Let $f:X\times \mathbb{R}\rightarrow X$ continuous, where $(X,d)$ is compact metric space. I wonder if this function is continuous:
$$s\rightarrow \sup_{(x,t)\in X\times \mathbb{R}}d(f(x,t),f(x,t+s)).$$
I see that the function is well defined because an upper bound is the diameter of the set $X$, but to show the continuity there is a process of change that limits I do not understand , appreciate if you could give me some suggestions on how to test that the function is continuous.
The function (call it $g$) is actually not continuous in general. Note that by definition $g(0) = 0$.
For example let $X = \mathbb S^1$ with the subspace metric of $\mathbb R^2$ and $f : \mathbb S^1 \times \mathbb R \to \mathbb S^1$ be given by $$f(e^{i\theta}, t) = e^{i(\theta + t^2)}.$$ For all $s\neq 0$, let $t = \frac{\pi-s^2}{2s}$. Then for all $x = e^{i\theta}$, we have $$f(x , t+s) = e^{i(\theta + (t+s)^2)} = e^{i(\theta + t^2 + \pi) }= - f(x, t)\Rightarrow d(f(x, t+s), f(x, t)) = 2.$$
Thus $g$ is given by
$$g(s) = \begin{cases} 0 & \text{if } s = 0 \\ 2 & \text{otherwise.} \end{cases}$$
and is not continuous.