Proving a function is constant when $f(x)f(y) + f(\frac{a}{x})f(\frac{a}{y}) = 2f(xy)$

Question

I've been working on the following homework problem:

Consider a function $f : (0,∞) → \mathbb{R}$ and a real number $a > 0$ such that $f(a) = 1$. Prove that if $f(x)f(y) + f(\frac{a}{x})f(\frac{a}{y}) = 2f(xy)$ for all $x, y ∈ (0,∞)$, then $f$ is a constant function.

I've gone through several steps, and been able to derive the following two relations:

$f(x) = f(\frac{a}{x}) = \frac{1}{f(\frac{1}{x})}$ (Conversely, $f(\frac{1}{x}) = \frac{1}{f(x)}$)

$f(1) = 1 = f(a)$

But now I'm stuck. I think I'm close, but I can't figure out how to make the jump to show how this implies the function is constant. Can anyone point me towards the next step?

Answer 1

You already have $f(x) = f(a/x)$, now use $f(x)f(a/x) + f(a/x)f(x) = 2f(a) = 2$ to obtain $f(x)f(a/x) = f(x)^2 = 1$. It follows that $f(x)=\pm 1$ for any $x$, so we only need to show $f(x) \ne -1$.

In fact we can show that $f(x) \ge 0$ with a nice trick: it is equal to the sum of two non-negative numbers. Can you see how? (Hint: squares are always non-negative.)

Answer 2

Hint

You already have $f(x)=f(\frac{a}{x})$(1) and $f(x^{-1})=f(x)^{-1}$ (3). Try to get $f(x)f(y)=f(xy)$(2) by(1). Hence it is easy to find $f(x)=\frac{1}{f(x)}$ by (1),(2),(3).

Therefore $f(x)=\pm 1$ for all $x$. Consequently $f$ is a homomorphism from $\langle(0,+\infty),\,\cdot\,,{}^{-1}\rangle$ into $\langle\{-1,1\},\,\cdot\,,{}^{-1}\rangle$. Then according to this discussion $f$ must be a constant valued function.