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I was given this problem. Been stuck on it for a while but I have an idea. The problem reads:

Call a function "multiplicatively periodic" if there is a positive number $c \neq 1$ such that $f(cx) = f(x)$ for all $x \in R$. Prove that if a multiplicatively periodic function is continuous at $0$, then it is a constant function.

What I know, a function is constant if $f(x)$ = $k$ for every $a$ $\in$ $(-\infty,\infty)$. We want to prove the statement that multiplicatively periodic functions that are continuous at $0$ have to be constant.

My idea or approach would be to assume that $c > 1$. Then I'll have $f(cx) = f(x)$. Then $f(c^2x)$ will give the same value as $f(x)$. And so on. Eventually I would have $f(c^nx)$ = $f(x)$. Since $c^n \rightarrow \infty$ as $n \rightarrow \infty$, this would imply $\lim_{n\to \infty}f(c^n x)$ = $f(x)$.

With this same $c$, I then want to go and state $f(c^{-1}x)$ = $f(x)$. Then $f(c^{-2}x)$ will give the same value as $f(x)$. Eventually $f(c^{-n}x)$. Since $c^{-n} \rightarrow 0$ as $n \rightarrow \infty$, this would imply $f(0x) = f(0) = f(x)$.

So not matter what $x$ you start with, you'll end up at $\infty$ and $0$. $0$ is our main point of interest. I want to show that it is continuous at zero; however I do not know how to go about it from here. Would I say something around the lines of $(0-\delta,0 + \delta)$ to prove it is continuous at zero?

Thanks for taking the time to read, and thanks in advanced for the feedback.

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  • $\begingroup$ My apologies, I didnt read it completely, you seem to have addressed that issue later on. $\endgroup$ – Vishesh Oct 16 '13 at 6:09
  • $\begingroup$ Also, your question at first seems to ask you to prove that the function is constant when the function is continuous at 0. So I dont see any need of proving the continuity or am I seeing something wrong?? And $f(\infty x)$ is not good notation, I think. $\endgroup$ – Vishesh Oct 16 '13 at 6:11
  • $\begingroup$ I could be wrong, my analysis of the problem is not the best. I am fairly new to this. However, I do think we need to prove it is continuous at zero. By proving it is continuous at zero, then the function is a constant function. I hope. That being said, would I just need to prove the continuity or would I also have to prove it is constant throughout the interval? $\endgroup$ – Kevin_H Oct 16 '13 at 6:31
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    $\begingroup$ @Kevin.H. Well I was just referring to the problem in the dot box at the very beginning. It says : " Prove that if a multiplicatively periodic function is continuous at 0, then it is a constant function." $\endgroup$ – Vishesh Oct 16 '13 at 6:35
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    $\begingroup$ @Kevin H and One and All: You have to assume continuity at zero, I think, otherwise there is not necessarily a limit to $f(c^nx)$ etc. as $n \to \infty$; then for example $f(x) = f(c^nx)$ may be undefined, perhaps jumping around wildly as $n$ increases $\endgroup$ – Robert Lewis Oct 16 '13 at 6:44
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I think this might work:

Note that $f(cx) = f(x)$ for all $x \in R$ implies $f(c^{-1}x) = f(x)$ as well, since we have $f(c^{-1}x) = f(c(c^{-1}x)) = f(x)$. Furthermore, for any positive integer $n$, $f(c^nx) = f(x)$; this is easy to see by a simple induction: if $f(c^kx) = f(x)$, then $f(c^{k + 1}x) = f(c^kx) = f(x)$; the given hypothesis $f(cx) = f(x)$ forms the base ($k = 1$) case. From this we conclude that $f(c^{-n}x) = f(c^nc^{-n}x) = f(x)$ as well.

Now if $c < 1$ we have $f(x) = f(c^nx)$ for all positive integers $n$, and $c^n x \to 0$ as $n \to \infty$. By continuity of $f(x)$ at $0$, $\lim_{n \to \infty} f(x) =\lim_{n \to \infty} f(c^nx) = f(0)$, whence $f(x) = f(0)$. If $c > 1$ we apply the analogous argument to $f(c^{-n}x)$ to obtain, once again, $f(x) = f(0)$ for all $x \in R$. QED.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

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  • $\begingroup$ Thank you for your appreciative remarks; glad to help out when I can! $\endgroup$ – Robert Lewis Oct 16 '13 at 6:49

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