Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1972 IMO Problems/Problem 5"
(→‎Solution)
Line 4: Line 4:
  
 
==Solution==
 
==Solution==
 +
 +
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 \geq |f(x+y)|+|f(x-y)| \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>.
 +
 +
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>.

Revision as of 09:47, 20 October 2014

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 $x,y$,

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

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, $|g(y)| \leq 1$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1972_IMO_Problems/Problem_5&oldid=65342"