If $x:[0,\infty)\to E$ is càdlàg and $\tau_0:=0$, show that $\inf\left\{t>\tau_{n-1}:\Delta x(t)\in B\right\}\xrightarrow{n\to\infty}\infty$

Question

Let $E$ be a normed $\mathbb R$-vector space and $x:[0,\infty)\to E$ be càdlàg with $x(0)=0$. Moreover, let $x(0-):=x(0)$, $$x(t-):=\lim_{s\to t-}x(s)\;\;\;\text{for }t>0$$ and $$\Delta x(t):=x(t)-x(t-)\;\;\;\text{for }t\ge0.$$

Given $B\subseteq E$ bounded below, let $\tau_0:=0$ and $$\tau_n:=\inf\underbrace{\left\{t>\tau_{n-1}:\Delta x(t)\in B\right\}}_{=:\:I_n}$$ for $n\in\mathbb N$.

How can we show that $\tau_1>0$ and $\tau_n\xrightarrow{n\to\infty}\infty$?

I know that $\left|\left\{t\in[a,b]:\Delta x(t)\ne0\right\}\right|\in\mathbb N_0$ for all $a,b\in\mathbb R$ and hence $\left\{t\ge0:\Delta x(t)\ne0\right\}$ is countable. Does the claim somehow follow from this fact?

For example, if $\tau_1=0$, then we would find a nonincreasing $(t_n)_{n\in\mathbb N}\subseteq I_1$ with $t_n\xrightarrow{n\to\infty}0$. Is this a contradiction?

I know how we can prove a similar statement for any right-continuous $y:[0,\infty)\to E$. Assuming that $B$ is closed and $I:=\{t\ge0:y(t)\in B\}$ is nonempty, we can easily show that $\tau:=\inf I\in I$. Moreover, if $y(0)\not\in B$, we can show that $\tau>0$.

Answer 1

We know from Show that if $x$ is càdlàg, $\left\{t\in[a,b]:d\left(x(t),x(t-)\right)>c\right\}$ is finite that for every $0 \le a < b$ the set $$ J(a, b) := \{t \in [a, b] :\Delta x(t)\in B \} $$ is empty or finite.

Now assume that $\tau_0< \tau_1 < \cdots < \tau_n$ have already been defined. There are two possibilities:

• The set $I_n = \{ t > \tau_n :\Delta x(t)\in B \}$ is empty. In that case, $\tau_{n+1}$ can not be defined, and the construction is finished.

• The set $I_n$ is not empty, so that $\tau_{n+1} = \inf I_n$ is well-defined. Since $J(\tau_n, \tau_n + 1)$ is at most finite, $\tau_{n+1}$ must be strictly greater than $\tau_n$.

So the construction gives a finite or infinite sequence of strictly increasing $\tau_n$. If the construction does not stop after finitely many steps, then necessarily $\lim_{n \to \infty} \tau_n = \infty$:

Otherwise $\tau_n \to c$. For every $n$ is $\tau_{n+1} = \inf I_n < c$, so that there is a $t_n \in (\tau_n, c)$ with $\Delta x(t_n) \in B$. But then $J(0, c)$ contains infinitely many elements, which is not possible.

Answer 2

So $\Delta x$ measures all the jumps of $x$ and you said you already know that any compact interval contains only finitely many jumps. As $B$ is bounded below, $0 \notin B$, so for any $a,b$ we also have $|t \in [a,b] : x(t) \in B| \in \mathbb{N}_0$. Now $\tau_n$ is just the $n$th jump of a size in $B$. So if there are only finitely many such jumps in any compact interval, it follows that $\tau_n \rightarrow \infty$.

Edit: To get $\tau_1>0$ consider say $[a,b]=[0,1]$. If there are no jumps in this interval, than $\tau_1>1$. If there are finitely many, the infimum among them is greater than zero as $\Delta x(0)=0$ by definition.

For the convergence to infinity, for any $N$ there exist only finitely many $n$ with $x(t) \neq 0$ and $t < N$, hence there exists a finite $n(N)$ such that $\tau_{n(N)}>N$ so the limit of the $\tau_n$ has to be infinity.