f is a real-valued function on the reals such that | f(x) | <= 1 and f(x + 13/42) + f(x) = f(x + 1/6) + f(x + 1/7) for all x. Show that there is a real number c > 0 such that f(x + c) = f(x) for all x.
Solution
We show that f(x + 1) = f(x) for all x.
Note that 13/42 = 1/6 + 1/7. Using the relation for x + k/7, x + k/7 + 1/6, ... , x + k/7 + (n-1)/6, we get:
f(x + (k+1)/7) + f(x + 1/6 + k/7) = f(x + k/7) + f(x + 1/6 + (k+1)/7)
f(x + 1/6 + (k+1)/7) + f(x + 2/6 + k/7) = f(x + 1/6 + k/7) + f(x + 2/6 + (k+1)/7)
f(x + 2/6 + (k+1)/7) + f(x + 3/6 + k/7) = f(x + 2/6 + k/7) + f(x + 3/6 + (k+1)/7)
...
f(x + (n-1)/6 + (k+1)/7) + f(x + n/6 + k/7) = f(x + (n-1)/6 + k/7) + f(x + n/6 + (k+1)/7)
Adding, most of the terms cancel and we get: f(x + (k+1)/7) + f(x + n/6 + k/7) = f(x + k/7) + f(x + n/6 + (k+1)/7). Taking x, x + 1/7, x + 2/7, ... , x + (m-1)/7, we get:
f(x + 1/7) + f(x + n/6) = f(x) + f(x + n/6 + 1/7)
f(x + 2/7) + f(x + n/6 + 1/7) = f(x + 1/7) + f(x + n/6 + 2/7)
f(x + 3/7) + f(x + n/6 + 2/7) = f(x + 2/7) + f(x + n/6 + 3/7)
...
f(x + m/7) + f(x + n/6 + (m-1)/7) = f(x + (m-1)/7) + f(x + n/6 + m/7)
Adding, most of the terms cancel and we get: f(x + m/7) + f(x + n/6) = f(x) + f(x + n/6 + m/7). Now putting m = 7, n = 6, we get 2 f(x+1) = f(x) + f(x+2). Hence f(x+2) - f(x+1) = f(x+1) - f(x). So if f(x+1) = f(x) + h, then f(x+2) = f(x) + 2h and by a trivial induction f(x + n) = f(x) + nh. But |f(x)| <= 1 for all x, so h = 0.
© John Scholes
jscholes@kalva.demon.co.uk
20 Dec 2002