Finding $\lambda$ for $u(x)= \lambda u(\frac{x}{2})$ 
Given:
  1. Function $u(x)$ is infinitely differentiable and $\lambda \in \Bbb{R}$
  2. Identity ①: $u(x)= \lambda \cdot u\left(\dfrac x2\right)$.
  Find all $\lambda$, for which ① is true for all x in R. 

Could you kindly check the solution and point out where it is wrong or unclear? My assumption was that all $\lambda$ are of a $2^n$ kind where $\space n \in \{\Bbb{Z^+} \cup 0\}$.
Here is an attempt for a solution so far: If $u(x)=\lambda u\left(\dfrac{x}{2}\right)$ then
\begin{align}
u(x)& = \lambda u\left(\frac{x}{2}\right)\\
\lambda u\left(\frac{x}{2}\right) 
& = \lambda ^2 u\left(\frac{x}{4}\right)\\
\lambda ^2 u\left(\frac{x}{4}\right)
& = \lambda ^3 u\left(\frac{x}{8}\right)\\
&\ \ \vdots\\
\lambda ^{n-1} u\left(\frac{x}{2^{n-1}}\right)
& = \lambda ^n u\left(\frac{x}{2^n}\right)\\
\implies u(x) &= \lambda ^n u\left(\frac{x}{2^n}\right) \tag*{②}
\end{align}
Now consider derivatives:
\begin{align}
u'(x)& = \frac{\lambda}{2} u'(\frac{x}{2})\\
u''(x)& = \frac{\lambda}{2^2} u''(\frac{x}{2})\\
u'''(x)& = \frac{\lambda}{2^3} u'''(\frac{x}{2})\\
& ...\\
u^{(n)}(x)& = \frac{\lambda}{2^n} u^{(n)} (\frac{x}{2^n})\\
\implies u^{(n)}(x)&= \left(\frac{\lambda}{2}\right)^n \space u^{(n)} \left(\frac{x}{2^n}\right) \tag*{③}\\
\end{align}
which we come to after repeating the last equation similarly to ②.
$\lambda=1$ from ② implies $u(x)=a$ for $a \in \Bbb{R}$, and $|\lambda| < 1$ from ② implies $u(x) \to 0$ and thus $u(x) = 0$ for all $x$. (Is this line  correct?)
I am a bit lost with the logic in the line below:
$u(0) \ne 0$,
$\dfrac{u(x)}{\lambda^n}
\to u(0)
$ when 
$|\lambda| > 1 \implies $,
$\dfrac{u(x)}{\lambda^n} \to 0 \quad$
contradicts
$\quad u(0) \ne 0$.
now we have:
$u(0) = u'(0) = u''(0) = \cdots = u^{(n+1)} = 0\implies$ using Taylor for $u(0)=0$
$u(x) = xu'(0)+O(x^2)$
$u\left(\frac{x}{2^n}\right) = x\frac{u'(0)}{2^n} + 
O\left(\frac{x^2}{4^n}\right)$
by using ②
$u(x)
=\lambda^n\left(x\frac{u'(0)}{2^n}+O\left(\frac{x^2}{4^n}\right)\right)$
$u(x)=\left(\frac{\lambda}{2}\right)^n xu'(0) + 
O\left(x^2\left(\frac{\lambda}{4}\right)^n\right)$.
in this equation
now let us apply 
$|\lambda| < 2 \implies u(x) \to 0$
$|\lambda| > 2 \implies u(x) \to \infty$.
$\therefore \lambda = 2$.
finally, applying the above procedure to the $(n)^{th}$ derivative of $u(x)$ and using ③ 
$u^{(n)}(x)=\left(\frac{\lambda}{2^n}\right)^n x\left(u^{(n+1)}(0)\right) + 
O\left(x^2\left(\frac{\lambda}{4^n}\right)^n\right)$.
we obtain that the $\lambda$ are all of a $2^n$ kind
Please feel free to point out the mistakes and\or missing parts.
P.S. If someone is interested there is also a more generalized version of this question on $K$ dimensions.
 A: Let $u : \mathbb{R} \to \mathbb{R}$ be infinitely differentiable and satisfies $u(x) = \lambda u(x/2)$ for some $\lambda \in \mathbb{R}$. We also assume that $u$ is not identically zero, to investigate non-trivial solutions.


*

*If $|\lambda| < 1$, then $u(x) = \lambda^k u(x/2^k) \to 0$ as $k \to \infty$, hence $u$ is identically zero, which contradicts our assumption. We henceforth assume that $|\lambda| \geq 1$.

*For any integer $N \geq 0$,
$$ u^{(N)}(x) = \frac{\lambda}{2^N} u^{(N)}(x/2). $$
Pick $N$ large that $|\lambda| < 2^N$. Then this implies $u^{(N)}(x) = (\lambda/2^N)^k u^{(N)}(x/2^k) \to 0$ as $k\to\infty$. So $u^{(N)}$ is identically zero, and therefore $u$ is polynomial.

*Since $u$ is a non-zero polynomial, we have
$$ \lambda = \lim_{x \to \infty} \frac{u(x)}{u(x/2)} = 2^d, $$
where $d$ is the degree of $u$.

*Finally, pick any $a \neq 0$ so that $u(a) \neq 0$. If we set $c = u(a)/a^d$, then
$$u(2^k a) = \lambda^k u(a) = c (2^k a)^d $$
for all $k \geq 0$. Since both $u(x)$ and $cx^d$ are polynomials which coincide for infinitely many values of $x$, it follows that $u(x) = cx^d$.
The converse is trivial. Summarizing:

Proposition. Let $u : \mathbb{R} \to \mathbb{R}$ be a $C^{\infty}$-function which is not identically zero and $$\forall x \in \mathbb{R}, \qquad u(x) = \lambda u(x/2)$$ for some $\lambda \in \mathbb{R}$. Then $u(x) = cx^d$ and $\lambda = 2^d$ for some $c \neq 0$ and $d \in \mathbb{N}\cup\{0\}$.

