Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 1 / 9F (Analysis I)

The function is continuous at $x$ when, for every $\epsilon >0$, there is $\delta >0$ such that $|y-x|<\delta$ implies $|f(y)-f(x)|<\epsilon$.

If this holds and $x_n\to x$, eventually $|x_n-x|<\delta$, so $f(x_n)\to f(x)$. Conversely, if continuity fails, some ${\epsilon }_0>0$ allows, for every $n$, a point $x_n$ with $|x_n-x|<1/n$ but $|f(x_n)-f(x)|\ge {\epsilon }_0$. Then $x_n\to x$ while $f(x_n)\nrightarrow f(x)$, a contradiction. This is the sequential characterization of continuity in metric spaces.

Let the nonconstant polynomial have degree $d$ and leading coefficient $c$. If $d$ is odd, its limits at $+\infty$ and $-\infty$ have opposite signs, so the intermediate value theorem shows that its image is $\mathbb{R}$. If $d$ is even and $c>0$, it tends to $+\infty$ at both ends. Outside a sufficiently large compact interval it exceeds $f(0)$, while on that interval the extreme value theorem gives a global minimum $a$. Continuity and the intermediate value theorem then give image $[a,\infty )$. For $c<0$, applying this argument to $-f$ gives $(-\infty ,a]$. These are exactly the three possible images.

Applying the functional equation at $x^{{\alpha }^{-1}}$ repeatedly givesSince $x^{{\alpha }^{-n}}\to 1$, continuity at $1$ gives $f(x)=f(1)$. Thus $f$ is constant.

Without continuity the claim is false. The function $f(1)=1$ and $f(x)=0$ for $x\ne 1$ satisfies $f(x)=f(x^{\alpha })$ because $x^{\alpha }=1$ for positive $x$ exactly when $x=1$, but it is not constant.