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?

Step $4)$ and $5)$ are not equivalent. If you distribute the derivative in the paranthesis at step $5)$ you will not get $4)$. — Zubzub, Dec 21 '18 at 14:08
For one thing, the function may not be differentiable. For another, $f(x) = x$ fulfills the criteria, so you can just check at which point that's no longer the case, and that's probably where you did the mistake. — Arthur, Dec 21 '18 at 14:08
Your step 1 assumes $f$ is differentiable. In fact there are discontinuous non-differentiable solutions. — kimchi lover, Dec 21 '18 at 14:08
One hint for another method. If you change variables $x=3^y$, that is $x=\exp(y\ln 3)$, then your new equation connects $y$ and $y+1$, which is a more common type of recurrence. — GEdgar, Dec 21 '18 at 14:18
I've edited the question with MathJax - here is the user guide — lioness99a, Dec 21 '18 at 14:30
Simply, the line that you correctly indicated as false assumes that $(f(3x))' = f'(3x)$, which is not correct — Damien, Dec 21 '18 at 15:14

score 1 · Answer 1 · answered Dec 21 '18 at 14:30

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, $$\frac{d(f(3x))}{dx}=\frac{d(f(3x))}{d(3x)}\frac{d(3x)}{dx}$$

answered Dec 21 '18 at 14:30

Sameer Baheti

3,239

Damien · Answer 2 · 2018-12-21T16:44:58.977

1

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 $f(x) = ax$

edited Dec 21 '18 at 16:44

answered Dec 21 '18 at 16:18

Damien

1,762

I would like to understand why this answer was downvoted ... Maybe the fact that I assume that $f'$ exists and is continue. It is implicit from OP question – Damien Dec 26 '18 at 07:45

score 1 · Answer 3 · answered Dec 21 '18 at 18:51

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 $x\in C$.

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

3 Answers3