frac{g^{\rho-1}{h{1}^{\rro}-h_2}^{(h{{\rho})}-\frac{\rho ^{\alpha-1}\varGamma(\alpha+1)}{(h{2}^{\rho2}-h_{1}^{)^{\α+1}}\cdot{}^{\ rho}I{h_{2}-}^{\alpha}\chi\bigl(g^{\rho}\bigr)\bigg|{g=h{1}}。\结束{对齐}$$
(18)
同样,我们可以证明
$$\开始{aligned}&\int_{0}^{1} 第页^{\rho\alpha}\cdot r^{\rho-1}\chi'\bigl)}+\frac{\rho^{\alpha-1}\varGamma(\alpha+1)}{(h{2}^{\rho}-h{1}^{\rho})^{\alpha+1}}\cdot{}^{\ rho}I{h{1}+}^{alpha}\chi\bigl(g^{\rro}\bigr)\bigg{g=h{2}}。\结束{对齐}$$
(19)
因此,从(18)和(19)我们得到(17). □
定理3.4
让
\(\alpha>0\)和
\(\rho>0\).让
\(\chi:[h{1}^{\rho},h{2}^{\ rho}]\子集\mathbb{右}_{+}\rightarrow\mathbb{R}\)是上的可微非负映射
\(((h{1}^{\rho},h{2}^{)\)这样的话
\(在L_{1}[h{1},h{2}]\中为chi'\)具有
\(0\leqh{1}<h{2}\).如果
\(|\chi'|^{q}\)是
\(tgs\)-凸的
\([h{1}^{\rho},h{2}^{\ rho}]\)对于一些固定的
\(q \ geq 1 \),那么下面的不等式成立:
$$\开始{aligned}&\biggl\vert\frac{\chi(h{1}^{\rho})+\chi{2}-\frac{\rho^{\alpha}\varGamma(\alpha+1)}{2(h{2}^{\rho}-h{1}^{\rho})_{2}-}\chi\bigl(h{1}^{\rho}\bigr)\bigr]\biggr\vert\\&\quad\leq\frac{(h_2}^{\ rho}-h_1}^{)}{2}\biggl(\ frac{2}{\alpha+1}\bighr)^{1-1/q}\\&\quad{}\times\biggl{(alpha+2)(\alpha+3)}\biggr]\bigl[\bigl\vert\chi'\bigl(h{1}^{\rho}\bigr)\bigr\vert^{q}+\bigl\ vert\chi'\bigl(h{2}^{\rho}\biger)\bigr\vert^{q}\bigr]\biggr)^{1/q}。\结束{对齐}$$
(20)
证明
使用引理3.1幂平均不等式和\(tgs\)-的凸性\(|\chi'|^{q}\),我们获得
$$开始{对齐}和\bigl\vert I{\chi}(\alpha,\rho,h{1},h{2})\bigr\vert\\&\quad=\biggl\vert\frac{\rho(h{2{{\rho}-h{1}^{\rhoS})}{2}\int_{0}^{1}\bigl\\bigl(1-r^{\rro}\bigr)^\alpha}-\bigl{\rho}\bigr)^{\alpha}\biger\}r^{\rho-1}\chi'\bigl(r^{\ rho}h{1}^{\hro}+\bigl-(1-r^{\\rho}\ bigr,dr\biggr\vert\\&\quad\leq\frac{\rho(h{2}^{\rho}-h_{1}^{\\rho})}{2}\biggl(int_{0}^{1}\bigl\vert\bigl(1-r^{\rro}\bigr)^{\alpha}-\bigl{1-1/q}\\&\qquad{}\times\biggl(\int_{0}^{1}\bigl\vert\bigl(1-r^{\rho}\bigr)^{\alpha}-\bigl\bigr\vert r^{\rho-1}\bigl\vert\chi'\bigl})}{2}\biggl(\int_{0}^{1}\bigl\{\bigl(1-r^{\rho}\bigr)^{\alpha}+\bigle(r^{\sho}\biger)^{\ alpha}\biger\}r^{\rho-1}\,dr\biggr)^{1-1/q}\\&\qquad{}\times\biggl(\int_{0}^{1}\bigl\{\bigl \bigr)\bigr\vert^{q}+\bigl\vert\chi'\bigl(h{2}^{\rho}\biger)\biger\vert|{q}\bigr]\,dr\biggr)^{1/q}。\结束{对齐}$$
(21)
通过使用变量的变化\(t=r^{\rho}\),我们得到
$$\开始{对齐}&\int_{0}^{1}\bigl\{\bigl(1-r^{\rho}\bigr)^{\alpha}+\bigle(r^{\sho}\biger)^{\ alpha}\biger\}r^{\rho-1}\,dr\\&\quad=\int_}0}^}\bigl(1-r ^{\rro}\bigr){0}^{1}\bigl(r^{\rho}\bigr)^{\alpha}r^{\sho-1}\,dr\\&\quad=\frac{2}{\rho(\alpha+1)},\end{aligned}$$
(22)
$$开始{对齐}&\int_{0}^{1}\bigl\{\bigl r)^{\alpha}r^{\rho-1}r^}\rho}\bigl(1-r^{\sho}\bigr)\,dr+\int_{0}^{1}\bigle(r^{\ rho}\ bigr\bigl(1-r^{\rho}\bigr)\,dr\\&\quad=\frac{1}{\rho}\beta(2,\alpha+2)+\frac}{\rro(\alpha+2)(\alfa+3)}。\结束{对齐}$$
(23)
因此使用(23)和(22)英寸(21)我们得到(20). □
推论3.5
考虑定理的类似假设3.4.
