Discontinuity of Greatest Integer Function

Question

Let $f(x)$ be an arbitrary function. Let $g(x) = \lfloor x\rfloor$ be the greatest integer function.

We know that $g(x)$ is discontinuous whenever $x$ is integer.

Can we say that $g(f(x)) = \lfloor f(x) \rfloor$ is discontinuous whenever $f(x)$ takes integer values?

Answer 1

There is a discontinuity when $f$ "crosses" an integer value, not if it reaches it and leaves it "from above".

$\lfloor x^2\rfloor$ is continuous where $x^2=0$.

Answer 2

No, we can't. If $f$ is constant then $g \circ f$ is constant and hence continuous.

By the way it's easy to characterize all functions $f:\mathbb R \to \mathbb R$ such that $g \circ f$ is continuous. We have $$g \circ f \quad \text{cont.}\quad \Leftrightarrow \quad g \circ f \quad \text{const.}\quad \Leftrightarrow \quad \exists k \in \mathbb Z, f(\mathbb R) \subseteq [k,k+1).$$

Answer 3

Well, you can say it, but that wouldn't be true in general.

Let $f(x)=\sin^2 x$, then $f$ is integer at all integer multiples of $\pi$.

However, $$(g\circ f)(x)=\begin{cases} 1 & \text{for }x=(2k+1)\pi, k\in\mathbb Z \\ 0 & \text{otherwise} \end{cases}$$ so it's discontinuous at odd multiples of $\pi$ only.