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 3 of 4: Obtain a new upper bound

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In plain words

If $|g(y)|$ were large, the entire range of $|f|$ would shrink.

|f(x)|\le\dfrac{k}{|g(y)|}\quad(g(y)\ne0)

Detailed analysis

If $g(y)=0$ , the conclusion is immediate. Otherwise, division yields $|f(x)|\le k/|g(y)|$ for every $x$ , so $k/|g(y)|$ is an upper bound for $|f|$ .