本文的目的是\((p,q)\)-一些经典积分不等式的类比。特别是,我们发现\((p,q)\)-的不等式的推广[2]以及一些涉及泰勒余数的新不等式[1].
定理2.1
假设
\(0<q<p \leq 1),\(测试版>0),\(n\in\mathbb{n}\)和F类,G公司定义为两个函数
\(F,G:[\alpha,\beta]\to\mathbb{R}\)具有
\(α=βφ{n}).让我们继续
\([\alpha,\beta]\)F类是
\((p,q)\)-减少和
\(0\leq G\leq 1\).此外,假设我们有任何数字
\(k,\ell\in \{0,1,2,\ldots\}\)和功能F类和G公司是这样的
$$\开始{对齐}&p\psi_{k}-\α\leq\int_{\alpha}^{\beta}G(x)\,\mathrm{d}_{p,q}x\leq\beta-p\psi_{\ell}\quad\textit{if}F\geq0\textit{on}[\alpha,\beta],\\&\beta-p\psi{\ell{leq\int_{\alpha}^{\beta}G(x),\mathrm{d}_{p,q}x\leqp\psi_{k}-\alpha\quad\textit{if}F\leq0\textit{on}[\alpha,\beta]。\结束{对齐}$$
然后
$$\int_{\psi_{\ell}}^{\beta}F(x)\,\mathrm{d}_{p,q}x\leq\int_{\alpha}^{\beta}F(x)G(x)\,\mathrm{d}_{p,q}x\leq\int_{alpha}^{psi_{k}}F(x)\,\mathrm{d}_{p,q}x$$
(2.1)
证明
我们证明了F类一\((p,q)\)-减少函数,当\(F\geq 0)以及\(F\leq 0\)。对于\(F\geq 0),我们只证明了左不等式(2.1). 我们有\(\ell\in\mathbb{N}\cup{0\}\).接受\(j=0,1,\ldots,\ell-1).在以下情况下\(F\geq 0),我们有\(F(\psi_{ell})\leq F(\psi _{j})\)对于\(psi{j}\leq\psi{ell}\)(\(j=0,1,\ldots,n-\ell\))和\(F(\psi_{\ell})\geq F(\psi _{\ell+j})\)对于\(psi{\ell}\leq\psi{\ell+j}\)同样,在这种情况下\(F\leq 0\),我们有\(F(\psi_{ell})\geq F(\psi _{j})\)对于\(psi{j}\geq\psi{ell}\)和\(F(\psi_{\ell})\leq F(\psi _{\ell+j})\)对于\(\psi_{\ell}\geq\psi_{\ell+j}\)。证明很简单,所以我们省略了它。现在
$$\开始{aligned}&\int_{\alpha}^{\beta}F(x)G(x)\,\mathrm{d}_{p,q}x-\int_{\psi_{\ell}}^{\beta}F(x)\,\mathrm{d}_{p,q}x\\&\quad=\int_{\alpha}^{\psi_{\ell}}F(x)G(x)\,\mathrm{d}_{p,q}x+\nint_{\psi_{\ell}}^{\beta}F(x)G(x)\,\mathrm{d}_{p,q}x-\nint_{\psi_{\ell}}^{\beta}F(x)\,\mathrm{d}_{p,q}x\\&\quad=\int_{\alpha}^{\psi_{\ell}}F(x)G(x)\,\mathrm{d}_{p,q}x-\int_{\psi_{\ell}}^{\beta}F(x)\bigl(1-G(x)\ bigr)\,\mathrm{d}_{p,q}x\\&\quad=\int_{\alpha}^{\psi_{\ell}}F(x)G(x)\,\mathrm{d}_{p,q}x-(p-q)\beta\sum_{j=0}^{\ell-1}\frac{q^{j}}{p^{j+1}}F(\psi_{j})\bigl(1-G(\psi.{j})\bigr)\&\quad\geq\int_{alpha}^{psi_}{\ell}}F(x)G(x)\,\mathrm{d}_{p,q}x-(p-q)\beta\sum_{j=0}^{\ell-1}\frac{q^{j}}{p^{j+1}}F(\psi{\ell})\bigl(1-G(\psi.