Problem 5

Let

f

and

g

be real-valued functions defined for all real

x,y

, satisfying

f(x+y)+f(x-y)=2f(x)g(y)

for all

x,y

. Prove that if

f

is not identically zero and

|f(x)|\le1

for all

x

, then

|g(y)|\le1

for all

y

Step 4 of 4: Invoke least-upper-bound minimality

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In plain words

A value of $|g(y)|>1$ contradicts the definition of $k$ .

k\le\dfrac{k}{|g(y)|}\Longrightarrow |g(y)|\le1

Detailed analysis

By minimality of the least upper bound, $k\le k/|g(y)|$ . Since $k>0$ , this gives $|g(y)|\le1$ . As $y$ was arbitrary, the result holds for every $y$ .