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Let $I$ be the subset $[-1,1]$ of $\Bbb R$. Consider a function $f:I\to I$ with properties:

(a) $f(x) = 0$ at all points $x_n = 2^{1-n}$ and $y_n = -2^{1-n}$ in $I$ ($n$ varies throughout all natural numbers).

(b) $f(x)$ is continuous at all these points.

(c) $f(x)$ is a bounded function.

(d) For all integers $n$, $f(x)$ with domain restricted to $[-2^{-n},2^{-n}]$ is equal to $\dfrac 1 {a^n}f(2^nx)$ for some fixed constant $a > 1$.

(e) $f(0) = 0.$

Is $f(x)$ necessarily continuous at the point $x = 0$?

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    yes, thanks for correction2012-11-28

1 Answers 1

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By hypothesis (d), $f(0)=a^{-1}f(0)$ hence $f(0)=0$. By hypothesis (c), there exists a finite $c$ such that $|f(x)|\leqslant c$ for every $|x|\leqslant1$. By hypothesis (d) again, $|f(x)|\leqslant ca^{-n}$ for every $|x|\leqslant2^{-n}$, for every $n\geqslant0$. Thus $f(x)\to0$ when $x\to0$ and $f$ is continuous at $0$.

(Hypotheses (a), (b) and (e) are not needed.)

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    hahahaha that seriously made me laugh. I mea$n$t to write (d) with each of those intervals not containing 0.2012-11-29