04Lp空间——Lp空间与范数

Lingfeng2024-06-11

04Lp空间——Lp空间与范数

定理1 当 $0< m(E) < +\infty$ 时, 有

\lim_{ p \to \infty } \left\Vert f \right\Vert _{p}= \left\Vert f \right\Vert _{\infty}

证明
一方面, 对于任意 $M<\left\Vert f \right\Vert_{\infty}$ , 显然 $M$ 不是 $f$ 的本性上界, 因此令

A=E(\lvert f \rvert >M)

显然有

m(A)>0

. 此时

\left\Vert f \right\Vert _{p}=\left( \int _{E}\lvert f(x) \rvert ^{p} \, dx \right)^{1/p}>M\cdot (m(A))^{1/p}

因此

\varliminf_{p \to \infty} \left\Vert f \right\Vert _{p}\geq M

再令

M\to \left\Vert f \right\Vert_{\infty}

, 故

\varliminf_{p \to \infty} \left\Vert f \right\Vert _{p}\geq \left\Vert f \right\Vert _{\infty}

另一方面, 显然有

\left\Vert f \right\Vert _{p}=\left( \int _{E}\lvert f(x) \rvert ^{p} \, dx \right)^{1/p}\leq \left( \int _{E} \left\Vert f \right\Vert _{\infty}^{p} \, dx \right)^{1/p}=\left\Vert f \right\Vert _{\infty}\cdot (m(E))^{1/p}

因此

\varlimsup_{p \to \infty} \left\Vert f \right\Vert _{p}\leq\left\Vert f \right\Vert _{\infty}

因此有

\lim_{ p \to \infty } \left\Vert f \right\Vert_{p}=\left\Vert f \right\Vert _{\infty}

定理2 若 $f$ , $g \in L^{p}(E)$ , $a$ , $b \in \mathbb{R}$ , 则 $af+bg \in L^{p}(E)$

证明
若 $1\leq p< +\infty$ , 注意到

\begin{align*} \lvert af(x)+bg(x) \rvert^{p} & \leq (\lvert af(x) \rvert+\lvert bg(x) \rvert )^{p} \\ & \leq (2\mathrm{max}\{ \lvert af(x) \rvert,\lvert bg(x) \rvert \})^{p} \\ & = 2^{p}(\mathrm{max}\{ \lvert af(x) \rvert,\lvert bg(x) \rvert \})^{p} \\ & \leq 2^{p}(\lvert a \rvert ^{p}\lvert f(x) \rvert^{p}+\lvert b \rvert^{p}\lvert g(x) \rvert^{p} ) \end{align*}

故

\int _{E}\lvert af(x)+bg(x) \rvert^{p} \, dx \leq 2^{p}\left( \lvert a \rvert ^{p}\int _{E} \lvert f(x) \rvert ^{p}\, dx + \lvert b \rvert ^{p}\int _{E}\lvert g(x) \rvert ^{p} \, dx \right) <+\infty

因此

af+bg \in L^{p}(E)

若 $p=+\infty$ , 由 $f$ , $g \in L^{\infty}(E)$ , 故存在 $M_{1}$ , $M_{2}$ , 使得

\lvert f(x) \rvert \leq M_{1} \quad \lvert g(x) \rvert \leq M_{2}

显然

\lvert af(x)+bg(x) \rvert \leq \lvert a \rvert \lvert f(x) \rvert +\lvert b \rvert \lvert g(x) \rvert \leq \lvert a \rvert M_{1}+\lvert b \rvert M_{2}<+\infty

故证毕.

引理设 $f \in L^{p}(E)$ , 则

\left\Vert f^{p} \right\Vert_{1} = \left\Vert f \right\Vert ^{p}_{p}

Young不等式 若 $a$ , $b>0$ , $p$ , $q>1$ , 且 $\frac{1}{p}+\frac{1}{q}=1$ , 则

ab \leq \frac{a^{p}}{p}+\frac{b^{q}}{q}

取等当且仅当

a^{p}=b^{q}

证明
先证明更广义的的Young不等式 设 $f(x)$ 单增, $a$ , $b>0$ , 有

ab\leq \int _{0}^{a}f(x) \, dx+\int _{0}^{b}f^{-1}(y) \, dy

取等当且仅当

b=f(a)

这是因为

\begin{align*} \int _{0}^{a}f(x) \, dx+ \int _{0}^{b} f^{-1}(y) \, dy & = \int _{0}^{a} f(x) \, dx + \int _{0}^{f^{-1}(b)} x\, df(x) \\ & = \int _{0}^{a}f(x) \, dx + b f^{-1}(b) - \int _{0}^{f^{-1}(b)} f(x)\, dx \\ & = bf^{-1}(b) + \int _{f^{-1}(b)}^{a}f(x) \, dx \end{align*}

