Limit of a sum of periodic functions

1. Let’s start with a simpler case, $T_1=T_2$. Then $h=f+g$ is also periodic with period $T_1$. But a periodic function which converges as $x\rightarrow +\infty$ is constant, as shown here (link to add, the thing is broken right now). This lemma will prove useful many times.

2. Now assume that $T_1$ and $T_2$ are distinct. The idea is to check how $h$ changes when we move according to, say, the period of $f$ :

$$h\left(x+T_1\right)=f\left(x+T_1\right)+g\left(x+T_1\right)=f(x)+g\left(x+T_1\right)$$This is almost $h(x)$. Let’s make this term appear :

$$h\left(x+T_1\right)=f(x)+g(x)-g(x)+g\left(x+T_1\right)=h(x)+g\left(x+T_1\right)-g(x)$$

We get the following interesting property :

$$h\left(x+T_1\right)-h(x)=g\left(x+T_1\right)-g(x)$$

which indicates that, when moving according to $T_1$, only the change $g$ is relevant, and not the change in $f$, which makes sense.
By assumption, $h(x)$ converges to a limit when $x\rightarrow +\infty$, so $h(x+T_1)$ converges to the same limit ; and so we have that :

$$g\left(x+T_1\right)-g(x) \underset{x \rightarrow+\infty}{\longrightarrow} 0$$

. But this shows that $g$ is periodic with period $T_1$, since $g(x+T_1)-g(x)$ is itself $T_2$-periodic, so by the same argument as earlier, it has to be 0.

In the end, we get that $f$ and $g$ share the same period $T_1$, so $f+g$ too ; by again the same argument, we deduce that $f+g$ is constant.

Voici une autre solution. Soit $l$ la limite quand $x$ tend vers $+\infty$ de $f(x) + g(x)$. Quitte à translater la fonction $g$ par $l$ sans changer $T_{2}$, on peut supposer que $l = 0$. Soit $x \in \mathbb{R}$. Comme $T_{1}$ et $T_{2}$ sont non nulles, on a alors

$\underset{n \rightarrow \infty}{\mathrm{lim}} f(x+n(T_{1}+T_{2}))+g(x+n(T_{1}+T_{2})) = 0$.

Or cette limite s’écrit aussi

$(*)\,\,\underset{n \rightarrow \infty}{\mathrm{lim}} f(x+nT_{2})+g(x+nT_{1}) = 0$.

Comme

$\underset{n \rightarrow \infty}{\mathrm{lim}}f(x+nT_{2}) + g(x+nT_{2}) = 0$,

on déduit que $\underset{n \rightarrow \infty}{\mathrm{lim}}f(x+nT_{2}) = -g(x)$. De même, on déduit que $\underset{n \rightarrow \infty}{\mathrm{lim}}g(x+nT_{1}) = -f(x)$. En injectant ces nouvelles égalités dans $(*)$, on trouve que $-f(x)-g(x) = 0$, donc que $f(x)+g(x) = 0$, d’où que $f+g$ est constante.