Let, for each $n\geq 1$, $x_{n}:[0,1]\longrightarrow X$ continuous, where $X$ stands for a banach space, with a norm $\|\cdot\|$. Assume that $X$ is reflexive and $x_{n}([0,1])\subset B(0,r)$ (the closed ball centered at zero and radius $r>0$). Denote by $\omega$ the weak topology of $X$.
As $B(0,r)$ is weakly compact, by the Eberlin-Smulian theorem, $B(0,r)$ is sequentially weakly compact. This means, that for each $t\in [0,1]$ we can pick a weakly convergent subsequence of $(x_{n}(t))_{n\geq 1}$ (for simplicity noted in the same way). Then, for each $t\in [0,1]$ define $x:[0,1]\longrightarrow X$ as $x(t):=\omega - \lim_{n} x_{n}(t)$ (the weak limit of $x_{n}(t)$, the picked weakly convergent subsequence of $x_{n}(t)$).
Is it true that $x$ is weakly continuous? That is, $x:[0,1]\longrightarrow (X,\omega)$ is continuous. I have tried the following.
Let any $t,t'\in [0,1]$ and $\varepsilon>0$. As $x_{n}(t)$ (resp. $x_{n}(t')$) congerves weakly to $x(t)$ (resp. $x(t')$) for each $x^{*}\in X^{*}$ (the topological dual of $X$) we have $$ |x^{*}(x(t)-x_{n}(t))|\longrightarrow 0,\quad |x^{*}(x(t')-x_{n}(t'))|\longrightarrow 0. \quad \textrm{whenever }n\rightarrow \infty\quad (*) $$ Also, as $x_{n}$ is continuous $$ \|x_{n}(t)-x_{n}(t')\|\longrightarrow 0,\quad \textrm{whenever } t\rightarrow t' .\quad (**) $$
Therefore, for each $x^{*}\in X^{*}$ we have $$ |x^{*}(x(t)-x(t'))|\leq |x^{*}(x(t)-x_{n}(t)|+\|x^{*}\| \|x_{n}(t)-x_{n}(t')\|+|x^{*}(x_{n}(t')-x(t')|, $$ which, in view of (*) and (**), tends to zero if $t\rightarrow t'$ and $n\rightarrow \infty$.
Many thanks in advance for your comments and suggestions.
Not true. Take $X=\mathbb R$, and $x_n(t) = \min( (2t)^n,1)$.