Functional Equation Different Substitution Different Result

Question

Okay Consider this g(x).g(y)=1 for all x,y ∈ ℝ Substitute x=y we get g(x)= 1 or -1 .Now Substitute y=1/x g(x).g(1/x)=1 we get g(x)=x^n where n∈N Different substitutions produced different solutions?

Answer 1

Note that when you do a substitution, you specialize. Not any function, which is a solution of the substituted equation, is also a solution of the originial one, you have to check. In your cases:

(1) $g(x) = 1$ or $g(x) = -1$ for all $x \in \mathbf R$ is also a solution of $g(x)g(y) = 1$, but for example $g(x) = 1$ for $x > 0$ and $g(x) = -1$ for $x \le 0$, which is a solution of $g(x)^2 = 1$ for all $x \in \mathbf R$ does not solve $g(x)g(y) = 1$. Hence, the solutions of the originial equation are $g(x) = 1$ for all $x$ or $g(x) = -1$ for all $x$.

(2) If $g(x) = \pm x^n$, then the original equation reads $x^ny^n =1$ for all $x,y \in \mathbf R$. This is only true for $n = 0$. Hence, the solutions of the orignal equation are $g(x) = \pm x^0 = \pm 1$, the same as above.