Difference between revisions of "1972 IMO Problems/Problem 5"
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Let <math>u>0</math> be the least upper bound for <math>|f(x)|</math> for all <math>x</math>. So, <math>|f(x)| \leq u</math> for all <math>x</math>. Then, for all <math>x,y</math>, | Let <math>u>0</math> be the least upper bound for <math>|f(x)|</math> for all <math>x</math>. So, <math>|f(x)| \leq u</math> for all <math>x</math>. Then, for all <math>x,y</math>, | ||
− | <math>2u | + | <math>2u \geq |f(x+y)+f(x-y)| = |2f(x)g(y)|=2|f(x)||g(y)|</math> |
Therefore, <math>u \geq |f(x)||g(y)|</math>, so <math>|f(x)| \leq u/|g(y)|</math>. | Therefore, <math>u \geq |f(x)||g(y)|</math>, so <math>|f(x)| \leq u/|g(y)|</math>. |
Revision as of 11:04, 11 June 2020
Let and be real-valued functions defined for all real values of and , and satisfying the equation for all . Prove that if is not identically zero, and if for all , then for all .
Solution
Let be the least upper bound for for all . So, for all . Then, for all ,
Therefore, , so .
Since is the least upper bound for , . Therefore, .
Borrowed from http://www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln725.html