Finding $\lambda$ for $u(x)= \lambda u(\frac{x}{2})$

Question

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.

Answer 1

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 $d \in \mathbb{N}\cup\{0\}$.