# Can a cube of discontinuous function be continuous?

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.

• The point is not that $x \longmapsto x^3$ is injective, as much as $x \longmapsto x^{1/3}$ is continuous. – Mindlack Jan 8 '19 at 1:03
• What is your domain? It matters really quite a lot. – user3482749 Jan 8 '19 at 1:04
• 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. – MJD Jan 8 '19 at 1:29
• No. But the cube of a non-differentiable function can be differentiable : $|x|^3$ – Eric Duminil Jan 8 '19 at 7:53
• 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). – user3482749 Jan 8 '19 at 15:19

If a function $$f(x)$$ is continuous, then its cube root $$g(x) = f(x)^{1/3}$$ is also continuous.
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.
• 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 $f = \phi^{-1}\circ \phi \circ f$. – Paul Sinclair Jan 8 '19 at 18:09