Show that if $f$ is bounded function on $[0;1]$ and satisfying $$f(ax) = b f(x)$$ with $0 \leq x \leq \frac{1}{a}$ and $a,b > 1$ then $$\displaystyle\lim_{x \rightarrow 0^{+}} f(x) = f(0)$$
Solution: There is exist $M \geq 0$ such that $|f(x)| \geq M$ with $x \in (0;1)$. From $f(ax) = b f(x)$, $x \in [0;\frac{1}{a}]$ implies $f(a^2 x) = b^2 f(x)$ with $x \in [0;\frac{1}{a^2}]$. Use induction on $n$, we have $$f(a^n x) = b^n f(x),\quad with \quad x \in \left[0;\frac{1}{a^n}\right], n \in \mathbb{N}$$ So, $$|f(x)| \leq M \frac{1}{b^n}, \quad with \quad x \in \left[0; \frac{1}{a^n}\right], n \in \mathbb{N} \qquad (*)$$ Other way, from $f(ax) = bf(x)$ implies $f(0)=0$. Associate with $(*)$ we have Q.E.D.
I do not understand why we conclude that $|f(x)| \leq M \frac{1}{b^n}$ after use the induction. Could you explain this to me. Thanks all!
As you said, $f(a^nx) = b^n f(x)$ with $x \in [0, 1/a^n]$. So you can also write $f(x) = \frac{f(a^nx)}{b^n}$, and then by upper bounding $f(a^nx) \leq M$ you get your result, i.e. : $$|f(x)| = \left|\frac{f(a^nx)}{b^n}\right| = \frac{ \left|f(a^nx)\right|}{b^n} \leq \frac{M}{b^n}$$