Iterative Nightmare

$\begin{aligned} g(x) & = x + f(x+f(x+f(x+f(\cdots)))) & \small \color{#3D99F6} \text{Note that }f(x) = \frac 12 f(2x) \\ & = x + \frac 12 f(2x+2f(x+f(x+f(\cdots)))) \\ & = x + \frac 12 f\left(2x+\frac 22 f\left(2x+\frac 22 f\left(2x+\frac 22 f\left(\cdots\right)\right)\right)\right) \\ & = \color{#3D99F6} x + \frac 12 f\left(2x+f\left(2x+f\left(2x+ f\left(\cdots\right)\right)\right)\right) \end{aligned}$

Now,

$\begin{aligned} g(2x) & = 2x + f(2x+f(2x+f(2x+f(\cdots)))) \\ & = 2\left({\color{#3D99F6}x + \frac 12 f\left(2x+f\left(2x+f\left(2x+ f\left(\cdots\right)\right)\right)\right)}\right) \\ & = 2{\color{#3D99F6}g(x)}\ \blacksquare \end{aligned}$

The only type of function define in $\mathbb{R}$ that satisfy the property $f(2x)=2f(x)$ is the linear function $f(x) =kx, \forall k \in \mathbb{R}$ .

Since $g(x)$ is construced by $f(x)$ with a recursive sequence of linear combination two linear functions ( $f(x)$ and $x$ ) , also $g(x)$ is linear.

Then $g(2x)=2g(x)$

Let $(g_n)_{n\in\mathbb{N}}$ the functional sequence such that $g_0(x) = x$ and $\forall n \in \mathbb{N}, g_{n+1}(x)=x+f(g_n(x))$ .

We have $\lim_{n\rightarrow\infty}g_n(x) = g(x)$ .

We have as well $g_0(2x) = 2x = 2g_0(x)$ .

Let's suppose that for $n\in\mathbb{N}, g_n(2x)=2g_n(x)$ , $\forall x \in \mathbb{R}$ .

We then have $g_{n+1}(2x)=2x+f(g_n(2x))=2x+f(2g_n(x))=2x+2f(g_n(x))=2(x+f(g_n(x))=2g_n(x)$

Hence $\forall n \in \mathbb{N}, g_{n}(2x)=2g_n(x)$ , $\forall x \in \mathbb{R}$ and thus $g(2x) = \lim_{n\rightarrow\infty}g_n(2x) = \lim_{n\rightarrow\infty}2g_n(x) = 2\lim_{n\rightarrow\infty}g_n(x) = 2g(x))$

Note that $g(x)=x+f(g(x))$ which yields $g(x)-f(g(x))=x$

Notice this means the function $h(y)=y-f(y)$ is the inverse function of $g$ , when its domain is restricted to $g(\mathbb{R})$ .

Clearly, $h$ has the property $h(2y)=2h(y)$ for all $y \in \mathbb{R}$ . Now choose $x \in \mathbb{R}$ and set $y=g(x)$ . Unfortunately, to proceed from this point, one must assume $2y \in g(\mathbb{R})$ , which is a weakened version of the conclusion we seek to prove. Nonetheless, from here we have

$g(2x)=g(2h(y))=g(h(2y))=2y=2g(x)$ as required.

If someone can complete this argument, it would be much appreciated.

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Edit: Thank you to Andrew Hayes and Thomas Ponweiser for continuing my investigation into this problem, eventually leading to the problem being edited.

Please could someone tell me if this is adequate: $g(x)=x+f(x+f(x+f(...)))$

$=x+f(g(x))$

$2g(x)=2x+2f(g(x))$

since: $g(2x)=2x+f(2x+f(2x+f(2x...)))$

$=2x+2f(x+f(x+f(x+f(...)))$

$=2x+2f(g(x))$

$2g(x)=g(2x)$

g(x) = x + f(x + f(x + f(...)))

g(x) = x + f(g(x))

2g(x) = 2 (x + f(g(x)))

2g(x) = 2x + 2f(g(x))

2g(x) = 2x + f(2g(x))

2g(x) = 2x + f(2x + f(2g(x)))

2g(x) = 2x + f(2x + f(2x + ....)) = g(2x)

The sequence terminates, right? For then it is a trivial exercise in induction.