I have a question:

Let $D$ be the space of cadlag functions defined on $[0,1]$ and $V$ be its subspace consisting of $x$ with finite variation and $x(0)=0$.

Define $TV(x)$ as the total variation of $x\in V$. Denote $I=\{0<t_1<t_2=1\}$. Now if we equip $V$ with the weak-$\ast$ topology (for $V$ can be indentified with a closed subspace of the space of finite measures on $[0,1]$), then $v_n\stackrel{\ast}{\to}v_0$ means for any continuous function $f$ on $[0,1]$ we have

$$\int_{[0,1]}f(t)dv_n\to\int_{[0,1]}f(t)dv_0(t)$$

We have in particular

$$v_n\stackrel{\ast}{\to}v_0\Rightarrow v_n(1)\to v_0(1)$$

and the boundedness of $\{TV(v_n)\}$ implies a weak-$\ast$ convergent subsequence $\{v_{n_k}\}$.

Now my question is whether we can find another convergence that is similar to weak-$\ast$ convergence such that

(i) The boundedness of $\{TV(v_n)\}$ implies a weak-$\ast$ convergent subsequence $\{v_{n_k}\}$.

(ii) This convergence implies $v_n(t_i)\to v_0(t_i)$ for $i=1,2$.

Does someone know this type of convergence? Many thanks for your help!