谢之题解：重积分的应用举例

7.设$u_i\in L^{p_i}(\Omega),p_i>0,i=1,2,\cdots,m$,且${\sum\limits_{i = 1}^{m } {\frac{1}{{{p_i}}}} }=1$.证明\[\iint\limits_\Omega {{u_1} \cdots {u_m}dxdy} \le {\left\| {{u_1}} \right\|_{{p_1}}} \cdots {\left\| {{u_m}} \right\|_{{p_m}}}.\]

解.利用P270的Holder不等式以及数学归纳法即可.假设$m-1$时不等式成立,则有

\begin{align*}&\iint\limits_\Omega {{u_1} \cdots {u_{m - 1}}{u_m}dxdy} \le \iint\limits_\Omega {\left| {{u_1} \cdots {u_{m - 1}}{u_m}} \right|dxdy} \\= &{\left\| {{u_1} \cdots {u_{m - 1}}{u_m}} \right\|_1} \le {\left\| {{u_1} \cdots {u_{m - 1}}} \right\|_{1/\sum\limits_{i = 1}^{m - 1} {\frac{1}{{{p_i}}}} }}{\left\| {{u_m}} \right\|_{{p_m}}}\\\le& {\left( {{{\left\| {u_1^{1/\sum\limits_{i = 1}^{m - 1} {\frac{1}{{{p_i}}}} }} \right\|}_{{p_1}\sum\limits_{i = 1}^{m - 1} {\frac{1}{{{p_i}}}} }} \cdots {{\left\| {u_{m - 1}^{1/\sum\limits_{i = 1}^{m - 1} {\frac{1}{{{p_i}}}} }} \right\|}_{{p_{m - 1}}\sum\limits_{i = 1}^{m - 1} {\frac{1}{{{p_i}}}} }}} \right)^{\sum\limits_{i = 1}^{m - 1} {\frac{1}{{{p_i}}}} }}{\left\| {{u_m}} \right\|_{{p_m}}}\\= &{\left\| {{u_1}} \right\|_{{p_1}}} \cdots {\left\| {{u_{m - 1}}} \right\|_{{p_{m - 1}}}}{\left\| {{u_m}} \right\|_{{p_m}}}.\end{align*}

另解:利用$n$元的Holder不等式

\[\prod\limits_{i = 1}^m {x_i^{{\theta _i}}} \le \sum\limits_{i = 1}^m {{\theta _i}{x_i}} ,\quad \text{其中}\sum\limits_{i = 1}^m {{\theta _i}} = 1,{\theta _i} \ge 0.\]

取\[{x_i} = \frac{{{{\left| {{u_i}} \right|}^{{p_i}}}}}{{\left\| {{u_i}} \right\|_{{p_i}}^{{p_i}}}},{\theta _i} = \frac{1}{{{p_i}}}\]再积分即可.

证明\[1 < \iiint\limits_{{{\left[ {0,1} \right]}^3}} {\left( {\cos \left( {xyz} \right) + \sin \left( {xyz} \right)} \right)dxdydz} < \sqrt 2 .\]

证.注意到\[1 < \cos \left( {xyz} \right) + \sin \left( {xyz} \right) = \sqrt 2 \sin \left( {xyz + \frac{\pi }{4}} \right) < \sqrt 2 \]即可.

证明

\[{\left\{ {{{\int_a^b {dx\left[ {\int_c^d {f\left( {x,y} \right)dy} } \right]} }^2}} \right\}^{1/2}} \le \int_c^d {dy{{\left[ {\int_a^b {{f^2}\left( {x,y} \right)dx} } \right]}^{1/2}}} ,\]

其中$f$是连续函数.

证.利用Cauchy-Schwarz不等式我们有

\begin{align*}&{\left\{ {\int_c^d {dy{{\left[ {\int_a^b {{f^2}\left( {x,y} \right)dx} } \right]}^{1/2}}} } \right\}^2}\\= &\int_c^d {dy{{\left[ {\int_a^b {{f^2}\left( {x,y} \right)dx} } \right]}^{1/2}}} \cdot \int_c^d {dz{{\left[ {\int_a^b {{f^2}\left( {x,z} \right)dx} } \right]}^{1/2}}} \\\ge& \int_c^d {dy} \int_c^d {dz} \int_a^b {f\left( {x,y} \right)f\left( {x,z} \right)dx} = {\int_a^b {dx\left[ {\int_c^d {f\left( {x,y} \right)dy} } \right]} ^2}.\end{align*}