因此

\begin{align*} \int _{0}^{a}f(x) \, dx+ \int _{0}^{b} f^{-1}(y) \, dy-ab & =\int _{f^{-1}(b)}^{a}f(x) \, dx-b(a-f^{-1}(b)) \\ & = \int _{f^{-1}(b)}^{a}(f(x)-b) \, dx\geq 0 \end{align*}

故令

f(x)=x^{p-1}

, 由

\frac{1}{p}+\frac{1}{q}=1

, 有

(p-1)(q-1)=1

因此

\begin{align*} ab & \leq \int _{0}^{a} x^{p-1} \, dx + \int _{0}^{b} y^{1/(p-1)} \, dy \\ & = \int _{0}^{a} x^{p-1} \, dx + \int _{0}^{b} y^{q-1} \, dy \\ & = \frac{a^{p}}{p}+\frac{b^{q}}{q} \end{align*}

Holder不等式 设 $p$ 与 $q$ 互为共轭指标, 则

\left\Vert fg \right\Vert_{1} \leq \left\Vert f \right\Vert _{p}\left\Vert g \right\Vert _{q}

即

\int _{E}\lvert f(x)g(x) \rvert \, dx \leq \left( \int _{E}\lvert f(x) \rvert^{p} \, dx \right)^{1/p}\left( \int _{E}\lvert g(x) \rvert ^{q} \, dx \right)^{1/q}

取等当且仅当

\frac{\lvert f \rvert ^{p}}{\lvert g \rvert ^{q}}= \frac{\left\Vert f \right\Vert ^{p}}{\left\Vert g \right\Vert ^{q}}

证明
只需设

a= \frac{\lvert f(x) \rvert }{ \left\Vert f \right\Vert _{p} } \quad b=\frac{\lvert g(x) \rvert }{\left\Vert g \right\Vert _{q}}

此时由Young不等式, 有

\begin{align*} \frac{\lvert f(x) g(x) \rvert}{\left\Vert f \right\Vert _{p}\left\Vert g \right\Vert _{q}}& \leq \frac{1}{p} \cdot\frac{\lvert f(x) \rvert ^{p}}{\left\Vert f \right\Vert _{p}^{p}} + \frac{1}{q} \cdot\frac{\lvert g(x) \rvert ^{q}}{\left\Vert g \right\Vert _{q}^{q} } \\ & = \frac{1}{p} \cdot\frac{\lvert f(x) \rvert ^{p}}{\left\Vert f^{p} \right\Vert_{1} } + \frac{1}{q} \cdot\frac{\lvert g(x) \rvert ^{q}}{\left\Vert g^{q} \right\Vert_{1} } \end{align*}

两边取积分即得

\frac{\left\Vert fg \right\Vert }{\left\Vert f \right\Vert _{p}\left\Vert g \right\Vert _{q}}\leq \frac{1}{p}+ \frac{1}{q}=1

故证毕.

Minkowski不等式 若 $1\leq p\leq+\infty$ , 则

\left\Vert f+g \right\Vert _{p}\leq \left\Vert f \right\Vert _{p}+\left\Vert g \right\Vert _{p}

证明
由Holder不等式, 有

\begin{align*} \left\Vert f+g \right\Vert ^{p}_{p} & = \left\Vert (f+g)^{p} \right\Vert_{1} \\ & \leq \left\Vert (f+g)^{p-1}f \right\Vert_{1}+ \left\Vert (f+g)^{p-1}g \right\Vert _{1} \\ & \leq \left\Vert (f+g)^{p-1} \right\Vert _{\frac{p}{p-1}}(\left\Vert f \right\Vert _{p}+\left\Vert g \right\Vert _{p}) \\ & = \left\Vert f+g \right\Vert _{p}^{p-1}(\left\Vert f \right\Vert _{p}+\left\Vert g \right\Vert _{p}) \end{align*}

故

\left\Vert f+g \right\Vert _{p}\leq \left\Vert f \right\Vert _{p}+\left\Vert g \right\Vert _{p}

广义Minkowski不等式 设 $f(x,y)$ 在 $\mathbb{R}^{n}\times \mathbb{R}^{n}$ 上可测, 若 $f(x,y) \in L^{p}(\mathbb{R}^{n})$ , $a.e.\forall y\in R$ , 且 $\int _{\mathbb{R}^{n}} \left\Vert f(x,y) \right\Vert_{p} \, dy<+\infty$ , 则

\left\Vert \int _{\mathbb{R}^{n}}f(x,y) \, dy \right\Vert_{p} \leq \int _{\mathbb{R}^{n}} \left\Vert f(x,y) \right\Vert _{p}\, dy

