作业帮英语翻译就是这段啊：More seriously:Note that if there is a weak limit of a sequence in $c_{0}$,it also is its weak$^{\ast}$-limit in $\ell^{\infty} = (c_{0})^{\ast\ast}$ (here $c_{0}$ is viewed as a subspace of $\ell^{\infty}$ via the canon
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/03 19:25:07
英语翻译就是这段啊：More seriously:Note that if there is a weak limit of a sequence in $c_{0}$,it also is its weak$^{\ast}$-limit in $\ell^{\infty} = (c_{0})^{\ast\ast}$ (here $c_{0}$ is viewed as a subspace of $\ell^{\infty}$ via the canon
英语翻译
就是这段啊：
More seriously:Note that if there is a weak limit of a sequence in $c_{0}$,it also is its weak$^{\ast}$-limit in $\ell^{\infty} = (c_{0})^{\ast\ast}$ (here $c_{0}$ is viewed as a subspace of $\ell^{\infty}$ via the canonical inclusion).Since the weak$^{\ast}$-topology is Hausdorff,it suffices to exhibit a subseries which is weak$^{\ast}$-convergent to some $s \in \ell^{\infty} \smallsetminus c_{0}$.The subseries $s_{k} = \sum_{n = 1}^{k} x_{2n} = (-1,1,-1,1,\cdots,-1,1,0,0,0,\cdots)$ weak$^{\ast}$-converges to $s = (-1,+1,\cdots)$:For all $(y_{n}) \in \ell^{1}$ and all $\varepsilon > 0$ there is $N$ such that $\sum_{n \geq N} |y_{n}| < \varepsilon$,therefore \[ |\langle s - s_{k},(y_{n}) \rangle_{\ell^{\infty},\ell^{1}}|< \varepsilon \quad \text{for $k \geq N$} \] by the Hölder inequality.
英语翻译就是这段啊：More seriously:Note that if there is a weak limit of a sequence in $c_{0}$,it also is its weak$^{\ast}$-limit in $\ell^{\infty} = (c_{0})^{\ast\ast}$ (here $c_{0}$ is viewed as a subspace of $\ell^{\infty}$ via the canon
这是latex源代码,有在线编译的网站,不过仅仅这一段代码是不能编译的.
把你的代码放在
\documentclass{article}
\usepackage{amsmath,amssymb}
\begin{document}
放在这里
\end{document}
编译结果,截图放在这里了.
如果不清楚,可以看我编译出来的结果（放大一下就清晰了）：