Continuous Functional Equation

Detailed Solution

The functional equation is $$ f(x)=f(x^3) $$ for every $x\in\mathbb R$. We show that continuity forces $f$ to be constant on the whole real line. For $0<x<1$, repeatedly taking cube roots gives $$ f(x)=f\left(x^{1/3}\right) =f\left(x^{1/3^2}\right) =\cdots =f\left(x^{1/3^n}\right). $$ Since $$ x^{1/3^n}\longrightarrow 1 \qquad\text{as }n\to\infty, $$ continuity of $f$ gives $$ f(x)=\lim_{n\to\infty}f\left(x^{1/3^n}\right)=f(1). $$ Thus, $$ f(x)=f(1)\qquad(0<x<1). $$ Similarly, if $-1<x<0$, then $$ x^{1/3^n}\longrightarrow -1, $$ and therefore $$ f(x)=f(-1). $$ Now consider $x=0$. Since $f$ is continuous at $0$ and $$ f(x)=f(1)\qquad(0<x<1), $$ letting $x\to0^+$ gives $$ f(0)=f(1). $$ Likewise, using $x\to0^-$, $$ f(0)=f(-1). $$ Hence $$ f(0)=f(1)=f(-1). $$ Next, let $x>1$. Again, $$ f(x)=f\left(x^{1/3}\right) =f\left(x^{1/3^2}\right) =\cdots =f\left(x^{1/3^n}\right). $$ But $$ x^{1/3^n}\longrightarrow1. $$ Therefore, by continuity, $$ f(x)=f(1)=f(0). $$ Finally, let $x<-1$. Then $$ x^{1/3^n}\longrightarrow-1, $$ so $$ f(x)=f(-1)=f(0). $$ Combining all cases, $$ f(x)=f(0) \qquad\text{for every }x\in\mathbb R. $$ Thus $f$ must be constant on $\mathbb R$. Therefore, option B is correct. Option A is false because $f$ cannot vary outside $[-1,1]$; the same functional equation and continuity force it to have the same constant value there. Option C is false because although $$ f(0)=f(1)=f(-1), $$ the function is necessarily globally constant. Option D is false because a constant function on $\mathbb R$ is not injective. Hence, $$ \boxed{\mathrm{B}}. $$