Weak convergence on nonseparable spaces

Question

In nonseparable normed vector spaces $Z$, the dual $Z'$ is also nonseparable. Intuitively there will be a lot of functionals. In case of $l^1$, weak convergence implies strong convergence.

What is an example where $x_n \xrightarrow{w} x $ but $x \not \rightarrow x$ where $Z$ is a nonseparable space?

Answer 1

In $\ell_1$ weak convergence of a sequence implies norm convergence.

Let $X$ be any Banach space where you have a sequence converging weakly but not in norm. Let $Z=X\oplus Y$ where $Y$ is any non-separable space.

Answer 2

I tried to give an example, but I'm not sure anymore, especially about my statement $$\mu(\mathbb{N})=\sum\limits_{k=1}^\infty \mu(\lbrace k \rbrace)$$ Someone should have a look at it...

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Choose $Z=\ell^\infty$ which is nonseparable. Let be $\mathfrak{m}$ the normed vector space of all functions $\mu: \mathcal{P}(\mathbb{N})\rightarrow \mathbb{C}$ with $||\mu||_\mathfrak{m}:= \sup_{A\in\mathcal{P}(\mathbb{N})}|\mu(A)|<\infty$ and $\mu(A \cup B)=\mu(A) + \mu(B)$ for $A\cap B=\emptyset$. It is possible to show that $(\ell^\infty)' \simeq \mathfrak{m}$ with the isomorphism $\mu(A)=\langle \varphi_\mu, 1_A\rangle$. Also notice that the series on the right hand side of $$\mu(\mathbb{N})=\sum\limits_{k=1}^\infty \mu(\lbrace k \rbrace)$$ converges for all $\mu$ since the given norm on $\mathfrak{m}$ implies the existence of the left hand side. With that, take a look at the following sequence in $\ell^\infty$: $$x_n=(\beta_{k,n})_{k\in\mathbb{N}} \text{ with } \beta_{k,n}=\begin{cases} 1 & \text{ for } k\leq n\\ 0 & \text{ else } \end{cases}$$ First of all, this sequence does not converge with respect to the $||.||_\infty$-norm, since $||x_n-x_k||_\infty=1$ for all $n\neq k$. But for all $\mu$ in $\mathfrak{m}$ we get $$\langle \varphi_\mu, 1_{\lbrace 1, ... , n\rbrace}\rangle=\mu(\lbrace 1, ... , n\rbrace)=\sum\limits_{k=1}^n \mu(\lbrace k \rbrace) \rightarrow \sum\limits_{k=1}^\infty \mu(\lbrace k \rbrace)=\mu(\mathbb{N})<\infty$$ for $n\rightarrow \infty$. Since $\mu(\mathbb{N})=\langle \varphi_\mu, (1)_{n\in \mathbb{N}}\rangle$, we have $$x_n \overset{w}{\rightarrow} (1)_{n\in \mathbb{N}}$$