# 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 $$a=t_0<\cdots. Set $$|\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 $$a ($$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 $$0 and $$(s,t]\cap I=\emptyset$$ implies $$\|x(s)-x(t)\|_{E_1}. 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?

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

• Thank you for your answer. I'm sorry for the late response. I wasn't able to think about this until now. How do we see that $t\mapsto x(t)-\sum_{s\le t}\Delta x(s)$ is continuous? And why does the sum $\sum_{s\le t}\Delta x(s)$ even exist? Mar 4, 2022 at 15:51
• I think instead of $\sum_{s\le t}\Delta x(s)$ we need to consider $\sum_{i=1}^m1_{[z_i,\:\infty)}(t)\Delta x(z_i)$. Using this, the function $\tilde x(t):=x(t)-\sum_{i=1}^m1_{[z_i,\:\infty)}(t)\Delta x(z_i)$ is still not continuous; but càdlàg. And we can show that there is a $\delta>0$ with $\|\tilde x(s)-\tilde x(t)\|_E<r$ for all $s,t\in[a,b]$ with $|s-t|<\delta$. Mar 4, 2022 at 16:05
• But please correct me, if I'm missing something. Mar 4, 2022 at 16:05