Today I Learned

April 3rd

Today I learned generalized Fourier series. Fix $G$ a locally compact abelian group. For $f\in L^1(G),$ we have a Fourier transform $\hat f.$ If we also have $\hat f\in L^1(\hat G),$ then we have an inverse Fourier transform satisfying\[f(g)=\int_{\hat G}\hat f(\chi)\chi(g)\,d\chi,\]which holds for all $g\in G$ if $f$ is continuous. In particular, for given Haar measure $dg$ of $G,$ there is a Haar measure $d\chi$ on $\hat G$ which makes the above true. I don't plan on showing any of this because I think it's quite involved; I will mention that $f\in L^1(G)\cap L^2(G)$ ensures that $\hat f\in L^1(G),$ so we can reduce this to a single condition on $f.$

To continue, we take the following lemma.

This is not terribly hard. Indeed, using the compact-open topology,\[\left\{\chi\in\hat G:\chi(G)\subseteq B(1,1/2)\right\}=\{1_G\}\]because $\chi(G)\subseteq S^1$ needs to be a multiplicative group, but the only multiplicative group contained in $B(1,1/2)$ is $\{1\}.$ Thus, that any character sending $G$ to $B(1,1/2)$ must be the trivial character.

It follows that $\{1_G\}$ is an open set in $\hat G,$ so by homogeneity, $\hat G$ is discrete. To be explicit, for any other character $\chi\in\hat G,$ multiplication by $\chi^{-1}$ induces a continuous function $\times_{\chi^{-1}}:\hat G\to\hat G.$ Then the preimage of $\{1_G\}$ under $\times_{\chi^{-1}}$ is $\{\chi\},$ so $\{\chi\}$ is also open. $\blacksquare$

We do not actually need the converse of this statement for where we're going, and its details are annoying. Endow a now compact abelian group $G$ with a Haar measure $dg.$ Now, for any sufficiently nice $f$ (satisfying the conditions we gave earlier), we write\[f(g)=\int_{\hat G}\hat f(\chi)\chi(g)\,d\chi.\]At a high level, because $\hat G$ is discrete, we expect this to be a series, and it will turn into a Fourier series. However, we need to contend with $d\chi$ a bit.

Note that the measure $d\chi$ on $\hat G$ needs to some multiple of the counting measure on $\hat G$ because, for any countable $S\subseteq\hat G,$\[\op{Vol}(S)=\sum_{\chi\in S}\op{Vol}(\{\chi\}).\]In particular, this is only legal because the $\{\chi\}$ are forming a disjoint open cover of $S,$ where $\{\chi\}$ is open because $\hat G$ is discrete by lemma. Because $d\chi$ is a Haar measure, the measure of $\{\chi\}$ needs to be equal to the measure of (say) $1_G,$ so\[\op{Vol}(S)=\sum_{\chi\in S}\op{Vol}(\{1_G\}).\]When $S$ is finite, this is $|S|\op{Vol}(\{1_G\}),$ and when $S$ is countably infinite, this is also infinite. Observe that when $S$ is uncountable, this is still infinite, so in fact our measure $d\chi$ is able to measure all subsets of $\hat G,$ albeit stupidly.

Now, keeping in mind the definition of integration with respect to a measure, we see\[f(g)=\op{Vol}(\{1_G\})\sum_{\chi\in\hat G}\hat f(\chi)\chi(g).\]This is our Fourier series. We will elaborate on this shortly, but for right now, we talk about measures. We would like to make $\op{Vol}(\{1_G\})=1$ so that we don't have to worry about it, so we take the following lemma.

This sounds impressive, but given the machinery that we have, it suffices to just check our Fourier series at a single $f$ to find out when $\op{Vol}(\{1_G\})=1.$ Indeed, this will imply and is in fact equivalent to the fact that, for $S\subseteq\hat G,$\[\op{Vol}(S)=|S|\]by plugging in $\op{Vol}(\{1_G\})=1$ in the argument from earlier. So we show that $\op{Vol}(G)=1$ if and only if $\op{Vol}(\{1_G\})=1.$

