Sufficiency of the condition $f(x) = f(x^3)$ for $f$ to be even or constant

Question

I've been playing around with some aspects of basic functions, and I reached a function that seemed a bit peculiar. Consider $\forall x \in \mathbb{R}$ a function $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $f(x) =f(x^3).$ I'm wondering if this is a constant function or not, or what assumption I would need in addition to make this a constant function. I am not sure what additional assumptions I would need for instance to make $f(x) = f(-x).$ Any help would be greatly appreciated.

Answer 1

As noted above in the comments, the condition that $f(x)=f(x^3)$ is not enough to guarantee that the function is constant. A simple example of such a function that is not constant would be the function $f: \mathbb{R}\rightarrow\mathbb{R}$ defined such that $f(1)=1$ and $f(x)= 0, x\neq 1$.

The problem is that there was no requirement of continuity, a condition which severely restricts the possibilities.

As another poster noted, if we add the condition that $f$ is continuous on the set $A=${$-1,0,1$}, we can see that this is enough to guarantee that $f$ then must be constant:

By definition of continuity, we know that for any $\epsilon > 0$, there exists a $\delta > 0$ such that $0< |x-y| < \delta \Rightarrow |f(x)-f(y)| < \epsilon$, where $y\in A$, and $x\in \mathbb{R}$.

Now assume that $f(0)= L$. Now the first task is to prove that $f(y)=L$ for any $y\in A$. Assume, for contradiction, that $f(1) = K\neq L$. Then, we know that $|K-L|>0$. Take $\epsilon = |K-L|/2 > 0$. We know that there is a $\delta_0 > 0$ such that $|f(x)-f(0)|< \epsilon$ whenever $0< |x-0|< \delta_0$. We also know that there is a $\delta_1 > 0$ such that $|f(x)-f(1)|< \epsilon$ whenever $0< |x-0|< \delta_1$.

Take $\delta = \min${$\delta_0,\delta_1$}. Pick a positive $x_1 < 1$, such that $|1-x_1|<\delta$. As $0< x_1 < 1$, we know that there is a $n\in\mathbb{N}$ such that $|x_1^{3n}-0|<\delta$. So, pick $x_0=x_1^{3n}$. Thus we definitely have that $|f(1)-x_1|=|K-f(x_1)|=|K-f(x_0)|<\epsilon$ and $|f(0)-f(x_0)|=|L-f(x_0)|<\epsilon$, which implies that $|K-L|< 2\epsilon = |K-L|$, which is obviously false. Thus we have, by contradiction, that $f(1)= f(0)= L$. You prove analougously that $f(-1)=L$, and after that, you can use similar reasoning to prove that $f(x)= L$ för any real number $x$.

Answer 2

For each real number, $r$, we can created a set $V_r = \{x| x = r^{3^i}; i \in Z\}$. $V_0 = \{0\}$ and $V_1 = \{1\}$ and $V_{-1} = \{-1\}$. But all other $V_x$ with by infinitely countable. However $V_r = V_s$ if $r = s^{3^i}$ for some integer i but $V_r$ and $V_s$ are completely disjoint otherwise.

For all $p \in V_r$ $f(p) = f(r)$. But we can set those values to anything at all in the world we would like and the values for all the terms in one $V_r$ can be utterly different for any other $V_p$.

The possibilities are absolutely endless.

If however the function must be continuous the $f(x) = c$ for some c is the only possibility. (Because every interval around 0 will contain a $y = x^{- \text {very large power of three}}$ for every possible value of $x$. So all $f(x)$ must be arbitrarily close to $f(0)$ in value.)