{j})\ biger){d}_{p,q}x-(p-q)βF(\psi_{\ell})}F(x)G(x)\,\mathrm{d}_{p,q}x-F(psi{ell}){d}_{p,q}x\\&\quad\geq\int_{\alpha}^{\psi_{\ell}}F(x)G(x)\,\mathrm{d}_{p,q}x-F(\psi_{\ell})\int_{\alpha}^{\beta}G(x)\,\mathrm{d}_{p,q}x+F(\psi_{\ell})\int_{\psi_{\ellneneneep}^{\beta}G(x)\,\mathrm{d}_{p,q}x\\&\quad=\int_{\alpha}^{\psi_{\ell}}F(x)G(x)\,\mathrm{d}_{p,q}x-F(\psi_{\ell})\biggl[\int_{\alpha}^{\beta}G(x)\,\mathrm{d}_{p,q}x-\nint_{\psi_{\ell}}^{\beta}G(x)\,\mathrm{d}_{p,q}x\biggr]\\&\quad=\int_{\alpha}^{\psi_{\ell}}F(x)G(x)\,\mathrm{d}_{p,q}x-F(\psi_{\ell})\ int _{\alpha}^{\psi_{\ell}}G(x)\,\mathrm{d}_{p,q}x\\&\quad=\int_{\alpha}^{\psi_{\ell}}\bigl(F(x)-F(\psi_{\ell{)\bigr)G(x)\,\mathrm{d}_{p,q}x\\&\quad=(p-q)\psi_{ell}\sum_{j=0}^{n-\ell-1}\frac{q^{j}}{p^{j+1}}\biggl[F\biggl(\psi_{ell}\frac{q^}}{p ^{j+1}}\biggr)-F{p^{j+1}}\biggr)\\&\quad=(p-q)\psi_{ell}\sum_{j=0}^{n-\ell-1}\frac{q^{j}}{p^}j+1}{bigl[F)\geq 0。\结束{对齐}$$
□
定理2.2
对于
\(0<q<p \leq 1),\(n\in\mathbb{n}\),我们有以下积分恒等式:
$$开始{aligned}&\int_{alpha}^{beta}\bigl(D_{p,q}^{mu+1}f\bigr)(x)\frac{(\gamma_{k} -qx)_{p,q}^{\mu+1}}{[\mu+1]{p,q}!}\,\马特姆{d}_{p,q}x\\&\quad=\int_{alpha}^{\gamma_{k}}R{(p,q),\mu,f}(\alpha,x)\,\mathrm{d}_{p,q}x+\int_{\gamma_{k}}^{\beta}R{(p,q),\mu,f}(\beta,x)\,\mathrm{d}_{p,q}x,\结束{对齐}$$
(2.2)
哪里μ是以下任意整数
\(\mu\geq-1),和
\(R_{(p,q),-1,f}(α,x)=f(x)).
此外,
$$\begin{aligned}&\int_{\alpha}^{\beta}\bigl(D_{p,q}^{\tu+1}f\bigr)(x)\frac{(v-qx){p,q}^{\t u+1}}{[\mu+1]{p,q}!}=\int_{\alpha}^{\beta}R_{(p,q),\mu,f}(\alpha,x),\end{aligned}$$
(2.3)
$$\begin{aligned}&\int_{\alpha}^{\beta}\bigl(D_{p,q}^{\tu+1}f\bigr)(x)\frac{(u-qx)_{p、q}^}{\tu+1}}{[\mu+1]{p,q}!}=\int_{\alpha}^{\beta}R{(p,q),\mu,f}(\beta,x)。\结束{对齐}$$
(2.4)
证明
我们用归纳法证明它。我们证明了这一点\(\mu=-1\)也就是说,\(\int_{\alpha}^{\beta}f(x)\,\mathrm{d}_{p,q}x=\int_{\alpha}^{\gamma_{k}}f(x)\,\mathrm{d}_{p,q}x+\int_{\gamma_{k}}^{\beta}f(x)\,\mathrm{d}_{p,q}x\)。假设这是真的μ.然后我们证明\(\mu+1\).
