Let $f$ be a continuous function in $\mathbb{R}$ and $f(x)=f(x+1)=f(x+\sqrt{2})$ $\forall$ $x \in \mathbb{R}$
Prove: By using Fourier series prove that $f$ must be constant
I finished above problem by using density of $A=\left\{a+b\sqrt{2}: a,b \in \mathbb{Z}\right\}$ in $\mathbb{R}$. But how can we use Fourier series to prove that? Or can we prove density of $A$ by Fourier series? Many thanks.
Let $g(x)=f(x+\sqrt{2}-2)$, $g$ is also $1$-periodic, and that $g=\tau^{2-\sqrt{2}}(f)$ (the translation), we know that $(\tau^{2-\sqrt{2}}f)^{\wedge}(m)=e^{-2\pi im(2-\sqrt{2})}\widehat{f}(m)$. But $g=f$ by the periodicity and assumption, if $\widehat{f}(m)\ne 0$, then we end up with $e^{-2\pi im(2-\sqrt{2})}=1$, this is impossible if $m\ne 0$. So $\widehat{f}(m)=0$ for every $m\ne 0$, and so the sequence of Fejer means which reduced to $\widehat{f}(0)$ converges to $f$ a.e., since $f$ is continuous, $f=\widehat{f}(0)$.