Difference between revisions of "1972 IMO Problems/Problem 5"
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==Solution== | ==Solution== | ||
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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>, | ||
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+ | <math>2u \geq |f(x+y)+f(x-y)| = |2f(x)g(y)|=2|f(x)||g(y)|</math> | ||
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+ | Therefore, <math>u \geq |f(x)||g(y)|</math>, so <math>|f(x)| \leq u/|g(y)|</math>. | ||
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+ | Since <math>u</math> is the least upper bound for <math>|f(x)|</math>, <math>u/|g(y)| \geq u</math>. Therefore, <math>|g(y)| \leq 1</math>. | ||
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+ | Borrowed from [http://www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln725.html] | ||
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+ | == See Also == {{IMO box|year=1972|num-b=4|num-a=6}} | ||
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+ | [[Category:Olympiad Algebra Problems]] | ||
+ | [[Category:Functional Equation Problems]] |
Latest revision as of 14:39, 29 January 2021
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 [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 |