Value of $f(2016),$ If $f(x) = f(2x)\forall x \in \mathbb{R}$ are given

Question

If $f(x) = f(2x)\forall \; x \in \mathbb{R}$ and $f(x)$ is continuous for all $x\in \mathbb{R}$ and $f(5) = 10,$ Then $f(2016) = $

$\bf{My\; Try::}$ Let $0<x<1$

Then $f(x)=f(2x) = f(2^2x) = f(2^3x)=......=\lim_{n\rightarrow \infty}f(2^nx) = f(0)=\bf{Constant}$

and when $x>1$

Then $\displaystyle f(x) = f\left(\frac{x}{2}\right) = f\left(\frac{x}{2^2}\right)=f\left(\frac{x}{2^3}\right)=.......=\lim_{n\rightarrow \infty}=f\left(\frac{x}{2^n}\right)=f(0)=\bf{Constant}$

But I did not understand How can we solve for $x<0$ and $x=0$ and for $x=1$

Help Required, Thanks

Answer 1

For any real number $x$, and any non-negative integer $n$, you have:

$f(x) = f\left(\dfrac{x}{2}\right) = f\left(\dfrac{x}{4}\right) = \cdots f\left(\dfrac{x}{2^n}\right)$. Hence, $f(x) = \displaystyle\lim_{n \to \infty}f\left(\dfrac{x}{2^n}\right)$.

But, since $f(x)$ is continuous, $\displaystyle\lim_{n \to \infty}f\left(\dfrac{x}{2^n}\right) = f\left(\lim_{n \to \infty}\dfrac{x}{2^n}\right) = f(0)$.

Therefore, $f(x) = f(0) = \textbf{constant}$ for all real numbers $x$.

Since we are also given that $f(5) = 10$, we get $\textbf{constant} = 10$. Therefore, $f(2016) = 10$.

Answer 2

I think your proof has already made sense.

$$f(x)=f\left(\frac{x}{2}\right)=...=f\left(\frac{x}{2^n}\right),$$

let $$n\rightarrow\infty,f(x)=f(0)$$

So $f(2016)=10.$

Answer 3

f is continuous. For any $\epsilon >0$, we can find $\delta$ so that $|x - y| < \delta \implies |f (x)-f (y)| < \epsilon$.

For any delta I can find an $m $ so that $2016/2^m - 5/2^m =(2011)/2^m <\delta $.

So $|f (2016/2^m) - f (5/2^m)| < \epsilon $. By induction $f (x/2^m)=f (x) $ so $|f (2016) - f (5)| < \epsilon $.

But $\epsilon$ was arbitrary and so $f (2016)=f (5)=10$.