通过各部分的集成,我们
$$\开始{aligned}&\int_{alpha}^{beta}\bigl(D_{p,q}^{mu}f\bigr)(x)\frac{(\gamma_{k} -qx)_{p,q}^{\mu}}{[\mu]{p,q}!}\,\马特姆{d}_{p,q}x\\&\quad=-\frac{1}{[\mu+1]{p,q}!}\int_{\alpha}^{\beta}\bigl(D_{p,q}^{f\bigr)(px)D_{p,q}(\gamma_{k} -qx)_{p,q}^{r+1}\,\mathrm{d}_{p,q}x\\&\quad=-\frac{1}{[\mu+1]{p,q}!}\biggl[\int_{\alpha}^{\beta}\bigl(D_{p,q}^{f\bigr)(v)(\gamma_{k} -v型)_{p,q}^{\mu+1}-\bigl(D_{p,q}^{\ mu}f\bigr)(u)(\gamma_{k} -u个)_{p,q}^{\mu+1}\\&\qquad{}-\int_{\alpha}^{\ beta}\bigl(D_{p,q}^{\tu+1}f\bigr)(x)(\gamma_{k} -qx)_{p,q}^{\mu+1}\,\mathrm{d}_{p,q}x\biggr]。\结束{对齐}$$
因此我们有
$$\begin{collected}[b]\int_{alpha}^{beta}(D_{p,q}^{mu+1}f)(x)\frac{(\gamma_{k} -qx)_{p,q}^{\mu+1}}{[\mu+1]{p,q}!}\,\马特姆{d}_{p,q}x\\\quad=\bigl(D_{p,q}^{\mu}f\bigr)(\beta)\ frac{(\gamma_{k} -v型)_{p,q}^{\mu+1}}{[\mu+1]{p,q}!}-\bigl(D_{p,q}^{\mu}f\bigr)(\alpha)\压裂{(\gamma_{k} -u个)_{p,q}^{\mu+1}}{[\mu+1]{p,q}!}\\\qquad{}+\int_{\alpha}^{\beta}\bigl(D_{p,q}^{f\bigr)(x)\frac{(\gamma_{k} -qx)_{p,q}^{\mu}}{[\mu]{p,q}!}\,\马特姆{d}_{p,q}x。\结束{已收集}$$
(2.5)
发件人(2.2),我们有
$$\boot{collected}[b]\int _{\alpha}^{\beta}(D_{p,q}^{\mu}f)(x)\frac{(\gamma_{k} -qx)_{p,q}^{\mu}}{[\mu]{p,q}!}\,\马特姆{d}_{p,q}x\\\quad=\int_{\alpha}^{\gamma_{k}R_{(p,q),\mu-1,f}(\alpha,x)\,\mathrm{d}_{p,q}x+\nint_{\gamma_{k}}^{\beta}R_{(p,q),\mu-1,f}(\beta,x)\,\mathrm{d}_{p,q}x。\结束{已收集}$$
(2.6)
发件人(2.5)和(2.6),我们有
$$\begin{collected}[b]\int_{alpha}^{beta}(D_{p,q}^{mu+1}f)(x)\frac{(\gamma_{k} -qx)_{p,q}^{\mu+1}}{[\mu+1]{p,q}!}\,\马特姆{d}_{p,q}x\\\quad=(D_{p,q}^{\mu}f)(v)\压裂{(\gamma_{k}-\β){p,q}^{\mu+1}}{[\mu+1]{p,q}!}-(D_{p,q}^{mu}f)(α)压裂{(γ_{k}-\α){p,q}^{\mu+1}}{[\mu+1]{p,q}!}\\\qquad{}+\int_{\alpha}^{\gamma_{k}}R{(p,q),\mu-1,f}(\alpha,x)\,\mathrm{d}_{p,q}x+\nint_{\gamma_{k}}^{\mathrm{beta}}R{(p,q),\mu-1,f}(\beta,x)\,\mathrm{d}_{p,q}x。\结束{已收集}$$
(2.7)
我们知道\((p,q)\)-积分
$$\int_{\alpha}^{\gamma_{k}}\frac{(x-\alpha){p,q}^{mu}}{[\mu]{p,q}!}=\压裂{(\gamma_{k}-\alpha){p,q}^{\mu+1}}{[\mu+1]{p,q}!},\qquad\int{\gamma{k}}^{\beta}\frac{(x-\beta){p,q}^{mu}}{[\mu]{p,q}!}=-\压裂{(\gamma_{k}-\β){p,q}^{\mu+1}}{[\mu+1]{p,q}!}$$
(2.8)
因此(2.7)意味着
