1

Let $f$ be a continuous function from $\Bbb R$ to itself.

If $ f(m\pm n\pi)=0$ for all $m,n\in \Bbb Z$ show that $f=O$

I am unable to proceed in this case.

Sorry I couldn't do anything positive to show you all.

I tried with some examples but don't know what to do.

Cameron Williams
  • 29,432
Learnmore
  • 31,062
  • 3
    Hint: show that for any irrational $\alpha$ the set ${m+n\alpha}_{m,n\in \mathbb Z}$ is dense in $\mathbb R$. – lulu Jun 05 '18 at 10:52
  • Are $m$ and $n$ supposed to be integers? If not, then you can show that for any $x$, $f(x) = 0$ by picking $m = x, n = 0$. – John Hughes Jun 05 '18 at 10:52
  • How to show that $m + n\pi$ is dense – Learnmore Jun 05 '18 at 10:59
  • That density result comes up a lot in number theory. As a first step to proving it, prove that it is dense near $0$. Hint: if it isn't dense near $0$ then there is a minimal non-zero element $\beta$. Show that the entire set consists of multiples of $\beta$. Deduce from that that $\beta$, and therefore $\alpha$, are rational. – lulu Jun 05 '18 at 11:02
  • 1
    Just to be clear, the statement you are trying to prove will need to use the non-trivial fact that $\pi$ is irrational. If you replace $\pi$ with, say, $\frac 12$ the claim is false. – lulu Jun 05 '18 at 11:03
  • @lulu,will you please explain what is your point in proving dat $m+n\pi$is dense – Learnmore Jun 05 '18 at 23:19
  • If a continuous function is constant on a dense set it is constant (easy exercise). – lulu Jun 05 '18 at 23:59
  • @lulu I know that,I am unable to understand how to show that $m+n \pi$ is dense – Learnmore Jun 06 '18 at 00:03
  • Why should we prove dense around 0 only – Learnmore Jun 06 '18 at 00:04
  • The links supplied in the comments to the posted solution are quite thorough. – lulu Jun 06 '18 at 00:09

1 Answers1

2

Hint: Here are the relevant facts:

  • If $\alpha \in \mathbb R$, then $\mathbb Z + \alpha\mathbb Z=\{m+n\alpha\ : m,n\in \mathbb Z\}$ is an additive subgroup of $\mathbb R$ .

  • An additive subgroup of $\mathbb R$ is either cyclic or dense.

  • $\mathbb Z + \alpha\mathbb Z$ is cyclic iff $\alpha$ is rational.

  • $\pi$ is irrational.

lhf
  • 216,483
  • See https://math.stackexchange.com/questions/136665/for-every-irrational-alpha-the-set-ab-alpha-a-b-in-mathbbz-is-den and https://math.stackexchange.com/questions/90177/subgroup-of-mathbbr-either-dense-or-has-a-least-positive-element for instance – lhf Jun 05 '18 at 11:37