[物理学与PDEs]第2章习题12 严格凸性的转换

设 $L=L(\xi_0,\xi_1,\cdots,\xi_n)$ 关于变量 $\xi_0>0,\xi_1,\cdots,\xi_n$ 为严格凸的. 证明函数 $$\bex M=\cfrac{1}{\xi_0}L(\xi_0,\xi_1,\cdots,\xi_n) \eex$$ 关于变量 $$\bex \eta_0=\cfrac{1}{\xi_0},\quad \xi_1=\cfrac{\xi_1}{\xi_0},\cdots,\eta_n=\cfrac{\xi_n}{\xi_0} \eex$$ 是严格凸的.

证明: 仅对 $n=1$ 的情形加以证明. 先给出 $$\bex M=\eta_0 L\sex{\cfrac{1}{\eta_0},\cfrac{\eta_1}{\eta_0}}. \eex$$ 于是 $$\beex \bea M_{\eta_0}&=L+\eta_0 \sez{L_{\xi_0}\sex{-\cfrac{1}{\eta_0^2}} +L_{\xi_1}\sex{-\cfrac{\eta_1}{\eta_0^2}}}\\ &=L-\cfrac{1}{\eta_0}L_{\xi_0} -\cfrac{\eta_1}{\eta_0}L_{\xi_1},\\ M_{\eta_1}&=\eta_0L_{\xi_1}\cfrac{1}{\eta_0} =L_{\xi_1}; \eea \eeex$$ $$\beex \bea M_{\eta_0\eta_1} &=L_{\xi_0}\sex{-\cfrac{1}{\eta_0^2}} +L_{\xi_1}\sex{-\cfrac{\eta_1}{\eta_0^2}}\\ &\quad+\cfrac{1}{\eta_0^2}L_{\xi_0}-\cfrac{1}{\eta_0}\sez{ L_{\xi_0\xi_0}\sex{-\cfrac{1}{\eta_0^2}} +L_{\xi_0\xi_1}\sex{-\cfrac{\eta_1}{\eta_0^2}} }\\ &\quad+\cfrac{\eta_1}{\eta_0^2}L_{\xi_1} -\cfrac{\eta_1}{\eta_0^2} \sez{ L_{\xi_0\xi_1}\sex{-\cfrac{1}{\eta_0^2}} +L_{\xi_1\xi_1}\sex{-\cfrac{\eta_1}{\eta_0^2}} }\\ &=\cfrac{1}{\eta_0^3}\sex{ L_{\xi_0\xi_0} +2\eta_1L_{\xi_0\xi_1}+\eta_1^2L_{\xi_1\xi_1} }\\ &=\cfrac{1}{\eta_0^3} \sex{\ba{cc}1& \eta_1 \ea} \sex{\ba{cc} L_{\xi_0\xi_0}&L_{\xi_0\xi_1}\\ L_{\xi_0\xi_1}&L_{\xi_1\xi_1} \ea} \sex{\ba{cc} 1\\ \eta_1 \ea}\\ &>0, \eea \eeex$$ $$\beex \bea M_{\eta_0\eta_1} &=M_{\eta_1\eta_0}=L_{\xi_1\xi_0}\sex{-\cfrac{1}{\eta_0^2}} +L_{\xi_1\xi_1}\sex{-\cfrac{\eta_1}{\eta_0^2}}\\ &=-\cfrac{1}{\eta_0^2}\sex{L_{\xi_0\xi_1}+\eta_1 L_{\xi_1\xi_1}},\\ M_{\eta_1\eta_1}&=\cfrac{1}{\eta_0}L_{\xi_1\xi_1}; \eea \eeex$$ $$\beex \bea M_{\eta_0\eta_0}M_{\eta_1\eta_1}-M_{\eta_0\eta_1}^2 &=\cfrac{1}{\eta_0^4} \sex{L_{\xi_0\xi_0}+2\eta_1L_{\xi_0\xi_1}+\eta_1^2L_{\xi_1\xi_1}}L_{\xi_1\xi_1}\\ &\quad -\cfrac{1}{\eta_0^4} \sex{L_{\xi_0\xi_1}^2+2\eta_1L_{\xi_0\xi_1L_{\xi_1\xi_1} +\eta_1^2L_{\xi_1\xi_1}^2}}\\ &=\cfrac{1}{\eta_0^4}L_{\xi_0\xi_0}L_{\eta_1\eta_1}\\ &>0. \eea \eeex$$