$$开始{aligned}&\int_{alpha}^{beta}\bigl(D_{p,q}^{mu+1}f\bigr)(x)\frac{(\gamma_{k} -qx)_{p,q}^{\mu+1}}{[\mu+1]{p,q}!}\,\马特姆{d}_{p,q}x\\&\quad=\int_{\alpha}^{\gamma_{k}}\biggl[R_{(p,q),\mu-1,f}(\alpha,x)-\bigl(D_{p,q}^{\mu}f\bigr)(\ alpha)\frac{(x-\alpha)_{p\大gr]\,\mathrm{d}_{p,q}x\\&&qquad{}+\int _{\gamma _{k}}^{\beta}\biggl[R_{(p,q),\mu-1,f}(β,x)-\bigl(D _{p,q}^{\mu}f\bigr)(β)\frac{(x-\β)_{p,q}^{\mu}}{[\mu]_{p,q}!}\大gr]\,\mathrm{d}_{p,q}x\\&\quad=\int_{\alpha}^{\gamma_{k}}R{(p,q),\mu,f}(\alpha,x){d}_{p,q}x,\结束{对齐}$$
哪里
$$\begin{aligned}&\int_{\alpha}^{\beta}\bigl(D_{p,q}^{\su+1}f\bigr)(x)\frac{(\beta-qx){p,q}^{\tu+1}}{[\mu+1]{p,q}!}=\int_{\alpha}^{\beta}R_{(p,q),\mu,f}(\alpha,x),\end{aligned}$$
(2.9)
$$\begin{aligned}&\int_{\alpha}^{\beta}\bigl(D_{p,q}^{\tu+1}f\bigr)(x)\frac{(\alpha-qx){p,q}^{\mu+1}}{[\mu+1]_{p、q}!}=\int_{\alpha}^{\beta}R{(p,q),\mu,f}(\beta,x)。\结束{对齐}$$
(2.10)
这就完成了证明。 □
定理2.3
假设
\(0<q<p \leq 1),\(测试版>0),\(n\in\mathbb{n}\)和(f)是由定义的函数
\(f:[\alpha,\beta]\rightarrow\mathbb{R}\)具有
\(α=βφ{n}).让,在
\([\alpha,\beta]\),\(D_{p,q}^{\mu}f\)要么是
\((p,q)\)-减少或
\((p,q)\)-在上增加
\([\alpha,\beta]\).此外,假设这些数字
\(k,\ell\in\{0,1,2,\ldots\}\)是这样的
$$\开始{对齐}&p\psi_{k}-\alpha\leq\frac{\beta-\alpha}{[\mu+2]_{p,q}}\leq\beta-p\psi _{\ell}\fquad\textit{if}D_{p,q}^{\mau}f\textit{is}(p,q)\textit{-递减},\\&&\beta-p\psi _{\ell}\leq\frac{\beta-p\psi}{[\mu+2]_{p,q}}\leq_{k}-\alpha\quad\textit{if}D_{p,q}^{f\textit{is}(p,q)\textit{-递增}。\结束{对齐}$$
然后
$$\begin{aligned}[b]\bigl(D_{p,q}^{mu}f\bigr)(p\psi_{k})-\ bigl{(p\beta-q\alpha){p,q}^{\mu+1}}\int_{alpha}^{\ beta}R{(p,q),\mu,f}(\alpha,x)\\&\leq\bigl(D_{p,q}^{f\bigr)。\结束{对齐}$$
(2.11)
证明
我们证明了结果\(D_{p,q}^{\mu}f\)一\((p,q)\)-递减函数,我们可以用类似的方法证明它\(D_{p,q}^{\mu}f\)是\((p,q)\)-增加功能。让\(F(x)=-(D_{p,q}^{mu+1}F)(x)\)。那么\(F(x)\)是一个\((p,q)\)-递减函数。自\(D_{p,q}^{\mu}f\)是\((p,q)\)-减少,\(D_{p,q}^{\mu+1}f\leq0\).因此\(F(x)\geq 0)并将成为\((p,q)\)-减少。假设
$$G(x)=\压裂{(\beta-qx){p,q}^{\mu+1}}{(p\beta-q\alpha){p、q}^}{\mu+1}}$$
然后
$$G(x)=压裂{(β-qx)(p\β-q^{2} x个)\cdots(p^{\mu}\beta-q^{\mu+1}x)}{(p\beta-q\alpha)$$