1如果
\(\rho=1\),然后
$$\begin{aligned}&\biggl\vert\frac{\chi(h{1})+\ch(h{2})}{2}-\frac{\varGamma(\alpha+1)}{2(h_{2} -小时_{1} )^{\alpha}}\bigl[J^{\alpha}_{h{1}+}\chi(h{2})+J^{\alpha}_{h_{2}-}\chi(h{1})\bigr]\biggr\vert\\&&quad\leq\frac{(h_{2} -小时_{1} )}{2}\biggl(\frac{2}{\alpha+1}\bigr)^{1-1/q}\\&\qquad{}\times\biggl(\biggl[\beta(2,\alpha+2)+\frac{1}{(\alpha+2)(\alfa+3)}\bighr]\bigl[\bigl\vert\chi'(h{1})\bigr\vert^{q}+\bigl\ vert\chi(h{2})^{q}\bigr]\biggr)^{1/q}。\结束{对齐}$$
(24)
2如果
\(\rho=\alpha=1\),然后
$$\begin{aligned}&\biggl\vert\frac{\chi(h{1})+\ch(h{2})}{2}-\压裂{1}{h_{2} -小时_{1} }\int_{h{1}}^{h{2}}\chi(g)\,dg\biggr\vert\&\quad\leq\frac{(h_{2} -小时_{1} )}{2}\biggl(\frac{2(\vert\chi'(h{1})\vert^{q}+\vert\chi'(h}2})\ vert^})}{3}\bigr)^{1/q}。\结束{对齐}$$
(25)
定理3.6
让
\(阿尔法>0)和
\(\rho>0\).让
\(\chi:[h{1}^{\rho},h{2}^{\ rho}]\子集\mathbb{右}_{+}\rightarrow\mathbb{R}\)是上的可微非负映射
\(((h{1}^{\rho},h{2}^{)\)这样的话
\(在L_{1}[h{1},h{2}]\中为chi'\)具有
\(0\leq h{1}<h _{2}\).如果
\(|\chi'|^{q}\)是
\(tgs\)-凸的
\([h{1}^{\rho},h{2}^{\ rho}]\)对于一些固定的
\(q \ geq 1 \),那么下面的不等式成立:
$$\开始{aligned}&\biggl\vert\frac{\chi(h{1}^{\rho})+\chi{2}-\裂缝{\rho^{\alpha}\varGamma(\alpha+1)}{2_{2}-}\chi\bigl(h{1}^{\rho}\bigr)\bigr]\biggr\vert\\&\quad\leq\frac{(h_{2}^{\ rho}-h_{1}^{\rro})}{2}\biggl ^{\rho}\bigr)\bigr\vert^{q}+\bigl\vert\chi'\bigl(h{2}^{\rro}\bigr)\ bigr\vert^{q}\biger]\biggr)^{1/q}。\结束{对齐}$$
(26)
证明
使用引理3.1幂平均不等式和\(tgs\)-的凸性\(|\chi'|^{q}\),我们获得
$$\开始{aligned}&\biggl\vert\frac{\chi(h{1}^{\rho})+\chi{2}-\frac{\rho^{\alpha}\varGamma(\alpha+1)}{2(h{2}^{\rho}-h{1}^{\rho})_{2}-}\chi\bigl(h{1}^{\rho}\bigr)\bigr]\biggr\vert\\&\quad=\biggl\vert\frac{\rho(h{2}^{\ rho}-h_{1}^{\rro})}{2}\int_{0}^{1}\bigl\{\bigr\ \bigr\}r^{\rho-1}\chi'\bigl(r^{\rho}h{1}^{\hro}+\bigl-(1-r^{\ rho}\bigr)h{2}^{\,dr\biggr\vert\\&\quad\leq\frac{\rho(h_2}^{\rho}-h_{1}^{\\rho})}{2}\biggl(\int_{0}^{1}r^{\rro-1}\,dr\bigr)^{1-1/q}\\&\qquad\}times\biggl ^{\alpha}-\bigl(r^{\rho}\bigr)^{\rho}\biger)\bigr\vert^{q}\,dr\biggr bigl(1-r^{\rho}\bigr)^{\alpha}+\bigl\bigl[\bigl\vert\chi'\bigl(h{1}^{\rho}\bigr)\bigr\vert^{q}+\bigl\fort\chi'\bigl(h{2}^{\ rho}\biger)\bigrar\vert^}\biger]\,dr\biggr)^{1/q}。\结束{对齐}$$
(27)
由于使用变量的变化\(t=r^{\rho}\),我们得到
$$开始{对齐}&\int_{0}^{1}\bigl\{\bigl r)^{\alpha}r^{\rho-1}r^}\rho}\bigl(1-r^{\sho}\bigr)\,dr+\int_{0}^{1}\bigle(r^{\ rho}\ bigr\bigl(1-r^{\rho}\bigr)\,dr\\&\quad=\frac{1}{\rho}\beta(2,\alpha+2)+\frac}{\rro(\alpha+2)(\alfa+3)}。\结束{对齐}$$
