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$C([0,1]_o)$ is the set of continuous functions $f:\mathbb R \rightarrow \mathbb C$ with period 1, with the inifinity norm. I proved that if $g_1 ,g_2$ are functionals in $C([0,1]_o)^*$ that satify $g_1 (e^{2 \pi i n x})=g_2 (e^{2 \pi i n x})$ for all $n \in \mathbb Z$ then $g_1=g_2$.

Now I am asked to show the following: if there exists irrational $a$ and a functional $g$ such that $g(f(x+a))=g(f)$ for all $f\in C([0,1]_o)$, then $g(f)=c\int _0 ^1 fdx$ for some complex $c$.

To do that, I tried to show that $g(e^{2 \pi i n x})=0$ and by that to deduce the required equality, because $ c\int _0 ^1 e^{2 \pi i n}dx=0$. My efforts were unsuccessful and I only managed to show that $g(e^{2 \pi i a})=0$, and it has nothing to do with the irrationality of $a$. Any suggestions?

MasterJ
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    I suppose that your functional are $\mathbb{C}$-linear. Then, taking $f(x)=\exp(2i\pi n x)$, we get that $(\exp(2i\pi na)-1)g(\exp(2i\pi nx))=0$. If $n\not = 0$, this imply $g(f)=0$ (We use here the irrationality of $a$). Then apply your first result to $g_1=g$ and $g_2(f)=g(1)\int_0^1f(t)dt$. – Kelenner May 18 '17 at 16:37
  • Thank you, that helped :-) – MasterJ May 18 '17 at 18:57

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