Fourier Analysis and Nonlinear Partial Differential Equations 阅读笔记 (第一章)

Markdown在展示公式的时候不能很好的处理一些密度很大的部分, 为了相对美观这里约定若无特殊说明, \(\parallel f\parallel_p\)就表示在\(L^p\)空间下的范数.

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

\[\parallel f\cdot g\parallel_r\le\parallel f\parallel_p\cdot\parallel g\parallel_q \]

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当\(p=1\)或\(p=\infty\)时, 命题易证. 现在设\(\mathbb{R}\ni p>1\), 沿用前置知识中的Holder不等式, 注意\(\frac{r}{p}+\frac{r}{q}=1\), 有:

\[\int_X|f\cdot g|^rd\mu=\int_X|f|^r\cdot|g|^rd\mu\le(\int_X|f|^{r\cdot\frac{p}{r}}d\mu)^{\frac{r}{p}}\cdot(\int_X|g|^{r\cdot\frac{q}{r}}d\mu)^{\frac{r}{q}}=\parallel f\parallel_p^r\cdot\parallel g\parallel_q^r \]

此即

\[\parallel fg\parallel_r^r=\parallel f\parallel_p^r\cdot\parallel g\parallel_q^r<+\infty \]

两边同取\(\frac{1}{r}\)次幂即得命题.

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下面这个引理说明Holder不等式在某些情况下是最优的.

引理

设\((X,\mu)\)是测度空间, \(p\in[1,\infty]\), \(p'\)是\(p\)的共轭指数, 设\(f\)是可测函数, 若

\[\underset{\parallel g\parallel_{p'}\le1}{\sup}\int_X|f(x)g(x)|d\mu(x)<\infty, \]

则\(f\in L^p\), 且

\[\parallel f\parallel_p=\underset{\parallel g\parallel_{p'}\le1}{\sup}\int_Xf(x)g(x)d\mu(x). \]

\(\blacklozenge\)(23.6.19: 这一部分在书上有些疑惑, 下面是自己的想法)

注意如果\(f\in L^p\), 则根据Holder不等式知当\(\parallel g\parallel_{p'}\le1\)时有:

\[\parallel f\cdot g\parallel_1\le\parallel f\parallel_p\cdot\parallel g\parallel_{p'}\le\parallel f\parallel_p \]

此即

\[\underset{\parallel g\parallel_{p'}\le1}{\sup}\int_Xf(x)g(x)dx\le\parallel f\parallel_p \]

故只需证明

\[\parallel f\parallel_p\le\underset{\parallel g\parallel_{p'}\le1}{\sup}\int_Xf(x)g(x)d\mu(x) \]

先考虑\(p=\infty\), 设\(\mathbb{R}\ni\lambda>0\)满足\(\mu(\{x\in E:|f(x)|\ge\lambda\})=0\). 记

\[E_{\lambda}=\{x\in E:|f(x)|\ge\lambda\} \]

考虑非负函数\(g_0\in L^1\), 其支集为\(E_{\lambda}\), 在\(X\)上的积分为1. 若设

\[g(x)=\frac{f(x)}{|f(x)|}g_0 \]

则\(g\in L^1\), 进而\(fg\)可积, 有

\[\int_Xf(x)\cdot g(x)d\mu(x)=\int_X|f(x)|\cdot|g(x)|d\mu(x)=\int_X|f(x)|\cdot g_0(x)d\mu(x)\ge\lambda\int_Xg_0(x)d\mu(x)=\lambda \]

进而

\[\parallel f\parallel_p\le\lambda\le\int_Xf(x)\cdot g(x)d\mu(x)\le\underset{\parallel g\parallel_{p'}\le1}{\sup}\int_Xf(x)g(x)d\mu(x). \]

此情况得证. 下设\(p\in(1,\infty)\), 考虑\(X\)中的非负可测递增列\(\{E_n\}_{n\in\mathbb{N}}\)满足\(\bigcup\limits_{n=1}^{\infty}E_n=X\), 设

\[f_n(x)=\boldsymbol{1}_{E_n\cap\{x\in X:|f(x)|\le n\}}(x)\cdot f(x),\quad g_n(x)=\frac{f_n(x)\cdot|f_n(x)|^{p-1}}{|f_n(x)|\cdot\parallel f_n\parallel_p^{\frac{p}{p'}}} \]

其中\(\boldsymbol{1}_A\)是\(A\)的示性函数, 显见\(f_n\in L^1\cap L^{\infty}\), 进而对任意\(p\)均有\(f\in L^p\). 更进一步, 知

\[\parallel g_n\parallel_{p'}^{p'}=\frac{1}{\parallel f_n\parallel_p^p}\int_X|f_n(x)|^{(p-1)\cdot\frac{p}{p-1}}d\mu(x)=1 \]

