Question

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`?

Answer 1

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