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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]]
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[[Category:Functional Equation Problems]]

Latest revision as of 15:39, 29 January 2021

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 $y$.

Solution

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
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