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Assume that $f,g:\mathbb{R}\to\mathbb{R}$ are continuous and periodc.

Then $f(x)+g(x)$ is periodic iff the quotient of periods of $f$ and $g$ is rational (well known, discussed many times, e.g. Sum of two periodic functions is periodic?).

What about the function $f(x)+g(x^n)$ with $n$ being a fixed positive integer?

For $n=1$ this problem becomes the known one mentioned above.

If $n>1$, is $g=const.$ the only example, when $f(x)+g(x^n)$ is periodic?

I could not find the answer myself...

I tried with many examples and watched the graphs, every time the result seems to be not periodic, hence I suppose that the answer is "yes", but the proof seems to be above my knowledge.

larry01
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tl; dr: Yes.

Proposition: If $g$ is continuous and periodic, and if there is an integer $n > 1$ such that the function $h(x) = g(x^{n})$ is periodic, then $g$ is constant.

Here's a proof that (perhaps unfortunately) relies on concepts and results from real analysis: A continuous, periodic function on the reals is uniformly continuous on the reals.

Proof: It suffices to prove the contrapositive, i.e., if $g$ is continuous, periodic, and non-constant, and $n > 1$, then $h(x) = g(x^{n})$ is not periodic.

Assume $g$ is non-constant, let $\ell$ denote the smallest positive period of $g$, and consider the intervals $[(k\ell)^{1/n}, ((k+1)\ell)^{1/n}]$ for $k$ a positive integer, which map to the period intervals $[k\ell, (k+1)\ell]$ under $x \mapsto x^{n}$. Intuitively, the length of these intervals decreases to $0$ as $k$ grows without bound, but "$h$ goes through a complete period of $g$ on each", so $h$ is not periodic unless $g$ is constant.

Rigorously, if $g$ is not constant, there exist points $a$ and $b$ in $[0, \ell]$ such that $g(a) < g(b)$. It follows that for each positive integer $k$, there exist points $a_{k}$ and $b_{k}$ in $[(k\ell)^{1/n}, ((k+1)\ell)^{1/n}]$ such that $h(a_{k}) = g(a)$ and $h(b_{k}) = g(b)$. Since the difference $|h(b_{k}) - h(a_{k})|$ is independent of $k$ but $|b_{k} - a_{k}|$ decreases to $0$, $h$ is not uniformly continuous, and therefore not periodic.

Andrew D. Hwang
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    thanks, but this is a partial solution (or it seems so for me): $h$ is not periodic, so my function is a sum of periodic and nonperiodic function, but this is not enough to conclude that the sum is not periodic. – larry01 May 24 '21 at 08:50
  • You're right. To finish, we can use that a sum of uniformly continuous functions is uniformly continuous. Since $f$ is uniformly continuous but $h$ is not, $f + h$ is not (otherwise $h = (-f) + (f + h)$ would be). That implies $f + h$ is not periodic either. – Andrew D. Hwang May 24 '21 at 11:08
  • ok, now I got it, thanks very much – larry01 May 24 '21 at 11:17