Can a cube (meaning $g(x) = f(x)^3 = f(x) \cdot f(x) \cdot f(x)$) of discontinuous function $f: D \to \mathbb{R}$ ($D$ is subset of $\mathbb{R}$) be continuous? I think it can't, since $x^3$ is injective, but I am not able to prove it or find a counterexample.

  • 11
    $\begingroup$ The point is not that $x \longmapsto x^3$ is injective, as much as $x \longmapsto x^{1/3}$ is continuous. $\endgroup$ – Mindlack Jan 8 at 1:03
  • 3
    $\begingroup$ What is your domain? It matters really quite a lot. $\endgroup$ – user3482749 Jan 8 at 1:04
  • 2
    $\begingroup$ I think the injectivity is very much to the point. $f(x) =\sqrt x$ is also continuous on $[0,∞)$ but there are any number of discontinuous functions $g$ on that interval with $g(x)\cdot g(x)$ continuous. $\endgroup$ – MJD Jan 8 at 1:29
  • $\begingroup$ No. But the cube of a non-differentiable function can be differentiable : $|x|^3$ $\endgroup$ – Eric Duminil Jan 8 at 7:53
  • 3
    $\begingroup$ I misspoke: it's the range that matters. If it's $\mathbb{C}$ and your domain is, say, connected, it's false (send some subset of your domain to $1$, and the rest to one of the other cube roots of unity). $\endgroup$ – user3482749 Jan 8 at 15:19

If a function $f(x)$ is continuous, then its cube root $g(x) = f(x)^{1/3}$ is also continuous.

So the contrapositive is also true, which is:

If a function $g(x)$ is not continuous, then its cube $f(x) = g(x)^3$ is not continuous either.

(Strictly speaking, the contrapositive is actually "if the cube root $f(x)^{1/3}$ of a function $f(x)$ is not continuous, then the function $f(x)$ is not continuous either". But this is equivalent.)


Since $\phi : \mathbb{R} \to \mathbb{R}, \phi(x) = x^3$, is a homeomorphism, you see that $f$ is continuous iff $\phi \circ f$ is continuous.

  • 9
    $\begingroup$ To lower the level of this answer, note that $\phi^{-1}(x) = \sqrt[3] x$ is also a continuous map ("homeomorphism" means a continuous, invertible map whose inverse is also continuous). So if $\phi\circ f$ is continuous, so is $f = \phi^{-1}\circ \phi \circ f$. $\endgroup$ – Paul Sinclair Jan 8 at 18:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.