If $f(3x)=3f(x)$ then what can we say about $f$?

Question

If $f(3x)=3f(x)$ for all $x \in \mathbb{R}$

Then what can we say about the function $f$?

Is it continuous? Is it differentiable?

I am trying to see if it is differentiable, If it is differentiable then It will be continuous.

$$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\lim_{h \to 0,n \to \infty} 3^n\left( \frac{f(\frac{x+h}{3^n})-f(\frac{x}{3^n})}{h}\right)$$

But It doesn't help me to go further. I do not think this would help because we have $3^n$ unbounded.

Any Ideas?

Answer 1

Here's a non-continuous $f$.

$$f(x)=\begin{cases}x&\text{if $x=3^n$}, n\in\Bbb Z,\\ 0&\text{otherwise}. \end{cases}$$

Answer 2

There are a multitude of functions $f$ with that property. Indeed, for each real number $x$ in the set $[-3,-1]\cup[1,3]$, you can choose the values of $f(x)$ arbitrarily and independently of one another, and there will be a function $f(x)$ satisfying the given functional equation: if $|x|>3$, then recursively define $f(x) = 3f(\frac x3)$, while if $0<|x|<1$, then recursively define $f(x) = \frac13f(3x)$. (If $x=0$ then we are forced to choose $f(0)=0$.)

In particular, it's not necessarily the case that $f$ is even continuous (or measurable), much less differentiable.

Answer 3

Here's a somewhat simpler discontinuous $f$:$$ f(x)=\begin{cases}x,&x\in\mathbb{Q} \\ 0,&x\notin\mathbb{Q}\end{cases}$$