1972 IMO Problems/Problem 5

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Let $f$ and $g$ be real-valued functions defined for all real values of $x$ and $y$, and satisfying the equation \[f(x + y) + f(x - y) = 2f(x)g(y)\] for all $x, y$. Prove that if $f(x)$ is not identically zero, and if $|f(x)| \leq 1$ for all $x$, then $|g(y)| \leq 1$ for all $y$.


Let $u>0$ be the least upper bound for $|f(x)|$ for all $x$. So, $|f(x)| \leq u$ for all $x$. Then, for all $x,y$,

$2u \geq |f(x+y)+f(x-y)| = |2f(x)g(y)|=2|f(x)||g(y)|$

Therefore, $u \geq |f(x)||g(y)|$, so $|f(x)| \leq u/|g(y)|$.

Since $u$ is the least upper bound for $|f(x)|$, $u/|g(y)| \geq u$. Therefore, $|g(y)| \leq 1$.

Borrowed from [1]

See Also

1972 IMO (Problems) • Resources
Preceded by
Problem 4
1 2 3 4 5 6 Followed by
Problem 6
All IMO Problems and Solutions