Consider $c_0 = \{ (x_n)_n\in \mathbb{R}^{\mathbb{N}}: \lim_{n\to\infty} x_n = 0\}\le (l_{\infty},\|\cdot\|_{\infty}) $.
Pointwise convergence and weak convergence are not equivalent in $c_0$, but I came up with the following reasoning:
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Suppose that $(x_n)_n$ is weakly convergent in $c_0$. Let $\phi\in c_0^*$, then $\phi((x_n)_n) = \sum_{n=1}^{\infty} x_na_n$ for some $(a_n)_n\in l_1$. In particular we can choose $(a_n)_n = e_n, n\in\mathbb{N}$ ($n$-th basis element), which would imply that $(x_n)_n$ converges pointwise in $c_0$.
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Suppose $\forall n: x_n\to y_n$ (pointwise convergence of $(x_n)_n$ to $(y_n)_n\in c_0$). Then $|x_n-y_n|\to 0,\forall n$, so that $\sup_n |x_n-y_n| = \| (x_n-y_n)_n\|_{\infty}\to 0$. By continuity of $\phi\in c_0^*$ we would have $\phi((x_n)_n)\to \phi((y_n)_n)$, which is the weak convergence.
Where is my mistake? How can I prove that there is a pointwise convergent sequence in $c_0$ which is not weakly convergent?
Best Answer
Pointwise or weak convergence applies to sequences of members of $c_0$, not to individual members of $c_0.$ Let $x^{(m)}=(x_{m,n})_{n\in\Bbb N}\in c_0$ where $x_{m,m}=m^2$ and $x_{m,n}=0$ if $m\ne n$. For each $n$ we have $\lim_{m\to\infty}x_{m,n}=0.$ But $\phi=(1/n^2)_{n\in\Bbb N}\in c_0^*=\ell_1$ and $\phi(x^{(m)})=1$ for every $m.$
If we regard each $x^{(m)}$ as a function with domain $\Bbb N$ (where $x^{(m)}(n)=x_{m,n}$) then $(x^{(m)})_m$ converges to $0$ at each point of the domain $\Bbb N. $But if we regard each $x^{(m)}$ as a function (a functional) with domain $c_0^*$ (where $x^{(m)}(\psi)=\psi(x^{(m)})$ for each $\psi\in c_0^*)$ then $(x^{(m)})_m$ does not converge to $0$ at each point of the domain $c_0^*.$