# Fourier transform of distributions on the circle

We can easily define Fourier transform of tempered distributions on the real line (or any Euclidean space $R^n$). My question is: can we define Fourier transform of distributions on the circle in a similar way? Here I think we can define distributions as the dual space of smooth functions on the circle (smooth periodic functions).

• I dont quite understand the question: in my understanding the notion of a Schwartz space over the unit circle does not make much sense...so how would you define tempered distributions on its dual, the integers? Jun 15, 2018 at 9:32
• @EduardTetzlaff Thanks for the comment. I understand it doesn’t make sense to define tempered distributions on the circle. What about general distributions (like the one I defined in the question)? Jun 15, 2018 at 9:38

In a word, yes this all works.

Yes, a distribution on $\Bbb T$ is an element of the dual of $C^\infty(\Bbb T)$. About "tempered distributions": One does not talk about tempered distributions in this context. But on the line, any distribution with compact support is tempered; since any distribution on $\Bbb T$ has compact support it turns out that any distribution on $\Bbb T$ is as nice as tempered distributions on the line, in particular it has a Fourier transform.

If $u$ is a distribution on $\Bbb T$ we define the Fourier coefficients by $$\hat u(n)=u(e_n)\quad(n\in\Bbb Z),$$where $$e_n(t)=e^{-int}.$$

Say the sequence $(a_n)$ has polynomial growth if there exist $c$ and $N$ so $$|a_n|\le c(1+|n|)^N.$$

Theorem If $u$ is a distribution on $\Bbb T$ then the sequence $(\hat u(n))$ has polynomial growth. Conversely, if $(a_n)$ is a sequence with polynomial growth then there exists a distribution $u$ on $\Bbb T$ such that $\hat u(n)=a_n$.

Proof: If $u$ is a distribution then the fact that $u$ is bounded on $C^\infty(\Bbb T)$ implies that $\hat u(n)$ has polynomial growth.

Conversely, if $(a_n)$ has polynomial growth we need to show that the sequence $$\sum a_n e^{int}$$converges in $(C^\infty(\Bbb T)'$. This is easy: If $\phi\in C^\infty(\Bbb T)$ then for every $N$ there exists $c$ with $$|\hat\phi(n)|\le c(1+|n|)^{-N};$$hence $\sum a_n\hat\phi(n)$ converges.

• I might quibble that we don't have to "define" the Fourier coefficients $\widehat{u}(n)$ as $u(e^{-int})$, but, rather, they are provably equal to that. So no whim or choice is involved, for example. Jun 15, 2018 at 17:20
• @paulgarrett What definition of $\hat u(n)$ do you have in mind? Jun 15, 2018 at 17:26
• I'm just quibbling about saying "definition", rather than that is the determinable, provable value of the Fourier coefficient. That is, we can prove that all distributions have Fourier expansions converging in a suitable Sobolev space (after proving a Sobolev imbedding/inequality on the circle), and then we deduce (not "define") the coefficients to be as you said. Maybe this is a distinction that matters more to me than to others... :) Jun 15, 2018 at 17:34
• @paulgarrett If I'm understanding you correctly then there's still a definition in what you're doing. Yes, we can prove that there exists a sequence $a_n$ such that $u=\sum a_n e^{int}$. And we can prove that $a_n=u(e_n)$. This is not a proof that $\hat u_n=u(e_n)$, unless we define $\hat u(n)=a_n$. Jun 15, 2018 at 17:39
• I agree, this may be about word-use more than mathematics per se... My use of words would be that if $u=\sum a_n e^{int}$ (in some Sobolev space) then this proves that $\widehat{u}(n)=a_n$, by uniqueness of Fourier expansions (even in Sobolev spaces), since the $\widehat{u}(n)$'s are the Fourier coefficients of $u$, if it has a Fourier expansion. For me, "definition" implies some element of volition or choice, which (from my viewpoint) is absent here (apart from normalizations). But, sure, maybe this is just quibbling... Jun 15, 2018 at 18:44