设$f(x)$在$(-\infty,+\infty)$可导,且$f(x)=f(x+2)=f(x+\sqrt3)$,证明:$f(x)$为常数.
证明:由题意,2为$f(x)$的周期,将$f(x)$展成周期为2的傅里叶级数,其傅里叶系数为:\begin{align*}
{\color{blue}a_n}=\int_{-1}^1 f(x)\cos(n\pi x)\mathrm d{x}\\
{\color{blue}b_n}=\int_{-1}^1 f(x)\sin(n\pi x)\mathrm d{x}
\end{align*}

\begin{align*}
\because &&f(x)&=f(x+\sqrt3)\\
\therefore &&a_n&=\int_{-1}^1 f(x)\cos(n\pi x)\mathrm d{x}\\
&&&=\int_{-1}^1 f(x+\sqrt3)\cos(n\pi x)\mathrm d{x}\\
&&&=\int_{-1+\sqrt3}^{1+\sqrt3} f(t)\cos[n\pi (t-\sqrt3)]\mathrm d{t}\\
&&&=\int_{-1+\sqrt3}^{1+\sqrt3} f(t)[\cos(n\pi t)\cos(\sqrt3n\pi)+\sin(n\pi t)\sin(\sqrt3n\pi)]\mathrm d{t}\\
&&&=\cos(\sqrt3n\pi){\color{red}\int_{-1+\sqrt3}^{1+\sqrt3} f(t)\cos(n\pi t)\mathrm d{t}}+\sin(\sqrt3n\pi){\color{red}\int_{-1+\sqrt3}^{1+\sqrt3} f(t)\sin(n\pi t)\mathrm d{t}}\\
&&&=\cos(\sqrt3n\pi){\color{red}\int_{-1}^{1} f(t)\cos(n\pi t)\mathrm d{t}}+\sin(\sqrt3n\pi){\color{red}\int_{-1}^{1} f(t)\sin(n\pi t)\mathrm d{t}}\\
&&&=\cos(\sqrt3n\pi){\color{blue}a_n}+\sin(\sqrt3n\pi){\color{blue}b_n}\\

&&b_n&=\int_{-1}^1 f(x)\sin(n\pi)\mathrm d{x}\\
&&&=\int_{-1}^1 f(x+\sqrt3)\sin(n\pi x)\mathrm d{x}\\
&&&=\int_{-1+\sqrt3}^{1+\sqrt3} f(t)\sin[n\pi (t-\sqrt3)]\mathrm d{t}\\
&&&=\int_{-1+\sqrt3}^{1+\sqrt3} f(t)[\sin(n\pi t)\cos(\sqrt3n\pi)-\cos(n\pi t)\sin(\sqrt3n\pi)]\mathrm d{t}\\
&&&=\cos(\sqrt3n\pi){\color{red}\int_{-1+\sqrt3}^{1+\sqrt3} f(t)\sin(n\pi t)\mathrm d{t}}-\sin(\sqrt3n\pi){\color{red}\int_{-1+\sqrt3}^{1+\sqrt3} f(t)\cos(n\pi t)\mathrm d{t}}\\
&&&=\cos(\sqrt3n\pi){\color{red}\int_{-1}^{1} f(t)\sin(n\pi t)\mathrm d{t}}-\sin(\sqrt3n\pi){\color{red}\int_{-1}^{1} f(t)\cos(n\pi t)\mathrm d{t}}\\
&&&=\cos(\sqrt3n\pi){\color{blue}b_n}-\sin(\sqrt3n\pi){\color{blue}a_n}\\
\end{align*}
\[\begin{equation*}
\left\{
\begin{split}
a_n=\cos(\sqrt3n\pi)a_n+\sin(\sqrt3n\pi)b_n\\
b_n=\cos(\sqrt3n\pi)b_n-\sin(\sqrt3n\pi)a_n
\end{split}
\right.
\end{equation*}\Rightarrow a_n=b_n=0\]
故$f(x)$为常数

这类题在周民强的《微积分专题论丛》里也有
15.设$f(x)$实值连续,且$f(x)=f(x+1)=f(x+\sqrt2)$,证明:$f(x)$为常数.
证明:由题意,1为$f(x)$的周期,将$f(x)$展成周期为1的傅里叶级数,其傅里叶系数为:\begin{align*}
{\color{blue}a_n}=2\int_0^1 f(x)\cos(2n\pi x)\mathrm d{x}\\
{\color{blue}b_n}=2\int_0^1 f(x)\sin(2n\pi x)\mathrm d{x}
\end{align*}

