I'm studying Pontryagin duality, and I read that the dual of a locally compact group $G$ is still locally compact.
EDIT: I got that this is indeed true.
Do you know how to prove it?
It seems to me that one have to prove that the closure of $V(K, \varepsilon) =\{f: f(K) \subset (-\varepsilon, \varepsilon) \} $ is compact for some $K, \varepsilon$. This would be a compact neighborhood of the constant character.
Infact, suppose that there a is a compact neighborhood $C$ of the constant character. Then there is some element of the basis $V(K, \varepsilon)\subset C $. Its closure would be closed in a compact, thus compact.
But I can't succed in showing the fact I mentioned. Thank you for the help, Andrea
This is a little silly, since there must be a proof out there in books.
The local compactness of $\hat G$ is going to come from equicontinuity.
Say $K$ is a compact neighborhood of the identity in $G$, $0<\alpha<1$, and $E$ is the set of all $\chi\in\hat G$ such that $|1-\chi|<\alpha$ on $K$. Then $E$ is equicontinuous at the origin: Given $\epsilon>0$, there exists $n$ such that if $z\in\Bbb T$ and $\epsilon\le|1-z|<\alpha$ then $|1-z^j|>\alpha$ for at least one integer $j$ with $1\le j\le n$ (the fact that $\alpha<1$ can't be omitted there.)
Edit: When I wrote this I was taking that last sentence as clear.. In answer to a comment, here's a formal proof. Note first that $|1-e^{is}|$ is a strictly increasing function of $s$, for $s\in[0,\pi]$. Now wlog $z$ has positive imaginary part. Hence there exists $t\in (0,\pi/2]$ with $z=e^{it}$. So $t\ge|1-e^{it}|>\epsilon$. Choose $\theta\in(0,\pi)$ with $|1-e^{i\theta}|=\alpha$. Since $t>\epsilon$ there exists $j$ with $1\le j\le \theta/\epsilon+1<\pi/(2\epsilon)+1$ such that $(j-1)t\le \theta<jt$. Hence $jt=(j-1)t+t\le \theta+\pi/2\le\pi$. Since $\theta<jt\le\pi$ the mmontonicity mentioned above shows that $|1-z^j|>|1-e^{i\theta}|=\alpha$.
Now choose a neighborhood of the origin $U$ such that $U+U+U\dots +U\subset K$. Then if $\chi\in E$ the fact that $\chi$ is multiplicative shows that $|1-\chi|<\epsilon$ on $U$.
So $E$ is equicontinuous at the origin, and hence uniformly equicontinuous at every point. Now a suitable version of Arzela-Ascoli(???) must show that $E$ is precompact in the compact-open topology.
Surely that's all there is to it.
Edit: Yes, that's all there is to it.
Details: Say a modulus of continuity on $G$ is a function $\omega:(0,\infty)\to\mathcal P(G)$ such that (i) $\omega(\epsilon)$ is a neighborhood of the identity for every $\epsilon>0$ and $(ii)$ $-\omega(\epsilon)=\omega(\epsilon)$. (Note: (ii) is just for convenience; it's no real restriction, since if $\omega$ satisfies (i) and $\omega'(\epsilon)=\omega(\epsilon)\cap(-\omega(\epsilon))$ then $\omega'$ satisfies (i) and (ii).)
Let $\hat G_\omega$ be the set of $\chi\in\hat G$ such that $x-y\in\omega(\epsilon)$ implies $|\chi(x)-\chi(y)|\le\epsilon$.
We're trying to show that $E$ is precompact. Above we've shown that there is a modulus of continuity $\omega$ such that if $\chi\in E$ and $x\in\omega(\epsilon)$ then $|1-\chi(x)|<\epsilon$. It follows that $\chi\in\hat G_\omega$: If $x-y\in\omega(\epsilon)$ then $$|\chi(x)-\chi(y)|=|\chi(x)(1-\chi(y-x)|=|1-\chi(y-x)|<\epsilon.$$
So $E\subset\hat G_\omega$, and hence we need only show this:
This is just Arzela-Ascoli. I don't recall seeing a version that applies directly, so we just prove it.
The proof is clearest in terms of nets. We need to show that every net has a convergent subnet. In a word, Tychonoff gies pointwise convergence and then equicontinuity shows that pointwise convegence implies uniform convergence on compact sets.
Say $(\chi_\alpha)$ is a net in $\hat G_\omega$. Regarding $\hat G_\omega$ as a subset of the product space $\Bbb T^G$, Tychonoff shows that there is a pointwise convergent subnet. So we need only show this:
Proof: The pointwise convergence makes it clear that $\chi\in\hat G_\omega$. Suppose $K\subset G$ is compact and $\epsilon>0$. Let $U=\omega(\epsilon)$ and choose $x_1,\dots,x_n\in K$ with $$K\subset\bigcup_{j=1}^n(x_j+U).$$
Now there exists $\beta$ such that if $\alpha>\beta$ then $$|\chi_\alpha(x_j)-\chi(x_j)|<\epsilon\quad(j=1,2,\dots,n).$$
Suppose $\alpha>\beta$ and $x\in K$. Choose $j$ so $x\in x_j+U$. Then $$|\chi_\alpha(x)-\chi(x)|\le|\chi_\alpha(x)-\chi_\alpha(x_j)|+|\chi_\alpha(x_j)-\chi(x_j)|+|\chi(x_j)-\chi(x)|<3\epsilon.$$