陈博士的二次型不等式问题

(二次型不等式)设$n$为正整数, $c_1,c_2,\cdots,c_n$是复数,满足$\sum_{j=1}^{n}c_j=0$, $x_1,x_2,\cdots,x_n$是实数.证明:
$$\sum_{j,k=1}^n{c_j\overline{c_k}\left| x_j-x_k \right|}\leqslant 0.$$

证明.利用
$$\int_0^{+\infty}{\frac{1-\cos \left( at \right)}{t^2}dt}=\left| a \right|$$
有
$$\begin{aligned} &\sum_{j,k=1}^n{c_j\overline{c_k}\left| x_j-x_k \right|} =\sum_{j,k=1}^n{c_j\overline{c_k}\int_0^{+\infty}{\frac{1-\cos \left( x_j-x_k \right) t}{t^2}dt}} \\ &=\int_0^{+\infty}{\frac{1}{t^2}\sum_{j,k=1}^n{c_j\overline{c_k}}dt}-\int_0^{+\infty}{\frac{1}{t^2}\sum_{j,k=1}^n{c_j\overline{c_k}\left( \cos x_jt\cos x_kt+\sin x_jt\sin x_kt \right)}dt} \\ &=\int_0^{+\infty}{\frac{1}{t^2}\left| \sum_{j=1}^n{c_j} \right|^2dt}-\int_0^{+\infty}{\frac{1}{t^2}\left( \left| \sum_{j=1}^n{c_j\cos x_j} \right|^2+\left| \sum_{j=1}^n{c_j\sin x_j} \right|^2 \right) dt} \\ &=-\int_0^{+\infty}{\frac{1}{t^2}\left( \left| \sum_{j=1}^n{c_j\cos x_jt} \right|^2+\left| \sum_{j=1}^n{c_j\sin x_jt} \right|^2 \right) dt}\leqslant 0. \end{aligned}$$

 

(二次型不等式)设$n$为正整数, $c_1,c_2,\cdots,c_n$是复数, $x_1,x_2,\cdots,x_n$为实数.证明:
$$\sum_{j,k=1}^n{c_j\overline{c_k}\frac{1}{1+\left( x_j-x_k \right) ^2}}\geqslant 0.$$

证明.利用
$$\int_0^{+\infty}{e^{-t}\cos \left( at \right) dt}=\frac{1}{1+a^2}$$
可得
$$\begin{aligned} &\sum_{j,k=1}^n{c_j\overline{c_k}\frac{1}{1+\left( x_j-x_k \right) ^2}}=\sum_{j,k=1}^n{c_j\overline{c_k}\int_0^{+\infty}{e^{-t}\cos \left( x_j-x_k \right) tdt}} \\ &=\int_0^{+\infty}{e^{-t}\sum_{j,k=1}^n{c_j\overline{c_k}\left( \cos x_jt\cos x_kt+\sin x_jt\sin x_kt \right)}dt} \\ &=\int_0^{+\infty}{e^{-t}\left( \left| \sum_{j=1}^n{c_j\cos x_jt} \right|^2+\left| \sum_{j=1}^n{c_j\sin x_jt} \right|^2 \right) dt}\geqslant 0. \end{aligned}$$

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(二次型不等式)设$n$为正整数, $c_1,c_2,\cdots,c_n$为复数, $x_1,x_2,\cdots,x_n$为实数.证明:
$$\sum_{j,k=1}^n{c_j\overline{c_k}e^{-\left( x_j-x_k \right) ^2}}\geqslant 0.$$

证.利用
$$e^{-a^2}=\frac{1}{2\sqrt{\pi}}\int_{-\infty}^{+\infty}{e^{iat}\cdot e^{-t^2/4}dt}$$
可得
$$\begin{aligned} \sum_{j,k=1}^n{c_j\overline{c_k}e^{-\left( x_j-x_k \right) ^2}} &=\sum_{j,k=1}^n{c_j\overline{c_k}\cdot \frac{1}{2\sqrt{\pi}}\int_{-\infty}^{+\infty}{e^{i\left( x_j-x_k \right) t}\cdot e^{-t^2/4}dt}} \\ &=\int_{-\infty}^{+\infty}{\frac{1}{2\sqrt{\pi}}e^{-t^2/4}\sum_{j,k=1}^n{c_j\overline{c_k}e^{i\left( x_j-x_k \right) t}}dt} \\ &=\int_{-\infty}^{+\infty}{\frac{1}{2\sqrt{\pi}}e^{-t^2/4}\left| \sum_{j=1}^n{c_je^{ix_jt}} \right|^2dt}\geqslant 0. \end{aligned}$$