\begin{align*}
\because &&f(x)&=f(x+\sqrt2)\\
\therefore &&a_n&=2\int_0^1 f(x)\cos(2n\pi x)\mathrm d{x}\\
&&&=2\int_0^1 f(x+\sqrt2)\cos(2n\pi x)\mathrm d{x}\\
&&&=2\int_{\sqrt2}^{1+\sqrt2} f(t)\cos[2n\pi (t-\sqrt2)]\mathrm d{t}\\
&&&=2\int_{\sqrt2}^{1+\sqrt2} f(t)[\cos(2n\pi t)\cos(2\sqrt2n\pi)+\sin(2n\pi t)\sin(2\sqrt2n\pi)]\mathrm d{t}\\
&&&=2\cos(2\sqrt2n\pi){\color{red}\int_{\sqrt2}^{1+\sqrt2} f(t)\cos(2n\pi t)\mathrm d{t}}+2\sin(2\sqrt2n\pi){\color{red}\int_{\sqrt2}^{1+\sqrt2} f(t)\sin(2n\pi t)\mathrm d{t}}\\
&&&=2\cos(2\sqrt2n\pi){\color{red}\int_{0}^{1} f(t)\cos(2n\pi t)\mathrm d{t}}+2\sin(2\sqrt2n\pi){\color{red}\int_{0}^{1} f(t)\sin(2n\pi t)\mathrm d{t}}\\
&&&=\cos(2\sqrt2n\pi){\color{blue}a_n}+\sin(2\sqrt2n\pi){\color{blue}b_n}\\

&&b_n&=2\int_0^1 f(x)\sin(2n\pi)\mathrm d{x}\\
&&&=2\int_0^1 f(x+\sqrt2)\sin(2n\pi x)\mathrm d{x}\\
&&&=2\int_{\sqrt2}^{1+\sqrt2} f(t)\sin[2n\pi (t-\sqrt2)]\mathrm d{t}\\
&&&=2\int_{\sqrt2}^{1+\sqrt2} f(t)[\sin(2n\pi t)\cos(2\sqrt2n\pi)-\cos(2n\pi t)\sin(2\sqrt2n\pi)]\mathrm d{t}\\
&&&=2\cos(2\sqrt2n\pi){\color{red}\int_{\sqrt2}^{1+\sqrt2} f(t)\sin(2n\pi t)\mathrm d{t}}-2\sin(2\sqrt2n\pi){\color{red}\int_{\sqrt2}^{1+\sqrt2} f(t)\cos(2n\pi t)\mathrm d{t}}\\
&&&=2\cos(2\sqrt2n\pi){\color{red}\int_{0}^{1} f(t)\sin(2n\pi t)\mathrm d{t}}-2\sin(2\sqrt2n\pi){\color{red}\int_{0}^{1} f(t)\cos(2n\pi t)\mathrm d{t}}\\
&&&=\cos(2\sqrt2n\pi){\color{blue}b_n}-\sin(2\sqrt2n\pi){\color{blue}a_n}\\
\end{align*}
\[\begin{equation*}
\left\{
\begin{split}
a_n=\cos(2\sqrt2n\pi)a_n+\sin(2\sqrt2n\pi)b_n\\
b_n=\cos(2\sqrt2n\pi)b_n-\sin(2\sqrt2n\pi)a_n
\end{split}
\right.
\end{equation*}\Rightarrow a_n=b_n=0\]故$f(x)$为常数.

设 $f:\mathbb R\to[0,2)$ 为连续周期函数.同时记$a_n=\lfloor f(n)\rfloor$.试问对任意0-1数列$\{a_n\}$ ,都是否有相应的函数$f$存在?求构造或反例.
找到反例了.比如数列$\{0,1,1,\ldots,1,1,\ldots\}$,可以不失一般性地令$a_0=0$,$a_1=a_2=\cdots =1$.

证明:显然周期$T$是无理数.构造映射$\pi:\mathbb R\to [0,T),x\mapsto x-T\lfloor\dfrac{x}{T}\rfloor$,则
$$f(\mathbb N^*)=f\circ\pi(\mathbb N^*)\in[1,2)$$
再由$\pi(\mathbb N^*)$在$(0,T)$中稠密与$f$的连续性知:
$$f(0)\in f([0,T))=f(\overline{\pi(\mathbb N^*)})\subseteq\overline{f(\pi(\mathbb N^*))}=\overline{f(\mathbb N^*)}\subseteq[1,2]$$
矛盾.
---------------------------------------------------
若将数列替换为$\{1,0,0,\ldots\}$,显然是能构造出$f$的(因为下取整函数是右连续而左间断的).如构造$f:x\mapsto 1-x(\sqrt2-x)$,$x\in[0,\sqrt2)$,周期$T=\sqrt 2$.

现在的新问题是:如何判断对哪些数列能构造出$f$?

观察函数$f=0.9\sin(kx)+1$,当$k$取遍$\mathbb R/\mathbb Q$时生成的数列互不相同.由于$\mathbb R/\mathbb Q$不可数,故能够够造出的数列是不可数的.而无法构造出的数列也至少是可数的(不知是否不可数?).