(28)
因此使用(28)英寸(27)我们得到(26). □
推论3.7
考虑定理的类似假设3.6.如果
\(\rho=1\),然后
$$\begin{aligned}&\biggl\vert\frac{\chi(h{1})+\ch(h{2})}{2}-\裂缝{varGamma(\alpha+1)}{2(h_{2} -小时_{1} )^{\alpha}}\bigl[J^{\alpha}_{h{1}+}\chi(h{2})+J^{\alpha}_{h_{2}-}\chi(h{1})\bigr]\biggr\vert\\&\quad\leq\frac{(h_{2} -小时_{1} )}{2}\biggl(\biggl[\beta(2,\alpha+2)+\frac{1}{(\alpha/2)(\alfa+3)}\bigr]\bigl[\bigl\vert\chi'(h{1})\bigr\vert^{q}+\bigl\ vert\chi(h{2})\ bigr\vert^{q}\bigr)^{1/q}。\结束{对齐}$$
(29)
定理3.8
让
\(chi{1}),\(chi{2})具有实际价值,对称于
\(压裂{h{1}^{\rho}+h{2}^{\ rho}}{2}),非负和
\(tgs\)-凸函数
\([h{1}^{\rho},h{2}^{\ rho}]\),哪里
\(\rho>0\).然后,为所有人
\(h{1},h{2}>0\)和
\(阿尔法>0),我们有
$$\frac{\rho^{\alpha}\^{\rho}I^{\alpha}_{h{1}+}1}^{\rho},h_{2}^{\tho})+N(h_{1}^},h2}^{)]}{\varGamma(\alpha+5)}$$
(30)
和
$$开始{对齐}和8\chi_{1}\biggl(\frac{h_{1{^{\rho}+h_{2}^{\rro}}{2}\bigr{α}{h{1}+},h_{2}^{\rho})+N(h_{1}^{\tho},h_}2}^})]}{\varGamma(\alpha+5)},\end{aligned}$$
(31)
哪里
\(M(h{1}^{\rho},h{2}^{\rho{)=\chi{1}和
\(N(h{1}^{\rho},h{2}^})=\chi{1}-(h{1})\chi{2}-(h{2})+\chi{1}-(h2}).
证明
自\(chi{1})和\(chi{2})是\(tgs\)-凸函数\([h{1},h{2}]\),我们可以
$$\chi_{1}\bigl(r^{\rho}h{1}^{\rro}+\bigr(1-r^{\\rho}\biger)h{2}^{\ rho}\ bigr)\leqr^{\ρo}\bigl \rho}\bigr)$$
和
$$\chi_{2}\bigl(r^{\rho}h_{1}^{\rro}+\bigl \rho}\biger)\bigr)$$
根据以上内容,我们获得
$$\begon{aligned}&\chi{1}\bigl(r^{\rho}h{1}^{\rho}+\bigl(1-r^{\rho}\bigr)h{2}^{\rho}\bigr)\chi{2}\bigl(r^{\rho}h{1}^{\rho}+\bigl(1-r^{\rho}\bigr)h{2}^{\rho}\bigl(1-r^{\ rho}\bigr)^{2}\bigl(\chi{1}\bigl(h{1}^{\rho}\bigr)+\chi{1}\bigl(h{2}^{\rho}\bigr)\bigl(\chi{2}\bigl(h{1}^{\rho}\biger)+\chi{2}\bigl(h{2}^{\rho}\bigr)。\结束{对齐}$$
(32)
将的两边相乘(32)由\(\frac{r^{\alpha\rho-1}}{\varGamma(\alpha)}\),\(阿尔法>0)然后将所得不等式与第页结束\([0,1]\),我们获得
$$\开始{aligned}&\frac{1}{\varGamma(\alpha)}\int_{0}^{1} 第页^{\alpha\rho-1}\chi_{1}\bigl&\quad\leq\frac{(\chi_1}(h_1}^{\rho})+\chi_1{(h_2}^{))^{1} 第页^{2\rho}\bigl(1-r^{\rho}\bigr)^{2}\,dr\end{aligned}$$
(33)
通过变量的变化\(t=r^{\rho}\),我们得到
$$\int_{0}^{1} 第页^{2\rho}\bigl(1-r^{\rho}\ bigr)^{2}\,dr=\frac{2\alpha(\alpha+1)}{\rho \varGamma(\alpha+5)}$$
(34)
也通过出租\(x^{\rho}=r^{\rro}h{1}^{\ρo}+(1-r^{\ρo})h{2}^{\ rho}\),我们获得
$$\开始{aligned}&\frac{1}{\varGamma(\alpha)}\int_{0}^{1} 第页^{\alpha\rho-1}\chi_{1}\bigl&\quad=\frac{\rho^{\alpha-1}\^{\rho}I^{\alpha}{h{1}+}。\结束{对齐}$$
(35)
因此从(33)–(35),我们得到(30).
