Let $(X,\|\cdot\|)$ a reflexive and separable Banach space, and note by $X^{*}$ its topological dual and $\omega$ its weak topology. Also, put $C_{\omega}(I,X)$ the space of the continuous mappings $f:I:=[0,1] \longrightarrow (X,\omega)$. Of course, the weak uniform convergence topology is determined by the basis
$$ V_{u}(\phi_{1}^{*},\ldots,\phi_{n}^{*},\epsilon):=\{f\in C_{\omega}(I,X): \sup_{t\in I}|\phi_{k}^{*}(f(t)-u(t))|<\epsilon,k=1,\ldots,n \} $$ where $u\in C_{\omega}(I,X) $, $\epsilon>0$, $f_{1}^{*},\ldots,f_{n}^{*}\in X^{*}$ and$n\in\mathbb{N}$. Thus, if $\psi_{k}^{*}$ is a separable sequence in $X^{*}$, a sequence $(f_{n})_{n\geq 1}\subset C_{\omega}(I,X)$ is weak uniformly convergent to a $f\in C_{\omega}(I,X)$ if, and only if, fixed $\epsilon >0$ and $k\geq 1$, there is $n_{0}\in\mathbb{N}$ such that
$$ |\psi_{k}^{*}(f_{n}(t)-f(t))|< \epsilon, \quad \forall t\in I, $$ for each $n\geq n_{0}$.
By the other hand, ¿somebody can show me an example "non-trivial" (not uniform convergent in norm, or not $\omega$-unif. conv. to the null mapping) of such weakly uniformly convergent sequence of $f_{n}$?
Thank you very much for your time!