Functional Funkiness

Let us test whether a linear function $f(z)$ satisfies this property. That is, it satisfies the superposition principle, so it is both homogenous and additive. Then we can write the identity given as $2f(x) + 2f(y) = f(f(x) + f(y)) \longrightarrow 2(f(x) + f(y)) = f(f(x) + f(y))$ Clearly, the function $f$ carries out a multiplication by 2, so $f(z) = 2z$ . It is also real, differentiable and continuous, while satisfying the identity for all $x,y\in \mathbb{R}$ , so this is a valid function. This is shown to be true below. $f(2x) + 2f(y) = 4x + 4y = 4(x+y)$ $f(f(x+y)) = 2(2(x+y)) = 4(x+y)$ However, since $f(z)$ is linear, $f(az) = af(z) = 2az$ for $a\in \mathbb{R}$ will also satisfy this property. Therefore, there are an infinite number of functions which satisfy these conditions.

Edit: I found a flaw in my last statement. It only works if $a = a^2$ so $a = 0$ or $a = 1$ . There must be other functions out there that satisfy this.

Differentiating with respect to $x$ :

$2f'(2x) = f'(f(x+y))f'(x+y)$

Differentiating with respect to $y$ : $2f'(y) = f'(f(x+y))f'(x+y)$

$\implies f'(y) = f'(2x)$

Plugging $x=y=0$ into the given equation: $f(0) + 2f(0) = f(f(0)) \implies 3f(0) = f(f(0))$

Let $f(0) = p$ .

Now: $2f'(2x) = f'(f(x+y))f'(x+y)$ $\implies 2f'(0) = f'(f(0))f'(0)$ $\implies f'(p) = 2$

Using: $\implies f'(y) = f'(2x)$

Plugging $x=p/2$ gives:

$f'(y) = f'(p) =2$ $\implies f(y) = 2y + p$

$\because f(0) = p$

And the condition $3f(0) = f(f(0))$ is also satisfied. Therefore, since $p$ can be any real number, an infinite number of functions satisfy the given equation.

Let's differentiate the equation:

$2f'(2x) = f'(f(x+y)) * f'(x+y)$ , for any $x$ and $y$ .

Consider $y = x$ :

$2f'(2x) = f'(f(2x)) * f'(2x)$ , for any $x$ .

If $f'(2x) \neq 0$ :

$f'(f(x)) = 2$

That would mean that initial differentiated equation looks like this:

$2f'(2x) = 2f'(x+y)$

$f'(2x) = 2f'(x+y)$

For $x = 0$ :

$f'(y) = f'(0)$ meaning $f'(x) = 2$ .

So $f(x) = 2x + C$ . And if we input this in initial equation we conclude that any real C fits, so we have infinite functions. (We didn't even consider $f'(2x) = 0$ case)