Functional Equations

We will try to extend the domain of $f$ by defining $g: \mathbb{Z} \rightarrow \mathbb{R}$ that satisfy $g(km)+g(kn)-g(k)g(mn) \geq 1$ (*) for all $k,m,n$ .

If there is a solution $g$ , that solution will also be a solution of $f$ with a restriction of domain (because $\mathbb{N} \subset \mathbb{Z}$ ). (**)

Substitute $k=m=n=1$ in (*) we get $2g(1) - g(1)^{2} \geq 1$

Or $(g(1)-1)^{2} \leq 0$ .

Therefore $g(1) = 1$ .

Now let $m=n=1$ in (*) we get $g(k) \geq 1$ . (1)

Substitute $k=m=n=0$ in (*) we get $g(0) = 1$ .

Let $m=n=0$ in (*) we get $g(k) \leq 1$ . (2)

From (1),(2); we get $g(k) = 1$ for all $k \in \mathbb{Z}$ .

From the statement (**); we get $f(k) = 1$ for all $k \in \mathbb{N}$ ~~~

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If you wonder why we can extend the domain, let's think an easy example:

• Solve $x^{3}-2x+1 = 0$ in $\mathbb{R}$ . (1)
• Solve $y^{3}-2y+1 = 0$ in $\mathbb{Z}$ . (2)

If we can solve (1), then we can solve (2) but restrict the domain of the variable.