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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?

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Try to see if you can compute $f(a^2)$. I promise that it will help! ;) – Ravi Jan 26 '13 at 5:38
Let x = y = a. $f(a)f(a) + f(\frac{a}{a})f(\frac{a}{a}) = 2f(a\cdot a)$. $f^2(a) + f^2(1) = 2f(a^2)$. $1^2 + 1^2 = 2f(a^2)$. $2 = 2f(a^2)$. $1 = f(a^2)$. Kind of helpful, but much more helpful to find $f^2(x)$. – SSumner Jan 26 '13 at 18:25
up vote 2 down vote accepted

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.)

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Is this valid? Let $f(x) = -1$ for all $x$. Then we have $-1\cdot-1 + -1\cdot -1 = 2\cdot-1$. Then we have $1 + 1 = -2$, or $2 = -2$, which is an obvious error. Thus $f(x)$ must equal $1$ for all $x$. – SSumner Jan 26 '13 at 18:38
@SSumner Sorry, not valid. We already know $f(a) = 1$ so your first assumption is an obvious strawman. The contradiction only means it is impossible for $f(x) = -1$ all the time. What if $f(x) = 1$ for some $x$ and $f(x) = -1$ for other $x$? – Erick Wong Jan 26 '13 at 19:32
Oops. You're right. How about this? Let $f(x) = -1$ for some x and $f(y) = 1$ for all other x. Then we have $-1\cdot1 + -1\cdot1 = 2\cdot f(xy)$. Then we have $-1 - 1 = 2f(xy)$, which yields $f(xy) = -1$. But $xy \neq x$ for all $y \neq 1$, and thus it is impossible for $f(x) = -1$ except when $y = 1$. But how do I show that when $y = 1$, $f(x) = 1$? – SSumner Jan 26 '13 at 20:30
@SSumner You are again ruling out only a tiny sliver of possibilities. There are (literally) countless ways that “$f(x) = 1$ for all $x$” might fail. You cannot hope to pick out one specific scenario at a time. Try assuming $f(x) = -1$ for some $x$ and nothing beyond this. You can still get a contradiction. – Erick Wong Jan 26 '13 at 20:38


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.

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I don't understand how $f(x)$ cannot be $-1$ in the linked discussion. Can you clarify that some? – SSumner Jan 26 '13 at 20:32
@SSumner Suppose we know $f(x) = f(\sqrt{x}) f(\sqrt{x})$. What can you conclude, even without knowing what $f(\sqrt{x})$ is? – Erick Wong Jan 26 '13 at 20:42
I see! $f(x) = f(\sqrt{x})f(\sqrt{x})$ implies $f(x) = f^2(\sqrt{x})$. This in turn implies $\sqrt{f(x)} = f(\sqrt{x})$. But we cannot have the square root of a negative number, so by definition $f(x) \geq 0$. Thus $f(x) = 1$. – SSumner Jan 26 '13 at 20:45

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