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I feel I'm overlooking something very silly.

The Bohr compactification of $\mathbb R$ has two equivalent definitions.

  1. The set of (possibly discontinuous) homomorphisms $\mathbb R \to \mathbb T$ under the pointwise topology.

  2. The maximal ideal space of the set $AP(\mathbb R)$ of almost periodic functions. i.e the set of multiplicative linear functionals under the pointwise topology.

To represent an element of 1 as a functional we use the fact that $AP(G)$ is the closure of functions of the form

$$ \sum c_i e^{i \lambda_i x}$$

Given a homomorphism $\phi: \mathbb R \to \mathbb T$ we can form the corresponding functional by sending each $e^{i \lambda x} \mapsto \phi(\lambda)$. Since exponentials with different $\lambda_i$ are linearly independent, we can extend the above to all linear combinations by linear independence, i.e

$$ \sum c_i e^{i \lambda_i x} \mapsto \sum c_i \phi(\lambda_i).$$

The fact that the above is multiplicative follows from the exponential multiplication rules and how $\phi$ commutes with multiplication.

Finally we extend the functional to all of $AP(\mathbb R)$ by continuity.

What confuses me is that the above makes perfect sense if $\phi$ is instead a homomorphism $\mathbb R \to \mathbb C^ \times$ (complex numbers under multiplication). For simplicity let $\phi(\lambda) = e^\lambda$. Then we can define the functional

$$ \sum c_i e^{i \lambda_i x} \mapsto \sum c_i e^{\lambda_i}.$$

This functional does not correspond to a homomorphism $\phi: \mathbb R \to \mathbb T$.

What am I missing?

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    $\begingroup$ I don't see why your functional (at the end) is bounded, and hence why it extends to all of AP(R) by continuity. $\endgroup$
    – Yemon Choi
    Commented May 27 at 12:39
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    $\begingroup$ I’m voting to close this question because it is based on an elementary oversight $\endgroup$
    – Yemon Choi
    Commented May 27 at 12:55
  • $\begingroup$ @YemonChoi I see now what the problem is. We need boundedness of the functional to extend it to the closure. And the homomorphisms $\mathbb R \to \mathbb C^\times$ with nonzero real part yield unbounded functionals. $\endgroup$
    – Daron
    Commented May 28 at 14:34
  • $\begingroup$ @YemonChoi Would you happen to know if it's easy to prove the functionals $ \sum c_i e^{i \lambda_i x} \mapsto \sum c_i \phi(\lambda_i)$ are continuous? It should be enough to prove the component functionals $ \sum c_i e^{i \lambda_i x} \mapsto \sum c_j$ are continuous but even that is escaping me for the moment. $\endgroup$
    – Daron
    Commented May 28 at 14:36

1 Answer 1

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This will not be continuous because the sequence of almost periodic functions $e^{-n} e^{i n x}$ converges to zero as an almost periodic function but the value of the functional on each element of the sequence is $e^{-n} e^n=1$.

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  • $\begingroup$ Ah, I see now. You need continuity (equivalent to boundedness) wrt the max-norm in order to extend to the closure. And the guys that don't map into the circle are all unbounded. $\endgroup$
    – Daron
    Commented May 28 at 13:25
  • $\begingroup$ If you don't mind, is it hard to show the function $ \sum c_i e^{i \lambda_i x} \mapsto \sum c_i \phi(\lambda_i)$ is continuous over the space of linear combinations? This amounts to proving each projection function $ \sum c_i e^{i \lambda_i x} \mapsto c_j$ is continuous. I am aware this will be proved in any harmonic analysis textbook in much greater generality. But I am just getting a feel for the subject at the moment and would like to keep things concrete. $\endgroup$
    – Daron
    Commented May 28 at 13:28