I have come across a question regarding the preservation of cadlag under supremum of a sequence of cadlag functions that converge uniformly from this post :https://almostsuremath.com/2010/03/25/preservation-of-the-local-martingale-property/ proof of theorem 5.
Assume that we have $Y^n \to Y$ uniformly on bounded intervals ($Y_t^n$, $Y_t$ are functions of time $t\ge 0$.)
Set $$M_t := \sup_n (Y_t^n)^+.$$ I am trying to show that $M$ is cadlag, i.e. continuous from the right and has left limits.
It is clear that $(Y^n)^+$ is cadlag.
Right Continuity: $M_t = \lim_{n\to \infty} \max_{m\le n} (Y_t^m)^+$.
I believe that if we have $Y^n \to Y$ uniformly on compacts, then so does $(Y^n)^+ \to Y^+$ uniformly on compacts. Is this true?
Assuming this, since we have $|\max_{m\le n} (Y^m_s)^+ - \max_{m \le n} (Y^m_t)^+|\le \max_{m\le n}|(Y^m_s)^+ - (Y_t^m)^+| \le \sup_n |(Y^n_s)^+ - (Y_t^n)^+|$,
and by taking limits $n\to \infty$, $|M_s - M_t| \le \sup_n |(Y_s^n)^+ - (Y_t^n)^+|$, we get from uniform convergence on compacts that for $s>t$ close enough, $\lim_{s\downarrow t} M_s = M_t$.
Left hand limits: Similarly, we can get that for any sequence $s_n \uparrow t$, $s_n<t$, $$|M_{s_n} - M_{s_m}| \le \sup_k |(Y_{s_n}^k)^+ - (Y_{s_m}^k)^+|$$and since the right hand side converges to $0$ as $n , m \to \infty$, $M_{s_n}$ is a cauchy sequence and has a limit.
I am not sure if the assumption on uniform convergence being preserved for the positive part of the functions is true. Does this hold and the rest of the proof follows as I claim? I would greatly appreciate some input on this.