0
$\begingroup$

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!

$\endgroup$
3
  • $\begingroup$ What does $v_n(t)$ mean? $v_n$ is a measure; its arguments are sets, not points. $\endgroup$ 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
    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.) $\endgroup$ Jan 21, 2014 at 18:06

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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
    Jan 27, 2014 at 23:15
  • $\begingroup$ $V\subset B$. Consider the weak$^*$ topology on $B=L^*$; restrict this topology to $V$. $\endgroup$ 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
    Jan 28, 2014 at 20:56
  • $\begingroup$ Moreover, could you give me the reference for $B$ and $L$? $\endgroup$
    – CodeGolf
    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 $0 \le a < b\le 1]$. $\endgroup$ Jan 31, 2014 at 15:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.