# 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#?

##### 1 Answer

Jun 24, 2018

We have:

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

Where (it is assumed) that

# 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