Difference between revisions of "1972 IMO Problems/Problem 5"

Latest revision as of 14:39, 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]

@@ Line 13: / Line 13: @@
 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>.
-Borrowed from http://www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln725.html
+Borrowed from [http://www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln725.html]
+== See Also == {{IMO box|year=1972|num-b=4|num-a=6}}
 [[Category:Olympiad Algebra Problems]]
 [[Category:Functional Equation Problems]]

1972 IMO (Problems) • Resources
Preceded by Problem 4	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 6
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Difference between revisions of "1972 IMO Problems/Problem 5"

Latest revision as of 14:39, 29 January 2021

Solution

See Also