$$\开始{aligned}\int_{\alpha}^{\beta}G(x)\,\mathrm{d}_{p,q}x=&\frac{1}{(\beta-q\alpha){p,q}^{mu+1}}\int{alpha}^{beta}(\beta-qx){p{d}_{p,q}x\\=&-\frac{1}{[\mu+2]{p,q}}\frac}1}{(p\beta-q\alpha){p,q}^{mu+1}}\int{\alpha}^{v} D类_{p,q}(\beta-x){p,q}^{\mu+2},\mathrm{d}{p,q}x\\=&\frac{1}{[\mu+2]{p,que}}\frac}{1}}{(p\beta-q\alpha){p,q}}\frac{(\β-\α)(p\β-q\α)\cdots(p^{\mu+1}\beta-q^{\mu+1}\alpha)}{(p\贝塔-q\阿尔法)\β-q^{\mu+1}\alpha)}\\=&\frac{\beta-\alpha}{[\mu+2]{p,q}}。\结束{对齐}$$
如果
$$磅/平方英寸_{k}-\alpha\leq\frac{\beta-\alpha}{[\mu+2]_{p,q}}\leq\beta-p\psi_{\ell}$$
然后
$$\开始{aligned}-\int_{p\psi_{ell}}^{beta}\bigl(D_{p,q}^{mu+1}f\bigr)(x)\,\mathrm{d}_{p,q}x\leq&-\int_{\alpha}^{\beta}\bigl(D_{p,q}^{\mu+1}f\bigr){d}_{p,q}x\\leq&-\int_{\alpha}^{p\psi_{k}}\bigl(D_{p,q}^{mu+1}f\bigr)(x)\,\mathrm{d}_{p,q}x.\结束{对齐}$$
发件人(2.9),我们可以写
$$\bigl(D_{p,q}^{\mu}f\bigr)(x)|_{x=\alpha}^{x=p\psi _{k}}\leq\frac{[\mu+1]_{p,q}!}{(p\beta-q\alpha){p,q}^{\mu+1}}\int_{\alpha}^{\ beta}R_{(p,q),\mu,f}(\alpha,x)\leq\bigl(D_{p,q}^{f\bigr)(x)|{x=p\psi_{\ell}}^{x=\beta}$$
□
定理2.4
假设
\(0<q<p \leq 1),\(测试版>0),\(n\in\mathbb{n}\)和(f)是由定义的函数
\(f:[\alpha,\beta]\rightarrow\mathbb{R}\)具有
\(α=βφ{n}).让,在
\([\alpha,\beta]\),\(D_{p,q}^{2} 如果\geq 0\)(f是
\((p,q)\)-凸面的).假设我们有任何数字
\(k,\ell\in\{0,1,2,\ldots\}\)这样的话
$$\开始{aligned}&p\psi_{\ell}\leq\frac{(p-1)\beta+q\beta+/alpha}{p+q}\quad\textit{和}\quad\\&p\psi{k}\leq \frac}{(p1)\alpha+q\alpha+/beta}{p+q},\quad\\textit{for}f\textit{is}1)β+q\β+\alpha}{p+q}\quad\textit{和}\quad\\&p\psi{k}\geq\frac{(p-1)\alpha+q\alpha+\beta}{p+q},\quad\textit{表示}f\textit{is}(p,q)\textit}-递增}。\结束{对齐}$$
然后
$$开始{对齐}f(p\psi_{k})+\frac{f(\alpha)}{p\beta-q\alpha}\bigl((1-p)\beta-(1-q)\alpha\bigr)\leq&\frac}{p\ beta-q \alphaneneneep \int_{alpha}^{beta}f(x)\,\mathrm{d}_{p,q}x\\{(β-\alpha)f(α)}{p\beta-q\alpha}+f(β)-f(p\psi_{ell})。\结束{对齐}$$
证明
我们证明了这一点(f)成为\((p,q)\)-递减函数,和(f)一\((p,q)\)-增加函数的证明是相似的。对于\(D_{p,q}^{2} 如果\geq 0\)在\([\alpha,\beta]\),\(f(x)\)是一个\((p,q)\)-凸函数和清晰的\(x,p)^{2} x个,q个^{2} x个\)和\(p,q,x\在{}[\alpha,\beta]\中),我们有\(qf(p^{2} x个)-(p+q)f(pqx)+pf(q^{2} x个)\geq 0\).如果\(f(x)\)在上是凸的\([\alpha,\beta]\),那么它也是q个-凸的\([\alpha,\beta]\)因此它也是一个\((p,q)\)-凸的\([\alpha,\beta]\).接受\(\mu=0\),然后\([\mu+2]{p,q}=p+q\).假设(f)是\((p,q)\)-减少,那么我们有\(磅/平方英寸_{k}-\α\leq\压裂{\beta-\alpha}{[\mu+2]{p,q}}\leq\beta-p\psi{\ell}\)这意味着,对于\(\mu=0\),\(p\psi{k}\leq\alpha+\frac{\beta-\alpha}{p+q}\leq \beta-p\psi})因此\(p\psi{\ell}\leq\frac{(p-1)\beta+q\beta+/alpha}{p+q}\),以及\(p\psi{k}\leq\frac{(p-1)\alpha+\alpha+/beta}{p+q}\).自\(D_{p,q}^{0}页)(x) =f(x)\),\(R_{(p,q),-1,f}(α,x)=f(x))来自定理2.2和