 

(二次型不等式)证明: $\displaystyle K\left( s,t \right) =\frac{1}{\ln \left( s+t \right)}$在$(1,+\infty)$上是正定核,也就是说,对任意正整数$n$,任意复数$c_1,c_2,\cdots,c_n$以及$(1,+\infty)$内的任意实数$x_1,x_2,\cdots,x_n$,有
$$\sum_{j,k=1}^n{\frac{c_j\overline{c_k}}{\ln \left( x_j+x_k \right)}}\geqslant 0.$$

证.利用
$$\int_1^{+\infty}{\frac{1}{a^t}dt}=\frac{1}{\ln a},\quad a>1$$
和
$$\frac{1}{x^t}=\frac{1}{\Gamma \left( t \right)}\int_0^{\infty}{u^{t-1}e^{-xu}du}$$
可得
$$\begin{aligned} \sum_{j,k=1}^n{\frac{c_j\overline{c_k}}{\ln \left( x_j+x_k \right)}} &=\sum_{j,k=1}^n{c_j\overline{c_k}\int_1^{+\infty}{\frac{1}{\left( x_j+x_k \right) ^t}dt}} \\ &=\sum_{j,k=1}^n{c_j\overline{c_k}\int_1^{+\infty}{dt\frac{1}{\Gamma \left( t \right)}\int_0^{+\infty}{u^{t-1}e^{-\left( x_j+x_k \right) u}du}}} \\ &=\int_1^{+\infty}{\frac{1}{\Gamma \left( t \right)}dt\int_0^{+\infty}{u^{t-1}\sum_{j,k=1}^n{c_j\overline{c_k}}e^{-\left( x_j+x_k \right) u}du}} \\ &=\int_1^{+\infty}{\frac{1}{\Gamma \left( t \right)}dt\int_0^{+\infty}{u^{t-1}\left| \sum_{j=1}^n{c_je^{-x_ju}} \right|^2du}}\geqslant 0. \end{aligned}$$

 

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(二次型不等式)证明: $\displaystyle K\left( z,w \right) =\frac{\Gamma \left( 1+\alpha \right)}{\left( z+\overline{w} \right) ^{\alpha}}$ ($\alpha$为大于$-1$的实数)
在开的右半复平面内是正定核,也就是说,对于任意正整数$n$,任意复数$c_1,c_2,\cdots,c_n$以及任意在右半复平面内的复数$z_1,z_2,\cdots,z_n$,有
$$\sum_{j,k=1}^n{c_j\overline{c_k}\frac{\Gamma \left( 1+\alpha \right)}{\left( z_j+\overline{z_k} \right) ^{\alpha}}}\geqslant 0.$$

证.当$\alpha >0$时,利用
$$\frac{1}{x^{\alpha}}=\frac{1}{\Gamma \left( \alpha \right)}\int_0^{+\infty}{t^{\alpha-1}e^{-xt}dt},\quad \alpha >0$$
可得
$$\begin{aligned} \sum_{j,k=1}^n{c_j\overline{c_k}\frac{\Gamma \left( 1+\alpha \right)}{\left( z_j+\overline{z_k} \right) ^{\alpha}}} &=\sum_{j,k=1}^n{c_j\overline{c_k}\frac{\Gamma \left( 1+\alpha \right)}{\Gamma \left( \alpha \right)}\int_0^{+\infty}{t^{\alpha -1}e^{-\left( z_j+\overline{z_k} \right) t}dt}} \\ &=\int_0^{+\infty}{\alpha t^{\alpha -1}\left( \sum_{j,k=1}^n{c_j\overline{c_k}e^{-\left( z_j+\overline{z_k} \right) t}} \right) dt} \\ &=\int_0^{+\infty}{\alpha t^{\alpha -1}\left| \sum_{j=1}^n{c_je^{-z_jt}} \right|^2dt}\geqslant 0. \end{aligned}$$

当$\alpha=0$时,
$$\sum_{j,k=1}^n{c_j\overline{c_k}\frac{\Gamma \left( 1+\alpha \right)}{\left( z_j+\overline{z_k} \right) ^{\alpha}}}=\sum_{j,k=1}^n{c_j\overline{c_k}}=\left| \sum_{j=1}^n{c_j} \right|^2\geqslant 0.$$

