Show that $\limsup_{s,\:t\:\to\:\tau}\left\|x(s)-x(t)\right\|_E\ge r$ implies $\left\|\Delta x(\tau)\right\|_E\ge r$ Let $E$ be a normed $\mathbb R$-vector space, and $x:[0,\infty)\to\mathbb R$ be right-continuous. Assume $x$ $$x(t-):=\lim_{s\to t-}x(s)$$ exists for all $t\in O$ and let $$\Delta x(t):=x(t)-x(t-)\;\;\;\text{for }t\ge0.$$

Let $\tau\ge0$ and $r>0$ with $$\ell:=\limsup_{s,\:t\:\to\:\tau}\left\|x(s)-x(t)\right\|_E\ge r\tag1.$$ How can we conclude that $\left\|x(\tau)\right\|_E\ge r$?

Intuitively, this should be easy to verify, but I'm unable to prove it rigorously. Since we can show that the assumptions imply that $x$ is bounded on each compact interval, we've clearly got $\ell<\infty$. From this we can deduce that for all $\varepsilon>0$, there is a $\delta>0$ with $$\left\|x(s)-x(t)\right\|_E<\ell+\varepsilon\tag2$$ for all $s,t\ge0$ with $\max(|s-\tau|,|t-\tau|)<\delta$. In particular, $$\left\|x(\tau)-x(t)\right\|_E<\ell+\varepsilon\tag3$$ for all $t\in(\tau-\delta,\tau)$.
On the other hand, we have $$\left\|\Delta x(\tau)\right\|_E=\lim_{t\to\tau-}\left\|x(\tau)-x(t)\right\|_E\tag4$$ and hence there is a $\tilde\delta$ with $$\left|\left\|x(\tau)-x(t)\right\|_E-\left\|\Delta x(\tau)\right\|_E\right|<\varepsilon\tag5$$ for all $t\in(\tau-\delta\tilde,\tau)$.
But how can we conclude?
 A: Assume that $d = \Vert \Delta x(\tau)\Vert < r$. Choose $\epsilon > 0$ so small that $d + 2\epsilon < r$.
From $x(\tau-) =\lim_{s\to \tau-}x(s)$ it follows that there is an $\delta_1 > 0$ such that
$$
 \forall x \in (\tau - \delta_1, \tau]:  \Vert x(s) - x(\tau-) \Vert < \epsilon \, ,
$$
and from the right continuity it follows that that there is an $\delta_2 > 0$ such that
$$
 \forall x \in [\tau, \tau + \delta_2):  \Vert x(s) - x(\tau) \Vert < \epsilon \, .
$$
Now let $s, t \in (\tau-\delta_1, \tau + \delta_2)$. We have to consider three cases:

*

*If both $s$ and $t$ are $\le \tau$ then
$$
 \Vert x(s) - x(t)\Vert \le \Vert x(s) - x(\tau-)\Vert + \Vert x(\tau-) - x(t)\Vert < 2 \epsilon \, .
$$


*If both $s$ and $t$ are $\ge \tau$ then also
$$
 \Vert x(s) - x(t)\Vert \le \Vert x(s) - x(\tau)\Vert + \Vert x(\tau) - x(t)\Vert < 2 \epsilon \, .
$$


*If $s < \tau < t$ then
$$
\Vert x(s) - x(t)\Vert \le \Vert x(s) - x(\tau-)\Vert + \Vert x(\tau-) - x(\tau)\Vert + \Vert x(\tau) - x(t)\Vert < d+2 \epsilon \, .
$$
So in any case we have
$$ 
\Vert x(s) - x(t)\Vert <d +2\epsilon
$$
for all $s, t \in (\tau-\delta_1, \tau + \delta_2)$, and it follows that
$$
r \le \ell = \limsup_{s, t \to \tau} \Vert x(s) - x(t)\Vert  \le d +2\epsilon < r \, 
$$
which is a contradiction.
A: I believe it would be easier to argue the contrapositive: $\left\|\Delta x\left(\tau\right)\right\|_E < r$ implies that $\limsup_{s,t \to \tau} \left\|x(s) - x(t)\right\|_E < r$.
We suppose that $\left\|\Delta x\left(\tau\right)\right\|_E < r$. In particular that implies that there exists an $\varepsilon > 0$ and a $R> 0$ such that
$$\ \left\|x\left(\tau\right) - x\left(s\right)\right\|_E \leq r - \varepsilon$$
for all $s < \tau$ such that $\tau - s < R$.
Secondly, as $x$ is right continuous, there exists a $R' > 0$ such that
$$ \left\|x\left(\tau\right) - x\left(t\right)\right\|_E < \varepsilon/2 $$
for all $t \geq \tau$ such that $t- \tau < R'$. This implies that if $s,t \in \left]\tau - R, \tau + R'\right[$,
$$\left\|x(s)-x(t)\right\|_E \leq \left\|x(s) - x(\tau)\right\|_E + \left\|x(s) - x(\tau)\right\|_E < r - \varepsilon/2 < r.$$
From this inequality we may conclude that
$$\limsup_{s,t \to \tau} \left\|x(s) - x(t)\right\|_E < r.$$
