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In El Fatini and Boukanjime "Stochastic analysis of a two delayed epidemic model incorporating Lévy processes with a general non-linear transmission" paper, can someone give a detailed explanation of the proof of Proposition 2.1 part (1)?

The highlighted part is what I'm having trouble with.

  1. What does the $U(u)$ mean? is it "the smallest value of $V$ such that $x_i\geq u$ or $x_i \leq \frac{1}{u}$"?

  2. I am confused with: $$V(S(0),I(0))+KT\geq \epsilon U(m),$$ where $\epsilon \in(0,1)$. They let $m\rightarrow \infty$, which leads to $\infty>V(S(0),I(0))+KT=\infty$. Shouldn't it be $V(S(0),I(0))+KT=\infty=\infty \geq \epsilon\cdot\infty = \infty$? So the inequality actually holds and isn't a contradiction?

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  • $\begingroup$ 1. Maybe this is what you mean, but I would rather phrase the description of $U(u)$ as follows: It is the infimum of $V$ on the set of $x$, which satisfy that $x_i \geq u$ or $x_1 \leq 1/u$ for $i=1$ or $i=2$. This is clearly related to the stopping time defined in (6), which is the stopping time for entering this set. $\endgroup$
    – Fabian Wirth
    Commented Oct 9, 2023 at 15:03
  • $\begingroup$ 2. The contradiction comes from two things: From the assumptions and the definitions we know that $V(S(0),I(0)) + KT$ is finite. From the derived inequalities on the other hand, if follows that it has to be larger than any bound. These two things do not go together. OK? $\endgroup$
    – Fabian Wirth
    Commented Oct 9, 2023 at 15:07
  • $\begingroup$ 1. is clear now. I still don't understand 2. as in the paper it states $$V(S(0),I(0)) +KT\geq U(m).$$ So when $m \rightarrow \infty$ then $V(S(0),I(0)) +KT = \infty$. But also when $m \rightarrow \infty$ then $U(m)=\infty$ so we have $\infty \geq \infty$, no? $\endgroup$
    – Leo
    Commented Oct 9, 2023 at 15:11
  • $\begingroup$ This is another way of saying what I tried to say. The inequality just above says that *for any $m$" we have $V(S(0),I(0)) + KT \geq \varepsilon U(m)$. If this inequality is true for all $m$, then we can let $m$ tend to $\infty$. This does not change the left hand side, as it does not depend on $m$. But the right hand side tends to $\infty$. So the only chance that all these inequalities are true is that $V(S(0),I(0)) + KT=\infty$. Note that it does not make sense to say $U(m) = \infty$. For each specific $m>0$, this value is finite. $\endgroup$
    – Fabian Wirth
    Commented Oct 9, 2023 at 15:16
  • $\begingroup$ I meant $\lim_{m\rightarrow \infty} U(m) = \infty$ as, of course, we are talking about limits. $\endgroup$
    – Leo
    Commented Oct 9, 2023 at 15:19

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