Given (non-zero) $f:[0,1]\to[-1,1]$ such that $f(2x) = 3f(x)$ on $[0,1/2]$, evaluate $\lim_{x\to 0^{+}} f(x)$

Question

Let $f : [0,1] \rightarrow [-1,1]$ be a non-zero function such that $$f(2x)=3f(x)\quad,\quad x\in \left[0, \frac{1}{2}\right]$$ Then $\lim_{x \to 0+} f(x)$ is equal to ... ?

Solution:

From the discussion above, we have $ x \in \bigg[0, \dfrac{1}{2}\bigg]$

We can assume that :

$\lim_{x\to 0+} f(x) = \lim_{x\to 0+} f(2x) = a,\\ $ $\quad a$ is the limit $\forall x \in [0, \dfrac{1}{2}] \quad ...(1)$

Now, we have, $f(x) =\dfrac{f(2x)}{3}$

From $(1)$ we have,

$\lim_{x\to 0+} f(x) = \lim_{x\to 0+} \dfrac{f(2x)}{3}\\ \implies \lim_{x\to 0+} f(x) =\dfrac{lim_{x\to 0+}f(2x)}{3} \\ \implies a = \dfrac{a}{3} \\ \implies a(3-1)=0$

Since, $2 \neq 0\ \implies a=0$

Therefore the limit, $a,$ equals $0$

Is the solution correct?

Answer 1

Suppose, by contradiction,that it is not true that $\lim_{x\to 0^+}f(x)=0.$ Then there exists $r>0$ such that for every $n\in \Bbb N$ there exists $x_n\in (0, 2^{-n})$ such that $|f(x_n)|>r.$

By induction on $n\in \Bbb N,$ if $x\in (0,2^{-n})$ then $f(2^nx)=3^nf(x).$

But if $n\in \Bbb N$ is large enough that $3^nr>1,$ we have $|f(2^nx_n)|=|3^nf(x_n)|>3^nr>1,$ contrary to $f(2^nx_n)\in [-1,1].$

BTW. An example of such a function is $f(x)=x^k$ where $2^k=3.$ (That is, $k=(\ln 3)/\ln 2\approx 1.59$).

Answer 2

You can define $f$ arbitrarily over $(\frac12,1]$ and the definition over $(0,1]$ follows. Hence $f$ exists (but $f(0)=0$ contrary to the initial statement.)

Then

$$x<\frac1{2^n}\implies|f(x)|=\left|\frac{f(2x)}{3}\right|=\cdots=\left|\frac{f(2^nx)}{3^n}\right|\le\frac1{3^n}.$$

Hence by the $\delta/\epsilon$ argument the limit exists and is zero.