Functional equation: $f(3x) = 3f(x)$

Question

Find $f(x)$ if: $$f: R \to R$$ $$f(3x) = 3f(x)$$

I've tried to find $f(x)$ by differentiating:

\begin{align}(f(3x))' &= (3f(x))'\\ 3f '(3x) &=3f '(x)\\ f '(3x) &= f '(x)\\ f'(3x) - f '(x) &= 0\\ (f(3x) - f(x))' &= 0\tag{$\dagger$}\\ (2f(x))' &= 0\\ &\Downarrow\\ f(x) &= c\end{align}

for $c$ constant.

I know, that I am wrong, but I don't know why. I think, that my mistake is in $(\dagger)$. But I can't explain why.

Could you please help me to solve this equation and explain, what's wrong?

Answer 1

In the second step $$f'(3x)=\frac{d(f(3x))}{d(3x)}$$ which can't be reconciled with what you've done in $(4)\to(5)$. Indeed, $$\frac{d(f(3x))}{dx}=\frac{d(f(3x))}{d(3x)}\frac{d(3x)}{dx}$$

Answer 2

Concerning your error: the line that you correctly indicated as false assumes that $(f(3x))′=f′(3x)$, which is not correct.

Now, to get a simple demonstration, I assume that $f'$ exists and is continue.

You have already shown that $f'(3x) = f'(x)$. Then,

$$f'(x) = f'(x/3) = f'(x/9) = \dots = f'(x/3^n)$$ By continuity ($n\to \infty$), $f'(x) = f'(0) := a$ and then $f(x) = ax$

Answer 3

Since no-one else has posted non differentiable solutions, I will give one here:

Let $f(x) = x$ for rational $x$ and let $f(x)=0$ for irrational $x$. This $f$ satisfies the functional equation.

More generally, define the equivalence relation $\sim$ on $\mathbb R$ by $x\sim 3x$. For each distinct coset $C$ pick a constant $a_C$, and let $f(x)=a_C x$ for $x\in C$.