当$-1<\alpha <0$时,利用
$$\frac{1}{x^{\alpha}}=\frac{\alpha}{\Gamma \left( 1+\alpha \right)}\int_0^{+\infty}{t^{\alpha -1}\left( e^{-xt}-1 \right) dt},\quad -1<\alpha <0$$
和分部积分可得
$$\begin{aligned} \sum_{j,k=1}^n{c_j\overline{c_k}\frac{\Gamma \left( 1+\alpha \right)}{\left( z_j+\overline{z_k} \right) ^{\alpha}}}&=\sum_{j,k=1}^n{c_j\overline{c_k}\alpha \int_0^{+\infty}{t^{\alpha -1}\left( e^{-\left( z_j+\overline{z_k} \right) t}-1 \right) dt}}\\ &=\int_0^{+\infty}{\alpha t^{\alpha -1}\left( \sum_{j,k=1}^n{c_j\overline{c_k}\left( e^{-\left( z_j+\overline{z_k} \right) t}-1 \right)} \right) dt}\\ &=\int_0^{+\infty}{\left( \sum_{j,k=1}^n{c_j\overline{c_k}\left( e^{-\left( z_j+\overline{z_k} \right) t}-1 \right)} \right) dt^{\alpha}}\\ &=\int_0^{+\infty}{t^{\alpha}\left( \sum_{j,k=1}^n{c_j\overline{c_k}\left( z_j+\overline{z_k} \right) e^{-\left( z_j+\overline{z_k} \right) t}} \right) dt}\\ &=2\mathrm{Re}\int_0^{+\infty}{t^{\alpha}\left( \sum_{j=1}^n{c_jz_je^{-z_jt}} \right) \left( \sum_{j=1}^n{\overline{c_j}e^{-\overline{z_j}t}} \right) dt}\geqslant 0.\\ \end{aligned}$$

(二次型不等式)设$n$为正整数, $c_1,c_2,\cdots,c_n$为复数, $x_1,x_2,\cdots,x_n$为实数.证明:
$$\sum_{j,k=1}^n\frac{c_j\overline{c_k}}{\cosh \left( x_j-x_k \right)}\geqslant 0.$$

证法一.利用
$$\frac{1}{\cosh \pi a}=\int_{-\infty}^{\infty}{\frac{e^{-2\pi iat}}{\cosh \pi t}dt}$$
可得
$$\begin{aligned} \sum_{j,k=1}^n{\frac{c_j\overline{c_k}}{\cosh \left( x_j-x_k \right)}} &=\sum_{j,k=1}^n{c_j\overline{c_k}\int_{-\infty}^{\infty}{\frac{e^{-2i\left( x_j-x_k \right) t}}{\cosh \pi t}dt}} \\ &=\int_{-\infty}^{\infty}{\frac{1}{\cosh \pi t}\sum_{j,k=1}^n{c_j\overline{c_k}e^{-2i\left( x_j-x_k \right) t}}dt} \\ &=\int_{-\infty}^{\infty}{\frac{1}{\cosh \pi t}\left| \sum_{j=1}^n{c_je^{-2ix_jt}} \right|^2dt}\geqslant 0. \end{aligned}$$

证法二.利用
$$\frac{1}{\cosh \frac{a}{2}}=2\int_0^{\infty}{\frac{\cos at}{\cosh \pi t}dt}$$
可得
$$\begin{aligned} \sum_{j,k=1}^n{\frac{c_j\overline{c_k}}{\cosh \left( x_j-x_k \right)}} &=\sum_{j,k=1}^n{c_j\overline{c_k}\cdot 2\int_0^{\infty}{\frac{\cos 2\left( x_j-x_k \right) t}{\cosh \pi t}dt}} \\ &=\int_{-\infty}^{\infty}{\frac{2}{\cosh \pi t}\sum_{j,k=1}^n{c_j\overline{c_k}\left( \cos 2x_jt\cos 2x_kt+\sin 2x_jt\sin 2x_kt \right)}dt} \\ &=\int_{-\infty}^{\infty}{\frac{2}{\cosh \pi t}\left( \left| \sum_{j=1}^n{c_j\cos 2x_jt} \right|^2+\left| \sum_{j=1}^n{c_j\sin 2x_jt} \right|^2 \right) dt}\geqslant 0. \end{aligned}$$