4. 设 $x,y,u\in\bbR^n$ 的分量都是递减的. 证明:

(1). 若 $x\prec y$ 则 $\sef{x,u}\leq \sef{y,u}$.

(2). 若 $x\prec_w y$ 且 $u\in\bbR^n_+$, 则 $\sef{x,u}\leq \sef{y,u}$.

证明:

(1). 由 $x\prec y$ 知若记 $$\bex s_k=\sum_{i=1}^k x_i,\quad t_l=\sum_{j=1}^l y_l, \eex$$ 则 $$\bee\label{3_4_decay} s_k\leq t_k,\quad k=1,\cdots,n-1;\quad s_n=t_n. \eee$$ 于是 $$\beex \bea \sef{x,u}&=\sum_{i=1}^n x_iu_i\\ &=s_1u_1+\sum_{i=2}^n (s_i-s_{i-1})u_i\\ &=s_1u_1+\sum_{i=2}^n s_iu_i -\sum_{i=1}^{n-1}s_iu_{i+1}\\ &=\sum_{i=1}^n s_iu_i -\sum_{i=1}^{n-1}s_iu_{i+1}\\ &=\sum_{i=1}^{n-1}s_i(u_i-u_{i+1}) +s_nu_n\\ &\leq \sum_{i=1}^{n-1}t_i(u_i-u_{i+1}) +t_nu_n\quad\sex{\eqref{3_4_decay}}\\ &=\sef{y,u}. \eea \eeex$$

(2). 记号同上, 有 $$\beex \bea \sef{x,u} &=\sum_{i=1}^{n-1}s_i(u_i-u_{i-1}) +s_nu_n\\ &\leq \sum_{i=1}^{n-1}t_i(u_i-u_{i+1}) +t_nu_n\\ &\quad\sex{ s_i\leq t_i,\ i=1,\cdots,n-1;\ s_n\leq t_n, u_n\geq 0 }\\ &=\sef{y,u}. \eea \eeex$$