$$\int_{\alpha}^{\gamma_{k}R_{(p,q),\mu,f}(\alpha,x)\,\mathrm{d}_{p,q}x=\int_{\alpha}^{\gamma_{k}}\biggl[R_{(p,q),\mu-1,f}(\alpha,x)-\bigl\biggr]\,\mathrm{d}_{p,q}x。 $$
因此,
$$\int _{\alpha}^{\gamma _{k}}R_{(p,q),0,f}(\alpha,x)\,\mathrm{d}_{p,q}x=\int_{\alpha}^{\gamma_{k}}\bigl[R_{(p,q),-1,f}(\alpha,x)-f(\alpha)\bigr]\,\mathrm{d}_{p,q}x。 $$
这意味着\(R_{(p,q),0,f}(α,x)=f(x)-f(α))。因此自(2.11)定理的2.3,我们有
$$f(p\psi_{k})-f(\alpha)\leq\frac{1}{p\beta-q\alpha}\int_{alpha}^{beta}\bigl[f(x)-f{d}_{p,q}x\leqf(β)-f(p\psi{ell})$$
这就完成了证明。 □
定理2.5
假设
\(0<q<p \leq 1),\(测试版>0),\(n\in\mathbb{n}\)和(f)是由定义的函数
\(f:[\alpha,\beta]\rightarrow\mathbb{R}\)具有
\(α=βφ{n}).假设,在
\([\alpha,\beta]\),\(D_{p,q}^{2} 如果\geq 0\)(f是
\((p,q)\)-凸面的).此外,假设我们有任何数字
\(k,\ell\in\{0,1,2,\ldots\}\)这样的话
$$开始{aligned}&p\psi_{ell}\geq\frac{(alpha+\beta)p}{p+q}\quad\textit{和}\quad p\psi.{k}\geq\frac},\quad\textit{和}\quad p\psi{k}\leq\frac{(alpha+\beta)q}{p+q},\quad\textit{代表}f\textit{is}(p,q)\textit}-递减}。\结束{对齐}$$
然后
$$f(\psi_{k})\leq\frac{1}{\beta-\alpha}\int_{alpha}^{beta}f(qx)\,\mathrm{d}_{p,q}x\leq f(\alpha)+f(\beta)-f(\psi_{\ell})$$
证明
我们证明了这一点(f)存在\((p,q)\)减少,以及(f)
\((p,q)\)-增加,证明是相似的。假设(f)是\((p,q)\)-凸的\([\alpha,\beta]\),\(D_{p,q}^{2} 如果\geq 0\)在\([\alpha,\beta]\)和\(D)_{p,q}f\)是\((p,q)\)-在上增加\([\alpha,\beta]\).假设\(F(x)=-D_{p,q}f\),\(G(x)=压裂{β-x}{β-α}),然后\(F(x)\)是\((p,q)\)-在上减少\([\alpha,\beta]\)同样,(f)是\((p,q)\)-在上减少\([\alpha,\beta]\),\(D)_{p,q}f\leq 0\).因此\(F(x)\geq 0)和\((p,q)\)-在上减少\([\alpha,\beta]\)因此,F类和G公司满足定理假设2.1.来自定理2.3,我们有
$$磅/平方英寸_{k}-\α\leq\压裂{\beta-\alpha}{[\mu+2]{p,q}}\leq\beta-p\psi{ell}$$
这意味着
$$\开始{对齐}&p\psi_{k}-\α\leq\int_{\alpha}^{\beta}G(x)\,\mathrm{d}_{p,q}x\leq\beta-p\psi_{\ell},\&p\psi_{k}-\α\leq\frac{q\beta-p\alpha}{p+q}\leq\β-p\psi{ell},\end{aligned}$$
以这种形式\(pc{\ell}\leq\frac{(alpha+\beta)p}{p+q}\),\(p\psi{k}\leq\frac{(alpha+\beta)q}{p+q}\).
