Let
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $\left(Y^{(n)}_k\right)_{k\in\mathbb N_0}$ be a time-homogeneous Markov chain on $(\Omega,\mathcal A,\operatorname P)$ and $$X^{(n)}_t=Y^{(n)}_{\lfloor nt\rfloor}\;\;\;\text{for all }t\in[0,1]$$
- $(N_t)_{t\ge0}$ be a Poisson process on $(\Omega,\mathcal A,\operatorname P)$ with intensity $1$ independent of $Y^{(n)}$ for all $n\in\mathbb N$ and $$Z^{(n)}_t:=\begin{cases}Y^{(n)}_{N_{nt}}&\text{for }t\in[0,1)\\ Z^{(n)}_{1-}\end{cases}$$ for $n\in\mathbb N$
Now, let $\tau_0:=0$, $$\tau_k:=\inf\left\{t>\tau_{k-1}:\Delta N_t=1\right\}\;\;\;\text{for }k\in\mathbb N$$ as well as $$\tau^{(n)}_k:=\frac{\tau_k}n\;\;\;\text{for }k\in\mathbb N$$ and $$\lambda^{(n)}_t:=\sum_{k=0}^\infty 1_{\left[\frac kn,\:\frac{k+1}n\right)}(t)\left(\tau^{(n)}_k+(nt-k)\left(\tau^{(n)}_{k+1}-\tau^{(n)}_k\right)\right)\;\;\;\text{for }t\ge0$$ for $n\in\mathbb N$.
How can we show that $$\sup_{t\ge0}\left|\lambda^{(n)}_t-t\right|=\sup_{t\in[0,\:1)}\left|\frac{N_{nt}}n-t\right|\tag1$$ for all $n\in\mathbb N$? (I'm unsure whether there isn't an error in the domain of one of the suprema. I could imagine that the left should be taken only over $[0,1)$ or the right actually over $[0,\infty)$.)
I've read that we obtain claim by noting that $N_{nt}=\lfloor n\frac{N_{nt}}n\rfloor$ for all $t\ge0$ and $n\in\mathbb N$, but while that's trivially true, I don't see how we can make use of it.
Moreover, I've read that the supremum is attained at one of the jump times (I guess the jump times of $X^{(n)}$, i.e. $\frac kn$ for $k\in\mathbb N$, are meant), but I don't know why.
However, we may observe the following: Let $t\in\left[\frac kn,\frac{k+1}n\right)$. Then, there is an $\alpha\in[0,1)$ with $$t=\frac{k+\alpha}n$$ and hence $$\lambda^{(n)}_t-t=\left(\tau^{(n)}_k-\frac kn\right)+\left(\tau^{(n)}_{k+1}-\tau^{(n)}_k\right)\alpha.\tag2$$ Since this is linear in $\alpha$, the supremum of $(2)$ is attained for $\alpha=0$ or $\alpha=1$. Thus, we should obtain $$\sup_{t\ge0}\left|\lambda^{(n)}_t-t\right|=\sup_{k\in\mathbb N_0}\left|\tau^{(n)}_k-\frac kn\right|=\frac1n\sup_{k\in\mathbb N_0}\left|\tau_k-k\right|\tag3.$$ Maybe we need to use $k=N_{\tau_k}$ almost surely such that $$\sup_{t\ge0}\left|\lambda^{(n)}_t-t\right|=\sup_{k\in\mathbb N_0}\left|\frac{N_{n\tau^{(n)}_k}}n-\tau^{(n)}_k\right|\tag4\;\;\;\text{almost surely}.$$ Now, clearly, $\left\{N_{n\tau^{(n)}_k(\omega)}(\omega):k\in\mathbb N_0\right\}=\left\{N_{\tau_k(\omega)}(\omega):k\in\mathbb N_0\right\}=\left\{N_t(\omega):t\ge0\right\}$, but this still doesn't yield $(1)$. However, it yields $$\left[\tau^{(n)}_k,\tau^{(n)}_{k+1}\right)\ni t\mapsto\frac{N_{nt}}n-t=k-t$$ which clearly attains its supremum at $t=\tau^{(n)}_k$ or $t=\tau^{(n)}_{k+1}$.
We may note that $\left(\frac{N_{nt}}n-t\right)_{t\ge0}$ is a martingale, but I'm not sure if we need this fact here.
New edit. An update has been introduced to highlight that equation (1) as stated does not hold unless the domain of the $\sup$ in both sides of the equation is $\mathbb{R}_{+}$.
Maybe I am missing something, but you have actually done everything. We have,
$\sup_{t\geq 0}\left|\lambda^{(n)}_t-t\right|=\frac{1}{n}\sup_{k\in\mathbb{N}_0} \left|\tau_k-k\right|$ (this is your equation (3))
$\sup_{t\geq 0} \left|\frac{N_{nt}}{n}-t\right|=\sup_{t\geq 0} \left|\frac{N_{nt}-nt}{n}\right|=\frac{1}{n}\sup_{t\geq 0} \left|N_{nt}-nt\right|=\frac{1}{n}\sup_{k\in\mathbb{N}_0} \left|N_{\tau_k}-\tau_k\right|$ (as before, without loss of optimality, we can restrict the domain to the jump moments of the Poisson process $N_t$, hence the last equality.)
As you pointed $N_{\tau_k}=k$, and the identity $\sup_{t\geq 0} \left|\frac{N_{nt}}{n}-t\right|=\sup_{t\geq 0}\left|\lambda^{(n)}_t-t\right|$ holds.
Remark. Identity (1) does not hold with the domains $t\geq 0$ on the left hand side and $t\in \left[\left.0,1\right)\right.$ on the right hand side. In fact, in this case $\sup_{t\in \left[\left.0,1\right)\right.} \left|\frac{N_{nt}}{n}-t\right|=\frac{1}{n}\sup_{k\in\left\{0,1,\ldots,N_n\right\}}\left|\tau_k-k\right|\vee \left|n-N_n\right|$ and we have that $\sup_{k\in\left\{0,1,\ldots,N_n\right\}}\left|\tau_k-k\right|\vee \left|n-N_n\right|<\sup_{k\in\mathbb{N}_0}\left|\tau_k-k\right|$ with positive probability, in particular, the strict inequality holds for all realizations $\omega\in\Omega$ so that $\tau_1(\omega)>n+\epsilon$ with $\epsilon>0$ -- which holds with positive probability.