If $U_{n} \to U$ weakly* in $BV[0,1]$ is it true that $U_{n}\mid_{[0,\frac{1}{2}]} \to U\mid_{[0,\frac{1}{2}]}$ weakly* in $BV[0,\frac{1}{2}]$?
I'll give some explication. Let $U_{n} \in BV[0,1]$ and $U_n \to U$ weakly*, i.e. for any $y \in C[0,1]$
$ \int\limits_{0}^{1} y(t)dU_{n}(t) \to \int\limits_{0}^{1}y(t)dU(t) $ Restrictions $U_{n} \mid _{[0,\frac{1}{2}]}, U \mid_{[0,\frac{1}{2}]} \in BV[0,\frac{1}{2}]$. For any $y \in C[0,\frac{1}{2}]$ such that $y(\frac{1}{2}) = 0$ we have $ \int\limits_{0}^{\frac{1}{2}} y(t)dU_{n}(t) \to \int\limits_{0}^{\frac{1}{2}}y(t)dU(t) $ But is it true for any $y \in C[0,\frac{1}{2}]$? In other words, it it true, that $U_{n}\mid_{[0,\frac{1}{2}]} \to U\mid_{[0,\frac{1}{2}]}$ weakly*?