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?

2 Answers
Feb 4, 2017

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#.

Feb 4, 2017

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.