Problem [B2 from IMO 1972]

f and g are real-valued functions defined on the real line. For all x and y,

f(x + y) + f(x − y) = 2f(x)g(y).

f is not identically zero and |f(x)| ≤ 1 for all x. Prove that |g(x)| ≤ 1 for all x.

Tom Hales’ Solution

Note first that |f(x)||g(y)|^{l} ≤ 1 for all l ≥ 0, by induction on l. For the induction step:

2|f(x)||g(y)|^{l+1} = |2f(x)g(y)||g(y)|^{l}
= |f(x + y) + f(x − y)||g(y)|^{l}
≤ |f(x + y)||g(y)|^{l} + |f(x − y)||g(y)|^{l}
≤ 2

Now suppose that |g(y)| > 1 for some y. We know f(z) ≠ 0 for some z, but then |f(z)||g(y)|^{l} → ∞ as l → ∞, contradicting the bound.