If $\xi:[0,1]\to[0,1)$ is decreasing, is $[0,\infty)\ni t\mapsto\xi(1)^{\lfloor t\rfloor}\xi(x-\lfloor x\rfloor)$ eventually nonincreasing?

Question

Let $$\iota:[0,\infty)\to[0,1)\;,\;\;\;x\mapsto x-\lfloor x\rfloor,$$ $\xi:[0,1]\to[0,1)$ be decreasing and $$\tilde\xi(t):=\xi(1)^{\lfloor t\rfloor}\left(\xi\circ\iota\right)(t)\;\;\;\text{for }t\ge0.$$

Are we able to show that $[t_0,\infty)\ni t\mapsto\tilde\xi(t)$ is nonincreasing for some sufficiently large $t_0\ge0$?

The first factor $$[0,\infty)\ni t\mapsto\xi(1)^{\lfloor t\rfloor}\tag1$$ is clearly nonincreasing.

Now let $t>s\ge0$. If $s=n+r_1$ and $t=n+r_2$ for some $n\in\mathbb N_0$ and $r_i\in[0,1)$ with $r_1<r_2$, then we clearly obtain $\tilde\xi(s)>\tilde\xi(t)$.

So, the crucial case is $s=m+r_1$ and $t=n+r_2$ for some $m,n\in\mathbb N_0$ and $r_i\in[0,1)$ with $m<n$. Maybe we can find $t_0\ge0$ so that as long as $t>s\ge t_0$, it still holds $\tilde\xi(s)>\tilde\xi(t)$.

Answer 1

If I don't miss something, it is almost trivially true that $\tilde{\xi}$ is decreasing without the additional restriction on domain you impose. For $n-1 \leq s < t \leq n$ one has $$\tilde{\xi}(t) = \xi(1)^{\lceil t\rceil - 1} \xi(t - \lceil t \rceil + 1) = \xi(1)^{\lfloor s\rfloor} \xi(t - \lceil t \rceil + 1) < \xi(1)^{\lfloor s \rfloor} \xi(s -\lfloor s\rfloor) = \tilde{\xi}(s)$$ since $t - \lceil t \rceil + 1 > s - \lfloor s \rfloor$.