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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!

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  • $\begingroup$ What does $v_n(t)$ mean? $v_n$ is a measure; its arguments are sets, not points. $\endgroup$
    – Nate Eldredge
    Commented Jan 21, 2014 at 15:52
  • $\begingroup$ Thanks for your reply. Since $v_n$ is a cadlag function with finite variation, so there is a one-to-one mapping between $v_n$ and some (Borel)measure $\nu_n$ such that $v_n(t)=\nu_n([0,t])$ for every $t\in [0,1]$ $\endgroup$
    – CodeGolf
    Commented Jan 21, 2014 at 17:39
  • $\begingroup$ Oh, I see now. This seems to me like a reasonable question and I don't understand the down votes. (I have a feeling the answer is going to be "no", however.) $\endgroup$
    – Nate Eldredge
    Commented Jan 21, 2014 at 18:06

1 Answer 1

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The dual of the space $L$ of left continuous functions on $[0,1]$ that have right limits and vanish at zero is the space $B$ of functions of bounded variation that vanish at zero, with the duality pairing given by the right Cauchy refinement integral. The integral of $1_{[0,t]}$ with respect to $v\in B$ is $v(t)$ for $t \in (0,1]$, so point wise evaluation on $B$ is weak$^*$ continuous.

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  • $\begingroup$ It is not so clear for me. I don't understand why you introduce $L$ and $B$, what is exactly the convergence for my space $V$? $\endgroup$
    – CodeGolf
    Commented Jan 27, 2014 at 23:15
  • $\begingroup$ $V\subset B$. Consider the weak$^*$ topology on $B=L^*$; restrict this topology to $V$. $\endgroup$
    – Bill Johnson
    Commented Jan 28, 2014 at 19:50
  • $\begingroup$ Thanks for your reply. Could you specify the details? $L^{\ast}=B$? Thus for $\{v_n\}\subset V$ and $v_n\to v\in V$ under this topology, can we conclude that $v_n(t)\to v(t)$? $\endgroup$
    – CodeGolf
    Commented Jan 28, 2014 at 20:56
  • $\begingroup$ Moreover, could you give me the reference for $B$ and $L$? $\endgroup$
    – CodeGolf
    Commented Jan 28, 2014 at 20:57
  • $\begingroup$ I don't know a reference, but I have assigned this duality as homework in undergraduate real analysis courses for students who did not have measure theory. The main lemma to prove is that $L$ is the closed linear span in the sup norm of the indicator functions of intervals of the form $(a,b]$ with $0 \le a < b\le 1]$. $\endgroup$
    – Bill Johnson
    Commented Jan 31, 2014 at 15:46

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