我们发现\(int_{\alpha}^{\beta}F(x)G(x)\,\mathrm{d}_{p,q}x\),通过使用按部件和\((1+qx){p,q}^{n-1}=\压裂{1}{[n]{p,q}}D{p,q}(1+x){p,q{^n}\)来自[7],我们有
$$\开始{aligned}&\int_{\alpha}^{\beta}G(x)\,\mathrm{d}_{p,q}x=\frac{1}{\beta-\alpha}\int_{\alpha}^{\beta}(\beta-x)\,\mathrm{d}_{p,q}x=\frac{(alpha-\beta){p,q}^{2}}{(\beta-\alpha)(p+q)}=\frac{q\beta-p\alpha}{p+q},\\&\int_{alpha}^{beta}F(x)G(x)\,\mathrm{d}_{p,q}x=\压裂{1}{\β-\α}\int_{\α}^{\β}(x-\β)(D_{p,q}f)\,\mathrm{d}_{p,q}x=\frac{-1}{\beta-\alpha}\int _{\alpha}^{\beta}f(qx)\,\mathrm{d}_{p,q}x+f(\alpha)。\结束{对齐}$$
发件人(2.1)我们有
$$\int_{\psi_{\ell}}^{\beta}F(x)\,\mathrm{d}_{p,q}x\leq\int_{\alpha}^{\beta}F(x)G(x)\,\mathrm{d}_{p,q}x\leq\int_{\alpha}^{\psi_{k}}F(x)\,\mathrm{d}_{p,q}x。 $$
因此,
$$-\int_{\psi_{\ell}}^{v}(D_{p,q}f)\,\mathrm{d}_{p,q}x\leq\frac{-1}{\beta-\alpha}\int_{\alpha{^{\beta}f(qx)\,\mathrm{d}_{p,q}x+f(\alpha)\leq-\int_{\alpha}^{\psi_{k}}(D_{p,q}f)\,\mathrm{d}_{p,q}x。 $$
这意味着
$$-f(x)|_{\psi_{\ell}}^{\beta}\leq\frac{-1}{\beta-\alpha}\int _{\alpha}^{\beta}f(qx)\,\mathrm{d}_{p,q}x+f(α)\leq-f(x){u}^{psi{k}}$$
因此
$$f(x)|_{\alpha}^{\psi_{k}}\leq\frac{1}{\beta-\alpha}\int_{\alpha}^}\beta}f(qx)\,\mathrm{d}_{p,q}x-f(\alpha)\leqf(x)|{\psi{\ell}}^{\beta}$$
这就完成了证明。 □
定理2.6
让
\(f:[\alpha,\beta]\rightarrow\mathbb{R}\)和
\(m\leq D_{p,q}^{mu+1}\leq m\),哪里
\(m<m\).假设我们有任何数字
\(k,\ell\in\{0,1,2,\ldots\}\)这样的话
$$磅/平方英寸_{k}-\alpha\leq\frac{1}{M-M}\bigl[\bigl(D_{p,q}^{\mu}f\bigr)-\bigl(D_}p,q{^{\mu}f\bigr)(\alpha)-M(\beta-\alpha)\bigr]\leq\ beta-p\psi_{\ell}$$
然后
$$开始{对齐}m+(m-m)\压裂{(\beta-\psi{\ell}){p,q}^{\mu+2}}{(\ beta-\ alpha){p,q{^{\mo+2}}\leq&\frac{}[\mu+2]{p,q}!}{(β-\alpha){p,q}^{mu+2}}\int_{alpha}^{beta}R{(p,q),\mu,f}(α,x)\,\mathrm{d}_{p,q}x\\\leq&M-(M-M)\frac{(\beta-\psi{k}){p,q}^{mu+2}}{(\ beta-\ alpha){p,q}^}{mu+2{}}。\结束{对齐}$$
证明
假设\(F(x)\)是\((p,q)\)-递减函数和\(F(x)\geq 0).我们接受\(F(x)=\压裂{(β-qx){p,q}^{\mu+1}}{[\mu+1]{p,q}!}\)和\(G(x)=D_{p,q}^{\mu+1}G\),其中\(g(x)=\压裂{1}{M-M}[f(x)-M\压裂{(x-\alpha){p,q}^{mu+1}}{[mu+1]{p,q}!}]\)因此\(G(x)=frac{1}{M-M}(D_{p,q}^{mu+1}f-M)因此,
