$\lim_{n\to\infty}\left\Vert f_n-f\right\Vert _{L^1}=0, \mbox{且} f_n \to f$?
本帖最后由 flyy 于 2023-5-22 15:33 编辑设 $\displaystyle \left\{f_n\right\}\subset L^1(\mathbb{R}^d)$, 且
$$\begin{aligned} \exists\ p\gt 1,\mathrm{ s.t.} \left\Vert f_{n+1}-f_n\right\Vert _{L^1}\leq\frac{1}{n^{2p}}, \forall\ n\geq 1. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
试证明:
$$\begin{aligned} \exists\ f\in L^1(\mathbb{R}^d),\mathrm{ s.t.} \lim_{n\to\infty}\left\Vert f_n-f\right\Vert _{L^1}=0, \mbox{且} f_n \to f, \mbox{a.e.} x\in\mathbb{R}^d. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
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(1)、 由
$$\begin{aligned} &\left\Vert f_{n+p}-f_n\right\Vert _{L^1}\leq \sum_{k=n+1}^{n+p} \left\Vert f_{k+1}-f_k\right\Vert _{L^1} \leq \sum_{k=n+1}^{n+p} \frac{1}{k^{2p}}\xrightarrow{n\to\infty}0 \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
知 $\displaystyle \left\{f_n\right\}$ 是 $\displaystyle L^1$ 中的 Cauchy 列. 由 $\displaystyle L^1$ 完备知
$$\begin{aligned} &\exists\ f\in L^1(\mathbb{R}),\mathrm{ s.t.} \lim_{n\to\infty}\left\Vert f_n-f\right\Vert =0\\ \Rightarrow& f_n\mbox{依测度收敛于}f\stackrel{\mbox{Riesz}}{\Rightarrow} \exists\ n_i,\mathrm{ s.t.} f_{n_i}\to f,\mbox{a.e.}.\qquad(I) \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
(2)、 设 $\displaystyle A_n=\left\{x\in\mathbb{R}^d; |f_{n+1}(x)-f(x)|\geq \frac{1}{n^p}\right\}$, 则
$$\begin{aligned} \frac{1}{n^{2p}}\geq \left\Vert f_{n+1}-f_n\right\Vert _{L^1}\geq \int_{A_n}|f_{n+1}-f_n|\mathrm{ d} x \geq \frac{1}{n^p}\cdot mA_n. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
从而 $\displaystyle mA_n\leq\frac{1}{n^p}$. 设 $\displaystyle A=\varlimsup_{n\to\infty}A_n$, 则
$$\begin{aligned} mA=&m\left(\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k\right) \leq m\left(\bigcup_{k=n}^\infty A_k\right) \leq \sum_{k=n}^\infty mA_k\leq \sum_{k=n}^\infty \frac{1}{k^p}, \forall\ n\geq 1. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
令 $\displaystyle n\to\infty$ 得 $\displaystyle mA=0$. 对
$$\begin{aligned} &\forall\ x\in\mathbb{R}^d\backslash A=\varliminf_{n\to\infty}A_n^c =\bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k^c,\\ &\exists\ n,\mathrm{ s.t.} \forall\ k\geq n, |f_{k+1}(x)-f_k(x)|\lt\frac{1}{k^p}. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
于是 $\displaystyle \forall\ k\gt l\geq n$,
$$\begin{aligned} |f_k(x)-f_l(x)|\leq\sum_{i=l}^{k-1}|f_{i+1}(x)-f_i(x)| \leq \sum_{i=l}^\infty \frac{1}{i^p}\xrightarrow{l\to\infty}0. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
这就证明了 $\displaystyle \left\{f_k(x)\right\}$ 是 $\displaystyle \mathbb{R}$ 中的Cauchy 列, $\displaystyle \lim_{k\to\infty}f_k(x)=g(x)$ 存在. 联合 $\displaystyle (I)$ 知
$$\begin{aligned} g(x)=f(x),\mbox{a.e.}\Rightarrow \lim_{n\to\infty}f_n(x)=f(x),\mbox{a.e.}. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
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