By J. Abraham


2019-01-08 00:58:34 8 Comments

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.

2 comments

@Tanner Swett 2019-01-08 05:51:24

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.)

@Paul Frost 2019-01-08 01:05:45

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 Sinclair 2019-01-08 18:09:44

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 $f = \phi^{-1}\circ \phi \circ f$.

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