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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?

Joe
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Gulshan Mishra
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    If $f=0$, then $g \circ f = 0$ and so $g \circ f$ is continuous everywhere. If $f(x)=x$ for all $x$, then $g \circ f = g$, and so $g \circ f$ is discontinuous at integer values. – Joe Jun 29 '21 at 11:00

3 Answers3

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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 $$g \circ f \quad \text{cont.}\quad \Leftrightarrow \quad g \circ f \quad \text{const.}\quad \Leftrightarrow \quad \exists k \in \mathbb Z, f(\mathbb R) \subseteq [k,k+1).$$

principal-ideal-domain
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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 $x^2=0$.

  • ...when $f$ "crosses" an integer value or at least "touches" it from below, like $x\mapsto 1/(1+x^2)$ at zero – CiaPan Jun 30 '21 at 10:05
  • @CiaPan: no, $\lfloor\frac1{1+x^2}\rfloor$ is not continuous at $x=0$. (If this is what you mean.) –  Jun 30 '21 at 10:09
  • Exactly. You said 'There is a _dis_continuity when...' and I expand the criterion for $f$ which makes the $\lfloor f\rfloor$ discontinuous, hence the example. This is a function which "touches" an integer $f=1$ from below but does not "cross" it, anyway it results in $g\circ f$ discontiniuous. – CiaPan Jun 30 '21 at 10:13
  • @CiaPan: ok, in fact my answer does not cover all cases, you are right. –  Jun 30 '21 at 10:18
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Nope, that isn't true in general.

Counterexample 1.

Let, for example, $$f(x)=\sin^2 \frac {\pi x}2,$$ then $f$ is integer at all integers, $x \in \mathbb Z$.

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

Counterexample 2.

Let $f(x)$ be the Conway base 13 function (described in this Wikpedia atrticle), or the Aksel Bergfeldt function (described here, at Math SE).

Those functions assume all real values on each open interval. This implies that $g(f(x))$ will have an uncountably infinite set of discontinuities on every open interval. As a result, the composed function will be discontinuous everywhere – also at points, where $f(x)$ is not integer.

CiaPan
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