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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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Do you mean $n$ varies throughout all natural numbers perhaps? –  Pedro Tamaroff Nov 28 '12 at 15:24
    
yes, thanks for correction –  Adam Rubinson Nov 28 '12 at 15:39

1 Answer 1

up vote 3 down vote accepted

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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Clearly you are correct. I was trying to make up a question which was trickier and which encorporated (a), (b) and (e) and trying to see when f is continuous at x = 0. My question above is too easy because of (c) and (d). –  Adam Rubinson Nov 28 '12 at 15:37
    
Got rid of hypothesis (e) as well. –  Did Nov 28 '12 at 18:08
    
hahahaha that seriously made me laugh. I meant to write (d) with each of those intervals not containing 0. –  Adam Rubinson Nov 29 '12 at 9:23

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