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Problem 5

Let ff and gg be real-valued functions defined for all real x,yx,y, satisfying f(x+y)+f(x−y)=2f(x)g(y)f(x+y)+f(x-y)=2f(x)g(y) for all x,yx,y. Prove that if ff is not identically zero and ∣f(x)∣≤1|f(x)|\le1 for all xx, then ∣g(y)∣≤1|g(y)|\le1 for all yy.
Step 1 of 4: Define the supremum
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In plain words

The nonzero assumption makes division by kk valid.

0<k=sup⁡x∈R∣f(x)∣≤10<k=\sup_{x\in\mathbb R}|f(x)|\le1
Detailed analysis

Since ff is not identically zero and ∣f(x)∣≤1|f(x)|\le1, the least upper bound k=sup⁡x∣f(x)∣k=\sup_x|f(x)| exists and satisfies 0<k≤10<k\le1.