, 2016

The identity function $f(x) = x$ clearly satisfies the functional equation. Now, let $f$ be a function satisfying the functional equation. Plugging $x = y = 1$ into (3) we get $2 f(f(z) + 1) = (z + 1) f(2)$ for all $z \in \mathbb{R}^{+}$. Hence, $f$ is not bounded above.

Lemma. Let $a, b, c$ be positive real numbers. If $c$ is greater than $1, a / b$ and $b / a$, then the system of linear equations
$$c u + v = a \quad u + c v = b$$
has a positive real solution $u, v$.
Proof. The solution is
$$u = \frac{c a - b}{c^{2} - 1} \quad v = \frac{c b - a}{c^{2} - 1}$$
The numbers $u$ and $v$ are positive if the conditions on $c$ above are satisfied.

We will now prove that
$$f(a) + f(b) = f(c) + f(d) \quad \text{ for all } a, b, c, d \in \mathbb{R}^{+} \text{ with } a + b = c + d .$$
Consider $a, b, c, d \in \mathbb{R}^{+}$ such that $a + b = c + d$. Since $f$ is not bounded above, we can choose a positive number $e$ such that $f(e)$ is greater than $1, a / b, b / a, c / d$ and $d / c$. Using the above lemma, we can find $u, v, w, t \in \mathbb{R}^{+}$ satisfying
$$\begin{array}{ll} f(e) u + v = a, & u + f(e) v = b \\ f(e) w + t = c, & w + f(e) t = d . \end{array}$$
Note that $u + v = w + t$ since $(u + v)(f(e) + 1) = a + b$ and $(w + t)(f(e) + 1) = c + d$. Plugging $x = u, y = v$ and $z = e$ into (3) yields $f(a) + f(b) = (e + 1) f(u + v)$. Similarly, we have $f(c) + f(d) = (e + 1) f(w + t)$. The claim follows immediately.
We then have
$$y f(x) = f(x f(y)) \quad \text{ for all } x, y \in \mathbb{R}^{+}$$
since by (3) and (4),
$$(y + 1) f(x) = f\left(\frac{x}{2} f(y) + \frac{x}{2}\right) + f\left(\frac{x}{2} f(y) + \frac{x}{2}\right) = f(x f(y)) + f(x) .$$
Now, let $a = f(1 / f(1))$. Plugging $x = 1$ and $y = 1 / f(1)$ into (5) yields $f(a) = 1$. Hence $a = a f(a)$ and $f(a f(a)) = f(a) = 1$. Since $a f(a) = f(a f(a))$ by (5), we have $f(1) = a = 1$. It follows from (5) that
$$f(f(y)) = y \quad \text{ for all } y \in \mathbb{R}^{+} .$$
Using (4) we have for all $x, y \in \mathbb{R}^{+}$ that
$$\begin{aligned} & f(x + y) + f(1) = f(x) + f(y + 1), \quad \text{ and } \\ & f(y + 1) + f(1) = f(y) + f(2) . \end{aligned}$$
Therefore
$$f(x + y) = f(x) + f(y) + b \quad \text{ for all } x, y \in \mathbb{R}^{+}$$
where $b = f(2) - 2 f(1) = f(2) - 2$. Using (5), (7) and (6), we get
$$4 + 2 b = 2 f(2) = f(2 f(2)) = f(f(2) + f(2)) = f(f(2)) + f(f(2)) + b = 4 + b .$$
This shows that $b = 0$ and thus
$$f(x + y) = f(x) + f(y) \quad \text{ for all } x, y \in \mathbb{R}^{+} .$$
In particular, $f$ is strictly increasing.
We conclude as follows. Take any positive real number $x$. If $f(x) > x$, then $f(f(x)) > f(x) > x = f(f(x))$, a contradiction. Similarly, it is not possible that $f(x) < x$. This shows that $f(x) = x$ for all positive real numbers $x$.