The answer to the final question of the mathematical analysis of the postgraduate examination in Nanjing University in 2013

The $f$ is a $c^1$ function with a period of $1$ on the $\MATHBB R $. If $f$ satisfies the condition $ $f (x) +f\left (x+\frac{1}{2}\right) =f (2x), X\IN\MATHBB r$$

Prove $f (x) \equiv0$.

Proof Set $ $f (x) =\frac{a_{0}}{2}+\sum_{n=1}^{\infty} (A_{n}\cos 2n\pi x+b_{n}\sin 2n\pi x) $$

(Because $f$ is a guide, so the above is actually an equal sign) so you can get \begin{align*}f (x) +f\left (x+\frac{1}{2}\right) &=a_{0}+\sum_{n=1}^{\infty}\left (A_{n}\cos2n\pi X+a_{n}\cos (2n\pi x+n\pi) +b_{n}\sin2n\pi x+b_{n}\sin (2n\pi x+n\pi) \right) \\&=a_{0}+2\sum_{n=1} ^{\infty}\left (a_{2n}\cos4n\pi x+b_{2n}\sin 4n\pi x\right) \tag{1}\end{align*}

and \begin{align*}f (2x) =\frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left (a_{n}\cos4n\pi x+b_{n}\sin 4n\pi x\right) \tag{2}\ end{align*}

According to the uniqueness of the Fourier series (or with the Parseval equation) it is known (1) that the coefficients of (2) can be known $a_{0}=0$ and $ $a _{n}=2a_{2n},b_{n}=2b_{2n}$$

So $ $a _{n}=2^2a_{2^2n}=\cdots=2^ka_{2^kn}\to0 (k\to+\infty) \tag{3}$$

The $f$ is due to the Fourier coefficient of $ $a _{n}=o\left (\frac{1}{n}\right), B_{n}=o\left (\frac{1}{n}\right) $$ when the derivative of the Guide function is integrable or absolutely integrable.

(This conclusion can be proved by using the Riemann-lebesgue lemma only once for partial integrals) similar to $b_{n}=0$. $ $f (x) \equiv0.$$

The proof process can be seen in the topic $c^1$ conditions too strong, just $f ' $ integrable or absolutely integrable can be.

The answer to the final question of the mathematical analysis of the postgraduate examination in Nanjing University in 2013