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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 $chiinhat G$ such that $|1-chi|<alpha$ on $K$. Then $E$ is equicontinuous at the origin: Given $epsilon>0$, there exists $n$ such that if $zinBbb T$ and $epsilonle|1-z|<alpha$ then $|1-z^j|>alpha$ for at least one integer $j$ with $1le jle n$ (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 $sin[0,pi]$. Now wlog $z$ has positive imaginary part. Hence there exists $tin (0,pi/2]$ with $z=e^{it}$. So $tge|1-e^{it}|>epsilon$. Choose $thetain(0,pi)$ with $|1-e^{itheta}|=alpha$. Since $t>epsilon$ there exists $j$ with $1le jle theta/epsilon+1<pi/(2epsilon)+1$ such that $(j-1)tle theta<jt$.
Hence $jt=(j-1)t+tle theta+pi/2lepi$. Since $theta<jtlepi$ the mmontonicity mentioned above shows that $|1-z^j|>|1-e^{itheta}|=alpha$.

Now choose a neighborhood of the origin $U$ such that $U+U+Udots +Usubset K$. Then if $chiin E$ 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 $omega:(0,infty)tomathcal P(G)$ 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 $chiinhat G$ such that $x-yinomega(epsilon)$ implies $|chi(x)-chi(y)|leepsilon$.

We're trying to show that $E$ is precompact. Above we've shown that there is a modulus of continuity $omega$ such that if $chiin E$ and $xinomega(epsilon)$ then $|1-chi(x)|<epsilon$. It follows that $chiinhat G_omega$: If $x-yinomega(epsilon)$ then $$|chi(x)-chi(y)|=|chi(x)(1-chi(y-x)|=|1-chi(y-x)|<epsilon.$$

So $Esubsethat G_omega$, and hence we need only show this:

If $omega$ is a modulus of continuity on $G$ then $hat G_omega$ is compact (in the compact-open topology).

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:

If $omega $ is a modulus of continuity on $G$ and $(chi_alpha)$ is a net in $hat G_omega$ such that $chi_alpha(x)tochi(x)$ for every $xin G$ then $chiin hat G_omega$ and $chi_alphatochi$ uniformly on compact sets.

Proof: The pointwise convergence makes it clear that $chiinhat G_omega$. Suppose $Ksubset G$ is compact and $epsilon>0$. Let $U=omega(epsilon)$ and choose $x_1,dots,x_nin K$ with $$Ksubsetbigcup_{j=1}^n(x_j+U).$$

Now there exists $beta$ such that if $alpha>beta$ then $$|chi_alpha(x_j)-chi(x_j)|<epsilonquad(j=1,2,dots,n).$$

Suppose $alpha>beta$ and $xin K$. Choose $j$ so $xin x_j+U$. Then
$$|chi_alpha(x)-chi(x)|le|chi_alpha(x)-chi_alpha(x_j)|+|chi_alpha(x_j)-chi(x_j)|+|chi(x_j)-chi(x)|<3epsilon.$$