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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.

J. Abraham
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    The point is not that $x \longmapsto x^3$ is injective, as much as $x \longmapsto x^{1/3}$ is continuous. – Aphelli Jan 08 '19 at 01:03
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    What is your domain? It matters really quite a lot. – user3482749 Jan 08 '19 at 01:04
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    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 08 '19 at 01:29
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    No. But the cube of a non-differentiable function can be differentiable : $|x|^3$ – Eric Duminil Jan 08 '19 at 07:53
  • @user3482749: Why does the domain matter? – TonyK Jan 08 '19 at 15:08
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    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 08 '19 at 15:19
  • I think you question could use a bit more context. Do you want a topological answer, a $\delta-\epsilon$ answer, a more general calculus answer, or what? – Steven Alexis Gregory Jan 08 '19 at 16:24

2 Answers2

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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.

Paul Frost
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    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 08 '19 at 18:09
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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.)

Tanner Swett
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