已知:F(X)+F(13/42)=F(1/6)+F(1/7);求证:F(X)是周期函数.

令x＝1／6得f（13／42）＝f（1／7）带入得f（x）＝f（1／6）为一常数函数

因为对任何x∈R,有
f(x+13/42)+f(x)=f(x+1/6)+f(x+1/7),
故f(x+7/42)-f(x)=f(x+13/42)-f(x+6/42)
````````````````=f(x+19/43)-f(x+12/42)
````````````````=……=f(x+49/42)-f(x+42/42).
即f(x+42/42)-f(x)=f(x+49/42)-f(x+7/42)……(1)
同样,有
f(x+7/42)-f(x+1/42)=f(x+14/42)-f(x+8/42)
```````````````````=f(x+21/42)-f(x+15/42)
```````````````````=……=f(x+49/42)-f(x+43/42)-f(x)
即f(x+49/42)-f(x)=f(x+43/42)-f(x+1/42)……(2)
由(1)(2),得
f(x+42/42)-f(x)=f(x+43/42)-f(x+1/42)
```````````````=f(x+44/42)-f(x+2/42)
```````````````=……=f(x+84/42)-f(x+42/42),
即f(x+1)-f(x)=f(x+2)-f(x+1).
因此,f(x+n)=f(x)+n[f(x+1)-f(x)]对所有n∈N成立.
又因为对所有x∈R,|f(x)|≤1,即f(x)有界,故只有f(x+1)-f(x)≡0.
因此对所有x∈R,f(x+1)=f(x),即f(x)为周期函数.