标签：Nonlinear Partial 不等式 Differential frac mu theta Holder parallel
实分析基础
Holder与卷积不等式
首先从经典的Holder不等式入手.
命题: 经典情况下的Holder不等式
设\((X,\mu)\)是测度空间, \((p,q,r)\in[1,\infty]^3\)满足
\[\frac{1}{p}+\frac{1}{q}=\frac{1}{r}
\]如果\((f,g)\in L^p(X,\mu)\times L^q(X,\mu)\), 则\(f\cdot g\in L^r(X,\mu)\), 且
\[\parallel f\cdot g\parallel_{L^r}\le\parallel f\parallel_{L^p}\cdot\parallel g\parallel_{L^q}
\]
\(\blacklozenge\)
当\(p=1\)或\(p=\infty\)时, 命题易证. 现在设\(\mathbb{R}\ni p>1\), 考虑Young不等式:
\[a^{\theta}b^{1-\theta}\le\theta a+(1-\theta)b,\quad\theta\in[0,1]
\]
标签：Nonlinear,
Partial,
不等式,
Differential,
frac,
mu,
theta,
Holder,
parallel
From： https://www.cnblogs.com/ununhappy/p/17487231.html