Let $f$ and $g$ be two real-valued functions such that $f(x+y) + f(x-y) = 2f(x).g(y) AA x,y in R$ . If $f(x)$ is not identically zero and $|f(x)| <=1 AA x in R$ , then prove that $|g(y)| <=1 AA y in R$ ?

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Answer 1 · 2018-06-24T21:45:23

We have:

$f(x+y) +f(x-y) =2f(x)g(y) AA x,y in RR$

Where (it is assumed) that $f$ and $g$ are function of one variable. By direct substitution we have:

$y=0 => f(x+0) +f(x-0) = 2f(x)g(y)$

$\ \ \ \ \ \ \ \ \ => 2f(x) = 2f(x)g(y)$

$\ \ \ \ \ \ \ \ \ => 1 = g(y)$ (because $f(x)!=0$ )

And trivially $g(y)=1 => |g(y)| le 1$ QED

Let ff and gg be two real-valued functions such that f(x+y) + f(x-y) = 2f(x).g(y) AA x,y in Rf(x+y) + f(x-y) = 2f(x).g(y) AA x,y in R. If f(x)f(x) is not identically zero and |f(x)| <=1 AA x in R|f(x)| <=1 AA x in R, then prove that |g(y)| <=1 AA y in R|g(y)| <=1 AA y in R?