从而根据\(f_n,g_n\)的定义有

\[\begin{aligned} \int_Xf(x)\boldsymbol{1}_{E_n\cap\{x\in X:|f(x)|\le n\}}(x)\cdot g_n(x)d\mu(x)&=\int_Xf_n(x)\cdot g_n(x)d\mu(x)=\int_X|f_n(x)|\cdot|g_n(x)|d\mu(x)\\ &=(\int_X|f_n(x)|^pd\mu(x))\cdot\parallel f_n\parallel_p^{-\frac{p}{p'}}=\parallel f_n\parallel_p \end{aligned}\]

这说明

\[\parallel f_n\parallel_p\le\underset{\parallel g\parallel_{p'}\le1}{\sup}\int_Xf(x)\cdot g(x)d\mu(x). \]

进而由单调收敛定理知

\[\parallel f\parallel_p\le\underset{\parallel g\parallel_{p'}\le1}{\sup}\int_Xf(x)\cdot g(x)d\mu(x) \]

最后, 当\(p=1\), 考虑函数列\(\{g_n\}_{n\in\mathbb{N}}\):

\[g_n(x)=\boldsymbol{1}_{\{x\in X:f_n(x)\ne0\}}(x)\cdot\frac{f_n(x)}{|f_n(x)|} \]

显见\(\parallel g_n\parallel_{\infty}=1\), 同时

\[\int_Xf(x)\cdot g_n(x)d\mu(x)=\int_X|f(x)|\cdot|g_n(x)|d\mu(x)=\int_X|f_n(x)|d\mu(x) \]

考虑单调收敛定理, 知

\[\int_X|f(x)|d\mu(x)<\infty,\quad\int_X|f(x)|d\mu(x)=\lim\limits_{n\rightarrow\infty}\int_X|f_n(x)|d\mu(x) \]

命题即证.

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下面介绍Minkowski不等式.

命题: Minkowski不等式

设\((X_1,\mu_1),(X_2,\mu_2)\)是两个测度空间, \(f\)是\(X_1\times X_2\)上的非负可测函数. 对任意\(1\le p\le q\le\infty\), 有:

\[\parallel\parallel f(\cdot,x_2)\parallel_{L^p(X_1,\mu_1)}\parallel_{L^q(X_2,\mu_2)}\le\parallel\parallel f(x_1,\cdot)\parallel_{L^q(X_2,\mu_2)}\parallel_{L^p(X_1,\mu_1)} \]

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当\(q=\infty\), 利用Fatou引理即得结果. 当\(q\)有限, 记\(r=(\frac{q}{p})'\), 考虑Fubini定理, 有:

\[\begin{aligned} \parallel\parallel f(\cdot,x_2)\parallel_{L^p(X_2,\mu_2)}\parallel_{L^q(X_2,\mu_2)}&=(\int_{X_2}(\int_{X_1}f^p(x_1,x_2)d\mu_1(x_1))^{\frac{q}{p}}d\mu_2(x_2))^{\frac{1}{q}}\\ &=(\underset{\parallel g\parallel_{L^r(X_2,\mu_2)}=1\atop g\ge0}{\sup}\int_{X_1\times X_2}f^p(x_1,x_2)g(x_2)d\mu_1(x_1)d\mu_2(x_2))^{\frac{1}{p}}\\ &\le(\int_{X_1}(\underset{\parallel g\parallel_{L^r(X_2,\mu_2)}=1\atop g\ge0}{\sup}\int_{X_2}f^p(x_1,x_2)g(x_2)d\mu_2(x_2))d\mu_1(x_1))^{\frac{1}{p}} \end{aligned}\]

特别注意上述过程中

\[(\int_{X_2}(\int_{X_1}f^p(x_1,x_2)d\mu_1(x_1))^{\frac{q}{p}}d\mu_2(x_2))^{\frac{1}{q}}=(\underset{\parallel g\parallel_{L^r(X_2,\mu_2)}=1\atop g\ge0}{\sup}\int_{X_1\times X_2}f^p(x_1,x_2)g(x_2)d\mu_1(x_1)d\mu_2(x_2))^{\frac{1}{p}} \]

这一步是因为对任意线性空间\(E\), 记\(E^*\)是其对偶, 总有

\[\parallel x\parallel=\underset{f\in E^*\atop\parallel f\parallel\le1}{\sup}|\langle f,x\rangle|=\underset{f\in E^*\atop\parallel f\parallel\le1}{\max}|\langle f,x\rangle|,\quad\forall x\in E \]

现在考虑Holder不等式, 知

\[\int_{X_2}f^p(x_1,x_2)g(x_2)d\mu_2(x_2)\le(\int_{X_2}f^{p\cdot\frac{q}{p}}(x_1,x_2)d\mu_2(x_2))^{\frac{p}{q}}\cdot(\int_{X_2}g^{r}(x_2)d\mu_2(x_2))^{1-\frac{p}{q}} \]

