If $x$ is a càdlàg function and $f$ has compact support, how can we approximate $\sum_{s\in(a,\:b]}f(\Delta x(t))$?

Let $E_i$ be a normed $\mathbb R$-vector space and $x:[0,\infty)\to E_1$ be right-continuous. Assume $$x(t-):=\lim_{s\to t-}x(s)$$ exists for all $t\ge0$ and let $\Delta x(t):=x(t)-x(t-)$ for $t\ge0$.

Let $f:E_1\to E_2$ be continuous with $0\not\in K:=\operatorname{supp}f$ and $b>a\ge0$. Moreover, let $$S_\varsigma:=\sum_{i=1}^kf(x(t_k)-x(t_{k-1}))$$ for $\varsigma=(t_0,\ldots,t_k)$, where $k\in\mathbb N$ and $a=t_0<\cdots<t_k$. Set $$|\varsigma|:=\max_{1\le i\le k}(t_i-t_{i-1}).$$

How can we show that $$S_\varsigma\to\sum_{t\in(a,\:b]}f(\Delta x(t))\tag1$$ as $|\varsigma|\to0$?

The idea is pretty clear. Since $0\not\in K$, $$r:=\operatorname{dist}(0,K)>0.$$ Moreover, $$I:=\{t\in(a,b]:\left\|\Delta x(t)\right\|_{E_1}\ge r\}$$ is finite and hence equal to $\{t_1,\ldots,t_n\}$ for some $n\in\mathbb N_0$ and $a<t_1<\cdots<t_n\le b$ ($n=0$, if no jump has size greater than or equal to $r$).

Now we can clearly choose $\delta>0$ such that $|\varsigma|<\delta$ implies $$\left|(t_{i-1},t_i]\cap I\right|\le1\tag2$$ for all $i\in\{1,\ldots,k\}$.

Next, I guess we need to choose $\delta$ even smaller to ensure that $s,t\in(a,b]$ with $0<t-s<\delta$ and $(s,t]\cap I=\emptyset$ implies $\|x(s)-x(t)\|_{E_1}<r$. Can we show this? (I've asked for that separately.)

But even when we are able to show this, I struggle to conclude. So, how do we need to argue?

Solution 1:

First, as you assumed your function is càdlag, for all $c>0$ there are a finite number of $t \in [a,b]$ such that $|\Delta x(t)| > c$. You can see a proof here.

Note that the function $t \mapsto x(t) - \sum \limits_{s \le t}\Delta x(s)$ is continuous, so it is uniformly continuous on $[a,b]$. In what follows, I'm going to keep your notation $r = \mbox{dist}(0, K)$. Denote $z_1, ..., z_m$ all $z \in [a,b]$ such that $|\Delta x(z)| > r$. Let $\varepsilon > 0$.

Because of the uniform continuity, there exists $\delta > 0$ such that for all $a \le s < t \le b$ such that $|t-s| < \delta$, $|x(t)-x(s) - \sum \limits_{s < z \le t} \Delta x(z)| < r$.

Also, since the function $(x_1, x_2) \mapsto f(x_2-x_1)$ is continuous at $(x(z_k-), x(z_k))$ for all $k$, and using that $x(t) \underset{t \to z_k-}{\to} x(z_k-)$, $x(t) \underset{t \to z_k+}{\to} x(z_k)$, the function $(t_1, t_2) \mapsto f(x(t_2)-x(t_1))$ is continuous at $(z_k-, z_k)$ for all $k$. Hence there exists $\delta' > 0$ such that for all $k \in [\![1,m]\!]$, for $t_1 \in [z_k-\delta', z_k[$ and $t_2 \in [z_k, z_k+\delta'[$, $|f\big(x(t_2)-x(t_1)) - f\big(x(z_k)-x(z_k-)\big)| < \frac{\varepsilon}{m}$.

Remark: at this point you should make sure that the two previous arguments convince you, because the conclusion follows from them.

Then for $(t_k^n)_{0 \le k \le n}$ with $a = t_0 < \cdots < t_n = b$, such that $\varsigma = \max\limits_{0 \le < n} |t_{k+1}^n - t_k^n| < \min(\delta, \delta')$, $$ \sum \limits_{i=1}^n f\big(x(t_i^n)-x(t_{i-1}^n)\big) = \sum\limits_{\overset{i=1,...,n}{\exists k: t_{i-1}^n < z_k \le t_i^n}} f\big(x(t_i^n)-x(t_{i-1}^n)\big)$$

Indeed, for those $i$ such that there is no $z_k \in ]t_{i-1}^n, t_i^n]$, by our assumption on $\delta$, $|x(t_{i-1}^n) - x(t_i^n) - 0| < r$, so by definition of $r$, $f\big(x(t_i^n)-x(t_{i-1}^n)\big) = 0$.

Then, for simplification we can also assume that $\varsigma$ is strictly less than $\min(z_1-a, \min z_k-z_{k-1}, b-z_m)$ so that there is exactly one $i_k$ such that $t_{i_k-1}^n < z_k \le t_{i_k}^n$ for every $k$. By our second assumption on $\delta'$, for $k=1,...,m$, $$\big|f\big(x(t_{i_k}^n)-x(t_{i_k-1}^n)\big) - f\big(x(z_k)-x(z_k-)\big)\big| < \frac{\varepsilon}{m}$$

Summing over all $k$, we get $$\Big| \sum \limits_{i=1}^n f\big(x(t_i^n)-x(t_{i-1}^n)\big) - \sum \limits_{k=1}^m f\big(\Delta x(z_k)\big) \Big| < m \frac{\varepsilon}{m} = \varepsilon$$