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Let f be a 2$\pi$ periodic, Riemann integrable function and let $\alpha$ be an irrational number. Suppose that $f(x+2\pi\alpha)=f(x)$ for all x. Show that f is constant almost everywhere.

I know that If f,g is $2\pi$ periodic function and the coefficients of Fourier series of f,g are same, then f=g

Finally, I want to prove that there exists a constant c such that $\{x:f(x)\neq c\} $is of measure zero.

How to prove that?

asfajaf
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2 Answers2

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Compute the $n$th Fourier coefficient of $f$: $$\begin{aligned} a_n &= \frac{1}{2\pi}\int_0^{2\pi}f(x) e^{-inx} dx \\ &= \frac{1}{2\pi}\int_0^{2\pi}f(x + 2\pi \alpha) e^{-inx} dx \\ &= \frac{1}{2\pi}\int_0^{2\pi}f(x)e^{-in(x - 2\pi \alpha)} dx \\ &= e^{in2\pi \alpha} \frac{1}{2\pi}\int_0^{2\pi}f(x) e^{-inx} dx \\ &= e^{in2\pi \alpha} a_n \\ \end{aligned}$$ (Note that we can still take the limits of integration as $0$ to $2\pi$ after the change of variable in the third line because both $f$ and $x \mapsto e^{-in(x-2\pi\alpha)}$ have period $2\pi$.)

Now, since $\alpha$ is irrational, we must have $e^{i n 2\pi \alpha} \neq 1$ whenever $n \neq 0$, which forces $a_n = 0$ for all $n \neq 0$. This means that the Fourier series for $f$ is simply $a_0$. Equivalently, the Fourier coefficients of the function $f - a_0$ are all zero. What can you conclude?

  • Why integral range is not change? – asfajaf Apr 08 '16 at 18:40
  • @asfajaf: I just edited to explain that - you may need to refresh to see the update. Since the integrand is $2\pi$-periodic, we can integrate over any interval of length $2\pi$ and the answer will be the same. –  Apr 08 '16 at 18:41
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As an additional exercise, you might like to try the following:

  • Show that for any $x\in[0,2\pi]$ the set $\{x+2n\pi\alpha\mod2\pi:n\in\mathbb Z\}$ is dense in $[0,2\pi]$
  • Show that if $f\equiv c_1$ on $C_1$ and $f\equiv c_2$ on $C_2$, where $C_1$ and $C_2$ are both dense subsets of $[0,2\pi]$, then $c_1=c_2$ (Hint: use Riemann integrability)
  • Conclude that $f$ is constant everywhere, not just almost everywhere.
Jason
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  • +1, nice observation. I thought that one would need the additional hypothesis that $f$ is continuous in order to make this conclusion, but this shows it's actually a consequence. –  Apr 08 '16 at 19:19