For this checking, we take $f=1_G.$ It is in $L^1(G)$ because the set $G$ is compact and so measurable. Computing Fourier transforms comes down to orthogonality of characters, for we have $\hat f:\hat G\to\CC$ by\[\hat f(\chi)=\int_Gf(g)\overline{\chi(g)}\,dg=\int_G\overline\chi(g)\,dg.\]If $\chi=1_G,$ then the integrand is trivial, and we get $\hat f(1_G)=\op{Vol}(G).$ Now, if $\chi\ne1_G,$ then $\overline\chi$ isn't either, so we can find some $g_0\in G$ for which $\overline\chi(g)\ne1,$ so\[\hat f(\chi)=\int_G\overline\chi(g)\,dg=\int_{g_0G}\overline\chi(g)\,dg=\int_G\overline\chi(gg_0)\,d(gg_0)=\overline\chi(g_0)\int_G\overline\chi(g)\,dg.\]But with $\chi(g_0)\ne1,$ this only way for this to be true is for $\hat f(\chi)=0.$ Thus, $\hat f=1_{1_G}\op{Vol}(G)$ is the $1_G$-indicator.

To finish, we check the inverse Fourier transform. Because of the above discussion, we know\[f(g)=\op{Vol}(\{1_G\})\sum_{\chi\in\hat G}\hat f(\chi)\chi(g).\]We know most of this: $f(g)=1$ because $f=1_G$; and $\hat f(\chi)=\op{Vol}(G)1_{1_G},$ so the summands vanish except for $\chi=1_G\in\hat G,$ in which case the summand will be $\op{Vol}(G)\cdot1$ in this case. Plugging everything in, we see\[1=\op{Vol}(\{1_G\})\op{Vol}(G).\]So we see that $\op{Vol}(G)=1$ if and only if $\op{Vol}(\{1_G\})=1,$ and we are done here. $\blacksquare$

This is nice, now, because the "natural'' Haar measure on $G$ giving it measure $1$ is also exactly the measure which makes $\hat G$ have the counting measure. In this case, we do have our series expansion\[f(g)=\sum_{\chi\in\hat G}\hat f(\chi)\chi(g)\]for sufficiently nice $f.$ In particular, the $\hat f(\chi)$ are constants not dependent on $g.$ This is our generalized Fourier series.

For effect, we derive standard Fourier series. Look at $\RR/\ZZ,$ whose dual is the discrete group $\ZZ.$ For a quick proof of this, we already know that $\RR\cong\widehat\RR,$ and surely we can extend any character on $\RR/\ZZ$ periodically. So all characters of $\RR/\ZZ$ look like\[x\mapsto e^{2\pi irx}\]for some $r\in\RR.$ Because we need to vanish on $\ZZ,$ it suffices for $e^{2\pi ir}=1,$ so $r\in\ZZ.$ Thus, all or characters actually look like $x\mapsto e^{2\pi inx}$ for $n\in\ZZ,$ which provides our correspondence between $\widehat{\RR/\ZZ}$ and $\ZZ.$

We ran through that proof to say that the inverse Fourier transform in $\RR/\ZZ$ endowed with measure giving $\RR/\ZZ$ volume $1$ has\[f(x)=\sum_{\chi\in\widehat{\RR/\ZZ}}\hat f(\chi)\chi(g)=\sum_{n\in\ZZ}\hat f(n)e^{2\pi inx}\]for sufficiently nice $f.$ Note that we are abusing notation a bit by writing $\hat f(n)$ because we actually mean $\hat f(x\mapsto e^{2\pi inx}),$ but we can redefine\[\hat f(n):=\int_{\RR/\ZZ}f(x)e^{-2\pi inx}\,dx.\]So yes, we are getting standard Fourier series out of this abstraction.

It is interesting that we can even carry this, practically verbatim, through to $\mathbb A_K/K,$ which has dual $K.$ Again, the correspondence here is that, for a standard $\psi:\mathbb A_K,$ we can associate $k\in K$ with $\psi_k(x):=\psi(kx).$ (We do not show this here.) Then our inverse Fourier transform (using the Haar measure which gives $\op{Vol}(\mathbb A_K/K)=1$) says\[f(a)=\sum_{\chi\in\widehat{\mathbb A_K/K}}\hat f(\chi)\chi(a)=\sum_{k\in K}\hat f(k)\psi(kx).\]Again we are abusing notation for a prettier statement (and emphasis) by writing $\hat f(k)$ for $\hat f(\psi_k).$ Anyways, the point is that this looks a lot like Fourier series with $\RR/\ZZ,$ entirely because we have a similarly nice classification of characters on $\mathbb A_K.$