I have an $\mathbb{R}_+ \to \mathbb{R}$ function $f(t)$, which is a combination of sums and products of $\sin$ and $\cos$ functions of incommensurable frequencies. Thus $f(t)$ is a quasiperiodic function, or, more generally, almost-periodic function. The goal is to show that if there exists $t_0$ such that $f(t_0)\ne 0$, then $$\int_0^\infty{f^2(s)ds}=\infty.$$
My idea is to do it by contradiction. Assume that $$\int_0^\infty{f^2(s)ds}=C,$$ then $$\lim_{t\to\infty}{f(t)}=0.$$ Then for any $\epsilon>0$ there exitis $t_\epsilon$ such that $|f(t)|<\epsilon$ for all $t>t_\epsilon$. Let us choose $\epsilon$ such that $|f(t_0)|>2\epsilon$. For alomost-periodic functions it is known that for any $T$ there exists $\tau>T$ such that $|f(t_0)-f(\tau)|<\epsilon$ and $|f(\tau)|>\epsilon$. This yields a contradiction. Thus if almost-periodic function is not identically zero, then it is not square-integrable.
Questions:
Q1. Is it correct that a combination of products and sums of periodic functions is almost-periodic?
Q2. Is the proof correct?
Q3. Most probably this is something very well-known. What is a good reference to cite?
Your proof is wrong. An $L^2$ function does not necessarily have limit $0$ at $\infty$.
However, what is true is that given $\epsilon > 0$, an almost periodic function $f$ has arbitrarily large "almost periods" $\tau$ such that $|f(t+\tau) - f(t)| < \epsilon$ for all $t \in \mathbb R$. You can use this to show that if $\int_a^b |f(t)|^2\; dt = c > 0$, there are infinitely many disjoint intervals $[a_n, b_n]$ with $\int_{a_n}^{b_n} |f(t)|^2 \; dt > c/2$.