Question

Suppose that `f:RR->RR` has the properties `(a)` `|f(x)| le 1, forall x in RR` `(b)` `f(x+13/42)+f(x)=f(x+1/6)+f(x+1/7), forall x in RR` Prove that `f` is periodic?

Answer 1

We are given

`f(x+13/42) + f(x) = f(x+1/6)+f(x+1/7)`

Substituting in `x+1/6` for `x`, we get

`f(x+20/42) + f(x+1/6) = f(x+2/6)+f(x+13/42)`

Adding these equations and canceling the terms which appear on each side, we get

`f(x+20/42) + f(x) = f(x+2/6) + f(x+1/7)`

Next, we can substitute `x+1/6` for `x` into the new equation, add the two equations, and cancel to obtain

`f(x+27/42) + f(x) = f(x+3/6)+f(x+1/7)`

Repeating this process three more times results in the equation

`f(x+48/42) + f(x) = f(x+1)+f(x+1/7)`

Next we substitute in `x = 1/7` for `x`, add, and cancel to obtain

`f(x+54/42)+f(x) = f(x+1)+f(x+2/7)`

Repeating this process five more times results in the equation

`f(x+2) + f(x) = f(x+1)+f(x+1)`

`=> f(x+2)-f(x+1) = f(x+1)-f(x)`

The above equation implies that the difference between `f(x)` and `f(x+1)` is constant for all `x`. If that difference were nonzero, then the sequence `f(x), f(x+1), f(x+2), ...` would tend to `-oo` or `oo`. However, we are given the condition that `|f(x)| < 1`, a contradiction. Thus the difference must be `0`, meaning `f(x) = f(x+1)" "AAx in RR`, i.e. `f(x)` is periodic with a period of `1`.

Answer 2

See below.

Explanation:

Here `f(x+a+b)+f(x)=f(x+a)+f(x+b)`

We have

`f(x+b+a)-f(x+a)=f(x+b)-f(x)`

so calling `p(x)=f(x+b)-f(x)` we have

`p(x+a)=p(x)` so `p(x)` is periodic with period `a`.

Also considering

`f(x+b+a)-f(x+b)=f(x+a)-f(x)`

making `q(x)=f(x+a)-f(x)` we have

`q(x+b)=q(x)` which is periodic with period `b`

and now we can write

`f(x)=alpha p(x)+beta q(x)+gamma p(x+a)+delta q(x+b)`

or

`f(x)=(alpha-delta)f(x+b)-(alpha+beta)f(x)+(gamma+delta)f(x+a+b)+(alpha-gamma)f(x+a)`

but

`f(x)=f(x+a)+f(x+b)-f(x+a+b)`

so

`{(alpha+beta=0),(alpha-gamma=1),(alpha-delta=1),(gamma+delta=-1):}`

solving for `alpha,beta,gamma,delta` we obtained

`(alpha = 1/2, beta = -1/2, gamma = -1/2, delta = -1/2)`

so `f(x)` is periodic because it can be composed as a finite sum of periodic functions and the distinct periods are relatively rational.