得到

\[\begin{aligned} &\parallel\parallel f(\cdot,x_2)\parallel_{L^p(X_2,\mu_2)}\parallel_{L^q(X_2,\mu_2)}\\ \le&(\int_{X_1}(\underset{\parallel g\parallel_{L^r}(X_2,\mu_2)=1\atop g\ge0}{\sup}\int_{X_2}f^p(x_1,x_2)g(x_2)d\mu_2(x_2))d\mu_1(x_1))^{\frac{1}{p}}\\ \le&(\int_{X_1}(\underset{\parallel g\parallel_{L^r}(X_2,\mu_2)=1\atop g\ge0}{\sup}(\int_{X_2}f^{p\cdot\frac{q}{p}}(x_1,x_2)d\mu_2(x_2))^{\frac{p}{q}}\cdot(\int_{X_2}g^{r}(x_2)d\mu_2(x_2))^{1-\frac{p}{q}})d\mu_1(x_1))^{\frac{1}{p}}\\ &=(\int_{X_1}(\int_{X_2}f^q(x_1,x_2)d\mu_2(x_2))^{\frac{p}{q}}d\mu_1(x_1))^{\frac{1}{p}}=\parallel\parallel f(x_1,\cdot)\parallel_{L^q(X_2,\mu_2)}\parallel_{L^p(X_1,\mu_1)} \end{aligned}\]

命题即证.

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下面引入卷积的概念, 注意卷积在实或复可测函数上, 乃至装备了左不变Haar测度\(\mu\)的局部紧拓扑群上都能定义. 两个函数\(f,g\)的卷积是指:

\[f*g(x)=\int_Gf(y)g(y^{-1}\cdot x)d\mu(y), \]

注意\(\mu\)是左不变Haar测度意味着\(\mu\)是\(G\)上的Borel测度, 满足对任意Borel集\(A\)与\(G\)的元素\(a\), 总有\(\mu(a\cdot A)=\mu(A)\).

现在介绍卷积的Young不等式.

引理

若\(G\)是装备了左不变Haar测度\(\mu\)的局部紧拓扑群, 且\(\mu\)满足

\[\mu(A^{-1})=\mu(A),\quad\forall\text{Borel集}A \]

则对任意的\((p,q,r)\in[1,\infty]^3\), 若它们满足

\[\frac{1}{p}+\frac{1}{q}=1+\frac{1}{r} \]

则对任意\((f,g)\in L^p(G,\mu)\times L^q(G,\mu)\), 有

\[f*g\in L^r(G,\mu),\quad\parallel f*g\parallel_r\le\parallel f\parallel_p\cdot\parallel g\parallel_q. \]

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首先注意到届于左不变性与逆不变性, 对任意的\(x\in G\)与\(G\)上的任一可测函数\(h\), 都有

\[\int_Gh(y)d\mu(y)=\int_Gh(y^{-1}\cdot x)d\mu(y) \]

进而当\(r=\infty\), 命题正是Holder不等式. 现在考虑\(r<\infty\), 显见不失一般性可设\(f,g\)非负且非零, 注意

\[\frac{r}{r+1}(\frac{1}{p}+\frac{1}{q})=1 \]

故可记

\[(f*g)(x)=\int_Gf^{\frac{r}{r+1}}(y)g^{\frac{1}{r+1}}(y^{-1}\cdot x)f^{\frac{1}{r+1}}(y)g^{\frac{r}{r+1}}(y^{-1}\cdot x)d\mu(y) \]

进而根据Holder不等式:

\[\begin{aligned} &\int_Gf^{\frac{r}{r+1}}(y)g^{\frac{1}{r+1}}(y^{-1}\cdot x)f^{\frac{1}{r+1}}(y)g^{\frac{r}{r+1}}(y^{-1}\cdot x)d\mu(y)\\ \le&(\int_Gf^{\frac{r}{r+1}\cdot\frac{(r+1)p}{r}}(y)g^{\frac{1}{r+1}\cdot\frac{(r+1)p}{r}}(y^{-1}\cdot x)d\mu(y))^{\frac{r}{r+1}\frac{1}{p}}\cdot(\int_Gf^{\frac{1}{r+1}\cdot\frac{(r+1)q}{r}}(y)g^{\frac{r}{r+1}\cdot\frac{(r+1)p}{r}}(y^{-1}\cdot x)d\mu(y))^{\frac{r}{r+1}\frac{1}{q}}\\ =&(\int_Gf^p(y)g^{\frac{p}{r}}(y^{-1}\cdot x)d\mu(y))^{\frac{r}{r+1}\frac{1}{p}}\cdot(\int_Gf^{\frac{q}{r}}(y)g^p(y^{-1}\cdot x)d\mu(y))^{\frac{r}{r+1}\frac{1}{q}} \end{aligned}\]

再对两个积分分别考虑: