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Let $f:\left[0,1\right]\to \mathbb{R}$ be a bounded function satisfying $f(2x)=3f(x)$ for $0\le x<\frac{1}{2}$

1)Show that $f(2^{n}x)=3^{n}f(x)$ for $0 \le x< \frac{1}{2^{n}}$ for all $n \in \mathbb{N}$

2)Prove that $\lim\limits_{x\to 0^{+}}f(x)=f(0)$

I could do the first part of this question by mathematical induction but with regards to the second part I have no idea about what to do. Maybe I can use the $\epsilon$ ,$\delta$ definition of the right limit

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Try applying what you have proven in the first part. Rewriting the equation, we have: $$f(x) = 3^{-n}f(2^nx)$$ for all $x \in [0, 2^{-n})$. Since $f$ is bounded, say $|f(x)| < M$ for some constant $M$ for all $x \in [0, \tfrac{1}{2})$, this implies that $|f(x)| < 3^{-n}M$ for all $x \in [0, 2^{-n})$. Can you proceed from here?

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  • $\begingroup$ Thank you Haran,in the end right side goes to zero hence absolute of $f$ goes to zero so this implies directly.I meant there always exists an $n$ such that $x$ $0<3^{-n}<x$ Hence $3^{-n}$ must go to zero hence $f(x)$ goes to and using $f(0)=0$ completes the task. $\endgroup$
    – maths and chess
    Commented Jun 13, 2023 at 8:00

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