There is a problem I encountered which has two functions $f(x)$ and $g(x)$ such that $g(f(x))=x$ and $f(\lfloor g(x) \rfloor)=x, \quad \forall x\geq 0$. I am not giving any other details here, because what I want to know is how it is possible that $f(\lfloor x \rfloor)=x$? All I know by graphical transformation of $f(x)\to f(\lfloor x \rfloor)$ that the graph obtained is piecewise, not continuous. So how is this case possible? One situation I could think of is $f(x)=x, x \in \mathbf Z$. But here, the domain is $\mathbf R$. I would like to know other simple cases when it could be possible, so that it won't be hard for me to imagine such a function.
Edit Here is the complete definition of function: $$f(x)=\begin{cases} \sqrt{1-x^2},& -1\leq x < 0\\ x^2+1,& 0\leq x < 1\\ \frac{(x-1)^2)}{4} +2, & x\geq 1 \end{cases} $$ $g(x)$ is defined such that $g(f(x))=x, x \geq -1$ and $f(\lfloor(g(x)\rfloor)= x, x\geq0$
Such a $g$ can't exist. Indeed, $S = \mathrm{Im}(x \mapsto f(\lfloor g(x) \rfloor)) \subset f(\mathbb{Z})$ because $\lfloor g \rfloor$ has value in $\mathbb{Z}$. The set $S$ is at most countable because $\mathbb{Z}$ is (whatever are $f$ and $g$, continuous or not). However, the image of the identity function is the whole $\mathbb{R}$, which is uncoutable.
With the same argument, you can prove that if $f(\lfloor g \rfloor)$ is continuous on a connected space, then its image is an interval, thus it is either uncountable either a point, and this image is at most countable (once more with no hypothesis on $f$ and $g$). It proves that in this case, $f(\lfloor g \rfloor)$ is constant.