1. ## Function

Hi people,

f is a continuous function in $\displaystyle \mathbb{R}$ such as:
$\displaystyle (\forall x \in \mathbb{R}) \ \ f(2x)=f(x)$

I showed that $\displaystyle (\forall \alpha \in \mathbb{R}) \ \ (\forall n \in \mathbb{N}) \\f(\frac{\alpha}{2^n})=f(\alpha)$, but idon't know how to deduct that
f is constant????

2. Originally Posted by bhitroofen01
Hi people,

f is a continuous function in $\displaystyle \mathbb{R}$ such as:
$\displaystyle (\forall x \in \mathbb{R}) \ \ f(2x)=f(x)$

I showed that $\displaystyle (\forall \alpha \in \mathbb{R}) \ \ (\forall n \in \mathbb{N}) \\f(\frac{\alpha}{2^n})=f(\alpha)$, but idon't know how to deduct that
f is constant????

Use continuity to evaluate $\displaystyle \lim_{n\to\infty}f\left(\frac{x}{2^n}\right)$ ... and you kill two birds with a single stone: you prove that the function is a constant and you find that constant.

Tonio

3. This is obviously just a corollary of a much larger theorem.

Theorem: Let $\displaystyle \varphi:\mathbb{R}\to\mathbb{R}$ be a continuous function such that $\displaystyle \varphi(ax+b)=\varphi(x)$ for all $\displaystyle x\in\mathbb{R}$. Then, $\displaystyle \varphi(x)=\varphi\left(\frac{b}{1-a}\right),\text{ }a>1 1$.

Proof: First suppose that $\displaystyle a\ne 1,0$ and define a sequence $\displaystyle \left\{x_n\right\}_{n\in\mathbb{N}}$ as $\displaystyle x_1=x$ and $\displaystyle x_{n+1}=\frac{x_n-b}{a}$. Clearly then $\displaystyle \varphi(x_{n+1})=\varphi\left(a\left(\frac{x_n-b}{a}\right)+b\right)=\varphi(x_n)$, and thus $\displaystyle \varphi(x_n)=\varphi(x_{n-1})=\cdots=\varphi(x_1)=\varphi(x)$. It is also clear that $\displaystyle x_n\to\frac{b}{1-a}$. Thus, $\displaystyle \varphi(x)=\lim_{n\to\infty}\varphi(x)=\lim_{n\to\ infty}\varphi\left(x_n\right)=\varphi\left(\frac{b }{1-a}\right)$ the last part gotten from continuity since $\displaystyle x_n\to y\implies f(x_n)\to f(y)$ for continuous $\displaystyle f$.

The conclusion follows $\displaystyle \blacksquare$