再次使用\(tgs\)-的凸性\(chi{1})和\(chi{2})在\([h{1}^{\rho},h{2}^{\ rho}]\),我们发现
$$\开始{对齐}和\chi_{1}\biggl(\frac{h{1}^{\rho}+h{2}^{\ rho}}{2}\bigr h{1}^{\rho}+(1-r^{\rro})h{2}^{\ rho}{2}+\frac{r^{\ρo}h{2{^{\ρo}+\biggl(\frac{r^{\rho}h{1}^{\rro}+(1-r^{\sho})h{2}^{\\rho}}{2}+\ frac{r ^{\ρ}h{2{{\rho}+(1-r^{\ rho})h _{1}^{\r o}}{2}\biggr)\\&\quad\leq\frac}{1}{4}\bigl[\chi_{1}\bigl\\&\qquad{}\ times\frac{1}{4}\bigl[\chi_{2}\bigr ^{\rho}\biger)\bigr]\\&\quad=\frac{1}{16}\bigl[\chi_{1}\bigle^{\rho}+\bigl r^{\rho}\bigr)h{1}^{\rro}\\&\qquad{}+\chi_{1}\bigl+\bigl(1-r^{\rho}\bigr)h{1}^{\rro}\\&\qquad{}+\chi_{1}\bigl.ρ}\bigr)h{2}^{\rho}\biger)\bigr]。\结束{对齐}$$
(36)
将的两边相乘(36)由\(\frac{r^{\alpha\rho-1}}{\varGamma(\alpha)}\),\(阿尔法>0)然后将所得不等式与第页结束\([0,1]\),我们获得
$$\begin{aligned}&&frac{1}{\rho\varGamma(\alpha+1)}\chi{1}\bigl(\frac{h{1}^{\rho}+h{2}^{\rho}{2}\biggr)\chi{2}\biggl(\frac{h{1}^{\rho}+h{2}^{\rho}}}{2}\biggr)&&quad\leq\frac{1}{16 \varGamma(\alpha)}\biggl[\int _{0}^{1} 对^{\alpha\rho-1}\chi_{1}\bigl(r^{\rho}h{1}^{\rro}+\bigr(1-r^{\\rho}\biger)h{2}^{\ rho}\ bigr)\chi_2}+\int_{0}^{1} 第页^{\alpha\rho-1}\chi_{1}\bigl(r^{\rho}h{2}^{\rro}+\bigr(1-r^{\ rho}\biger)h{1}^{\ρo}\bigr)\chi_2}+\int_{0}^{1} 第页^{\alpha\rho-1}\chi_{1}\bigl(r^{\rho}h{1}^{\rro}+\bigr(1-r^{\\rho}\biger)h{2}^{\ rho}\ bigr)\chi_2}+\int_{0}^{1} 第页^{\alpha\rho-1}\chi_{1}\bigl克]。\结束{对齐}$$
那就是,
$$\开始{对齐}和8\chi_{1}\biggl(\frac{h{1}^{\rho}+h{2}^{\ rho}}{2}\bigr)\chi_2}\bigl{2}^{\rho}-h_{1}^{\\rho})^{\alpha}}\bigl(h{2}^{\rho}\bigr)\chi_{2}\bigl chi{1}\bigl(h{1}^{\rho}\bigr)\chi{2}\bigle。\结束{对齐}$$
经过一些计算,我们得到了所需的不等式(31). □
备注3.2
1.通过出租\(\rho=1\)在定理中3.8不平等(30)和(31)给出不等式\((3.11)\)和\((3.12)\)分别在的定理3.2中[18].
2.通过出租\(\rho=\alpha=1\)在定理中3.8不平等(30)成为定理中的不等式\((2.2)\)第页,共页[18].