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Let $G$ be a locally compact topological group and $A(G)$ be the Fourier algebra. For $f\in A(G)$ we define $H(x) = f(x^2)$. Could we say that $H\in A(G)$? What about discrete group $G$?

For definition of Fourier algebra see this link.

Hamid Shafie Asl
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1 Answers1

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It is false!

I get the following answer from Prof. Ross Stokke

Consider the group $\mathcal{Z}_2$. Suppose that $G=\Pi_1^\infty\mathcal{Z}_2$ where its topology is discrete. Set $f=\delta_{e_G}$ (point mass at $e_G$. Then $f\in A(G)$. Also for all $x\in G$, $x^2=e_G$. Hence $$H(x)= f(x^2)=f(e_G)=\delta_{e_G}(e_G)=1$$ for all $x\in G$. So $H= 1_G$. Sice $G$ is not compact we conclude that $1_G\not\in A(G)$ and consequently $H\not\in A(G)$.

user51514
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