$$\开始{aligned}\int_{\alpha}^{\beta}G(x)\,\mathrm{d}_{p,q}x=&\frac{1}{M-M}\int_{alpha}^{beta}D_{p,q}^{mu+1}f\,\mathrm{d}_{p,q}x-\压裂{m(\beta-\alpha)}{m-m}\\=&\frac{1}{m-m}\bigl[\bigl。\结束{对齐}$$
这意味着
$$磅/平方英寸_{k}-\α\leq\frac{1}{M-M}\bigl[\bigl(D_{p,q}^{\mu}f\bigr)(\beta)-\bigl-(D_}p,q{{\mu{f\biger)(\alpha)-M(\beta-\alpha)\bigr]\leq\ beta-p\psi{\ell}$$
而且
$$\开始{aligned}\int_{\psi_{\ell}}^{\beta}F(x)\,\mathrm{d}_{p,q}x=\int_{\psi_{\ell}}^{\beta}\frac{(\beta-qx){p,q}^{\mu+1}}{[\mu+1]{p,q}!}\,\数学{d}_{p,q}x=&-\frac{1}{[\mu+2]{p,q}!}\int _{\psi _{\ell}}^{\beta}D _{p,q}(β-x)_{p,q}^{\mu+2}\,\mathrm{d}_{p,q}x\\=&\压裂{(\beta-\psi{\ell}){p,q}^{\mu+2}}{[\mu+2]{p,q}!}。\结束{对齐}$$
类似地
$$\begin{aligned}&&int _{\alpha}^{\psi _{k}}F(x)\,\mathrm{d}_{p,q}x=\int_{\alpha}^{\psi_{k}}\frac{(\beta-qx){p,q}^{mu+1}}{[\mu+1]{p,q}!}\,\马特姆{d}_{p,q}x=\frac{1}{[\mu+2]{p,q}!}\bigl[(beta-\alpha){p,q}^{\mu+2}-(beta-\psi{k}){p,q}^{\mu+2}\bigr],\\&\int_{\alpha}^{\ beta}F(x)G(x)\,\mathrm{d}_{p,q}x=\int_{alpha}^{beta}\frac{(\beta-qx){p,q}^{mu+1}}{[\mu+1]{p,q}!}D_{p,q}^{\mu+2}g\,\mathrm{d}_{p,q}x\\&\hphantom{\int_{\alpha}^{\beta}F(x)G(x)\,\mathrm{d}_{p,q}x}=\int_{\alpha}^{\beta}R_{(p,q),\mu,g}(\alpha,x)\,\mathrm{d}_{p,q}x\\&\hphantom{\int_{\alpha}^{\beta}F(x)G(x)\,\mathrm{d}_{p,q}x}=\int_{\alpha}^{\beta}\biggl[\frac{1}{M-M}右_{(p,q),\mu,f}(\alpha,x)-\frac{m}{m-m}\frac{(x-\alpha)_{p,q}^{\mu+1}}}{[\mu+1]_{p,q}!}\大gr]\,\mathrm{d}_{p,q}x\\&\hphantom{\int_{\alpha}^{\beta}F(x)G(x)\,\mathrm{d}_{p,q}x}=\frac{1}{M-M}\int_{alpha}^{beta}R{(p,q),\mu,f}(\alpha,x)\,\mathrm{d}_{p,q}x-\frac{m}{m-m}\frac{(β-α)_{p,q}^{\mu+2}}}{[\mu+2]_{p,q}!}。\结束{对齐}$$
通过替换积分\(int_{\psi_{\ell}}^{\beta}F(x)\,\mathrm{d}_{p,q}x\),\(int_{\alpha}^{\psi_{k}}F(x)\,\mathrm{d}_{p,q}x\)和\(int_{\alpha}^{\beta}F(x)G(x)\,\mathrm{d}_{p,q}x\)英寸(2.1)定理的2.1,我们得到了预期的结果。 □