Using topology for proving periodicity

Question

Let $f\in\mathcal{C}\big(\mathbb{R},\mathbb{C}^\ast\big)$ a continuous function with modulus $r$ satisfying: $f(t)=r(t)e^{it}$. Assume that the image of $f$ is homeomorphic to the unit circle.

Question: Is $f$ a $2\pi$-periodic function?

Answer 1

Yes, it must be $2\pi$-periodic.

We show first that $t\mapsto r(t+2\pi)-r(t)$ must hit zero. If it never does, then it is either always positive or always negative; we assume it is always positive, the other case being completely analogous. We now know that $k\in\mathbb Z\mapsto r(2k\pi)$ is an increasing function of $\mathbb Z$, but its image is the intersection of the image of $f$ with $[0,\infty)$, so it is compact. This is impossible (say because it has a largest element, which cannot be).

Secondly, we show that if $r(t+2\pi)=r(t)$ vanishes for a given $t$, then in fact $r(s+2\pi)=r(s)$ for all $s\in t+[-\pi/2,\pi/2]$. This will conclude, by connectedness of $\mathbb R$. If it is not the case, then there exists $s\in [-\pi/2,\pi/2]$ such that $r(s+2\pi)\neq r(s)$. Suppose $s>t$ and $r(s+2\pi)>r(s)$, the other cases being similar. Consider the largest time $\sigma<s$ such that $r(\sigma+2\pi)=r(\sigma)$; it must exist by continuity of $r$, and we must have $r(u+2\pi)>r(u)$ for all $\sigma<u\leq s$. We use these to construct a subspace of the image of $f$ that cannot be embedded in a circle.

Namely, consider the set defined as the union of the two arcs $\{f(u),u\in t+[-\pi/2,\pi/2]\}$ and $\{f(u), u\in [\sigma+2\pi,s+2\pi]\}$. Since $f$ is injective over sets of diameter less than $2\pi$, these are injective images of line segments; by compactness, they are topological arcs. Moreover, by construction, they have a single intersection point, namely $f(\sigma)=f(\sigma+2\pi)$. In other words, this set is homeomorphic to a capital T shape (if some readers feel unconvinced, they can write this set explicitly as the continuous injective image of a compact, canonical T, say $[-1,1]\cup(-\mathsf i[0,1])\subset\mathbb C$, which shows this is indeed topologically what we expect).

But it is a known fact that a circle does not contain a T shape. One elementary argument is given by Pietro Majer in the comments, through the characterization of connected subsets of $\mathbb S^1$. My original argument used the (subjectively) fun fact that one can fit at most countably many T's but countinuum many circles in the plane, see this question. This contradicts the fact that this T is in the image of $f$, hence the existence of $s$, and as discussed above we find that $r$ is $2\pi$-periodic.

Answer 2

Not quite a proof but also too long for a comment. The following approach seems convincing to me:

1. Prove that $f$ may not be injective; it seems reasonable that, given the very special form of $f$, that if such an $f$ were injective it would induce a homeomorphism onto its image, contradicting the premise; there therefore exists an $x_0$ and $k_0 \in \mathbb{N}\setminus \{ 0 \}$ with $r(x_0 + 2k_0\pi) = r(x_0)$.
2. Argue that actually there must exist a $y$ with $r(y + 2\pi) = r(y)$; an attempt to argue the above would be by contradiction: consider the set of all positive $k$'s such that for some $x\in\mathbb{R}$ it holds that $r(x + 2k\pi) = r(x)$; pick its minimum element $m$ and suppose by contradiction that $m \geq 2$ corresponding to some $x_m$; from a drawing it seems reasonable that there should be $1 \leq k < m$ and $x$ with $x_m < x < x + 2k\pi < x_m + 2m\pi$ with $r(x + 2k\pi) = r(x)$; this contradicts the minimality of $m$ whence $m = 1$;
3. the restriction of $f$ to $[x_m, x_m + 2\pi]$ induces a homeomorphism $[x_m, x_m + 2\pi]\;/\sim$ onto its image $T \subset f(\mathbb{R})$, where $\sim$ identifies $x_m$ and $x_m + 2\pi$
4. the circle admits no proper subset that is homeomorphic to the circle and thus $T = f(\mathbb{R})$

From $T = f(\mathbb{R})$ it follows that $r$ is $2\pi$-periodic.