A function $f(x)$ not continuous at 0 such that $\left[f(x)\right]^3$ is continuous at 0.

Question

The exercise 4.3.6 e) from Abbott's "Understanding Analysis 2nd edition" asks to provide an example of a real function $f(x)$ not continuous at 0 such that $\left[f(x)\right]^3$ is continuous at 0 or disproof the existence of such a function.

I would like to check if my reasoning is correct:

If $\left[f(x)\right]^3$ is continuous at $0$, then for every $\epsilon>0$ there exists a $\delta>0$ such that $|x-0|<\delta$ implies $|f^3(x)-f^3(0)|<\epsilon$.

Hence $$ \epsilon>|f^3(x)-f^3(0)|=|f(x)-f(0)|\cdot |f^2(x)+f(x)f(0)+f^2(0)|\ge \\ \ge|f(x)-f(0)|\cdot |f^2(0)| $$ and this would imply that $f$ is continuous at $0$ and the request is impossible to satisfy.

Note: I have used the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$ and also the fact that if $x$ and $0$ are close enough $f(x)$ and $f(0)$ have the same sign.

Answer 1

For a different argument, the function $g(x)=x^{1/3}$ is continuous everywhere. So if $x\longmapsto f(x)^3$ is continuous, then so is $f(x)=g(f(x)^3)$. Thus, if $f$ is not continuous at $0$, neither is $f(x)^3$.

Answer 2

This is the right idea but not quite complete. First, how do you know that $f(x)$ and $f(0)$ have the same sign as long as $x$ and $0$ are close enough? That would follow if you knew $f$ was continuous at $0$, but that's exactly what you're trying to prove! This assertion is still correct (assuming that you consider $0$ to have the "same sign" as either positive or negative numbers in the case $f(0)=0$), but you should provide some justification for it.

Second, your argument does not work if $f(0)=0$, since then $|f(x)-f(0)|\cdot|f(0)^2|<\epsilon$ does not give you any bound on $|f(x)-f(0)|$. So you will need to handle that case separately.

Answer 3

Hint: the function $g(x) = \sqrt[3]{x}$ is continuous and $f(x) = g(f(x)^3)$.

Answer 4

Does it have to be a real function? Because defining:

$$ f(x) = (x+1) e^{\frac{2\pi i}{3}}, \, (x<0) $$ $$ f(x) = (x+1) e^{\frac{4\pi i}{3}}, \, (x>0) $$

Then $f(x)$ is not continuous at zero, but: $$ f^3(x) = (x+1)^3 e^{2\pi i}= (x+1)^3 e^{4\pi i} $$