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Given a function $f:\mathbb{R}\to\mathbb{R}$ that is continuous at $0$, and such that there exists a non-zero natural number $k$ (not equal to $1$) for which $f(kx)=f(x)$, show that $f$ is constant.

kandb
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xichrome
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  • I attempted to prove the constancy of the function using limits, by considering the fact that f(kx) and f(x) have the same limit as x approaches infinity. I constructed a sequence k^nx, which diverges to infinity, but unfortunately, I was unable to make any progress with this method. Any help or suggestions would be greatly appreciated – xichrome Jan 26 '23 at 03:24
  • Hint: we have $f(x/k) = f(x)$ for all $x$. – Mark Saving Jan 26 '23 at 03:32
  • Interesting question! You might consider in future providing some of your own thoughts on the problem as a show of good faith and effort – kandb Jan 26 '23 at 03:42
  • Thank you so much for your response, it was very helpful. Your suggestion of using a sequence was not far from my own thoughts, but I realized that I had overlooked the simplest solution which is making the limit zero. Your help is greatly appreciated. Thanks to all the folks in this forum as well. – xichrome Jan 26 '23 at 03:55

1 Answers1

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First, let's build intuition about this problem. We're trying to prove $f$ is constant. What happens if it's not? What if $f(x_1) = y_1$ and $f(x_2) = y_2 \neq y_1$?

If you think about this possibility, you should start to become suspicious: that means $y_1 = f(x_1/k) = f(x_1/k^2) = \ldots$ and $y_2 = f(x_2/k) = f(x_2/k^2) = \ldots$ and that seems impossible since it suggests $f(0)$ is equal to both $y_1 $ and $y_2$.

So, now that we have a rough outline of an argument, we can look for a proof. Suppose, for contradiction, that $f(x_1) \neq f(x_2)$ for some $x_1, x_2$. Can you think of how to use the $\epsilon,\delta$ definition of continuity to turn the above informal argument into a rigorous proof?

user7530
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