## New answers tagged topological-groups

4

To complement Christian Remling's nice concrete explanation, let me just add that more is known.
Your 2nd question can be rephrased as asking: does the function $x\mapsto |x|$ ``operate in'' the Fourier algebra? This kind of question used to be of interest to people working on Banach algebras of functions. For the Fourier algebra of a LCA group the answer ...

1

This won't work. I want to show that we can't take square roots in $A(\mathbb R)$. My function will be of the type
$$
f(x) = \sum h_n \varphi\left( \frac{x-a_n}{L_n}\right) ,\quad\quad\quad\quad (1)
$$
and here the individual summands will have disjoint supports. I will take $h_n\in\ell^2$, $h_n\notin\ell^1$. Since the $L^1$ norm of the Fourier transform of ...

2

I'm not sure about the proof I gave, But as I checked, it didn't use full AC, but as Asaf mentioned in a comment it uses DC. The following theorem is due to Pontrjagin. See Book by Montgomery and Zippin(Page 29).
I will give a sketch of proof.
Note: Maybe it's needed to add some separation axiom to the following theorem. please edit it, if needed.
Theorem: ...

5

No, contrary to what you said, it is not always possible, even if $G$ is a semidirect product. Let us fix some $g \in N$, so $\sigma(gx) = \sigma(x)$ for all $x$. Then $\sigma(x)^{-1} g \sigma(x) = \omega(g,x)$ is a bounded function of $x$. It is easy to construct a counterexample to this, by arranging for the conjugacy class of $g$ to be unbounded. For ...

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