证明
首先注意到, 当

p=1

时, 由Fubini定理显然

\begin{align*} \left\Vert \int _{\mathbb{R}^{n}}f(x,y) \, dy \right\Vert_{1} & = \int _{\mathbb{R}^{n}} \left\lvert \int _{\mathbb{R}^{n}}f(x,y) \, dy \right\rvert \, dx \\ & \leq \int _{\mathbb{R}^{n}} \left( \int _{\mathbb{R}^{n}} \lvert f(x,y)\rvert \, dy \right) \, dx\\ & = \int _{\mathbb{R}^{n}} \left( \int _{\mathbb{R}^{n}} \lvert f(x,y)\rvert \, dx \right) \, dy \\ & = \int _{\mathbb{R}^{n}}\left\Vert f(x,y) \right\Vert_{1} \, dy \end{align*}

因此

\begin{align*} \left\Vert \int _{\mathbb{R}^{n}}f(x,y) \, dy \right\Vert_{p}^{p} & = \left\Vert \left( \int _{\mathbb{R}_{n}}f(x,y) \, dy \right)^{p} \right\Vert_{1} \\ & =\left\Vert \int _{\mathbb{R}^{n}} f(x,y)\left( \int _{\mathbb{R}_{n}}f(x,y) \, dy \right)^{p-1}\, dy \right\Vert _{1} \\ & \leq \int _{\mathbb{R}^{n}} \left\Vert f(x,y)\left( \int _{\mathbb{R}^{n}}f(x,y) \, dy \right)^{p-1} \right\Vert_{1} \, dy \\ & \leq \int _{\mathbb{R}^{n}}\left\Vert f(x,y) \right\Vert_{p}\left\Vert\left( \int _{\mathbb{R}^{n}}f(x,y) \, dy \right)^{p-1}\right\Vert_{\frac{p}{p-1}} \, dy \\ & = \int _{\mathbb{R}^{n}}\left\Vert f(x,y) \right\Vert _{p}\left\Vert \int _{\mathbb{R}^{n}}f(x,y) \, dy \right\Vert_{p}^{p-1} \, dy \\ & = \left\Vert \int _{\mathbb{R}^{n}}f(x,y) \, dy \right\Vert _{p}^{p-1}\int _{\mathbb{R}^{n}}\left\Vert f(x,y) \right\Vert _{p} \, dy \end{align*}

故

\left\Vert \int _{\mathbb{R}^{n}}f(x,y) \, dy \right\Vert_{p} \leq \int _{\mathbb{R}^{n}} \left\Vert f(x,y) \right\Vert _{p}\, dy

Jenson不等式 若 $\varphi$ 为凸函数, 则

\varphi\left( \frac{1}{m(E)}\int _{E}f(x) \, dx \right)\leq \frac{1}{m(E)}\int _{E}\varphi(f(x)) \, dx

证明
由积分中值定理, 存在 $x_{0} \in E$ , 使得

f(x_{0})=\frac{1}{m(E)}\int _{E}f(x) \, dx

由凸函数定义, 有

\varphi(f(x))\geq \varphi(f(x_{0}))+\varphi'(f(x_{0}))(f(x)-f(x_{0}))

两边取积分得到

\begin{align*} \int _{E}\varphi(f(x)) \, dx & \geq \int _{E}(\varphi(f(x_{0}))+\varphi'(f(x_{0}))(f(x)-f(x_{0}))) \, dx \\ & = \left( \int _{E}\varphi(f(x_{0})) \, dx \right) + \varphi'(f(x_{0}))\left( \int _{E}(f(x)-f(x_{0})) \, dx \right) \\ & =\varphi(f(x_{0})) m(E) \end{align*}

因此

\varphi\left( \frac{1}{m(E)}\int _{E}f(x) \, dx \right)\leq \frac{1}{m(E)}\int _{E}\varphi(f(x)) \, dx

推论1 在设 $X$ 为随机变量, 有 $\varphi(E(X))\leq E(\varphi(X))$ .

证明
这是因为显然

\varphi\left( \int_{X} x \, dF(x) \right)\leq \int _{X}\varphi(x) \, dF(x)

推论2 若

m(E)<+\infty

1<p<q<+\infty

, 则

L^{\infty}\subset L^{q} \subset L^{p} \subset L^{1}

证明
取 $\varphi(t)=t^{q/p}$ , 显然 $\varphi$ 是凸函数, 则

\begin{align*} \int _{E}\lvert f \rvert ^{q} \, dx & = \int _{E}\lvert f \rvert ^{p\cdot q/p} \, dx \\ & = \int _{E}\varphi(\lvert f \rvert ^{p}) \, dx \\ & \geq m(E)\cdot \varphi\left( \frac{1}{m(E)}\int _{E}\lvert f \rvert ^{p} \, dx \right) \\ & = m(E)^{1-q/p}\left( \int _{E}\lvert f \rvert ^{p} \, dx \right)^{q/p}<+\infty \end{align*}

故证毕.