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).

  • $\begingroup$ 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? $\endgroup$ Jun 15, 2018 at 9:32
  • $\begingroup$ @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)? $\endgroup$
    – Cohen Lu
    Jun 15, 2018 at 9:38

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ Jun 15, 2018 at 17:20
  • $\begingroup$ @paulgarrett What definition of $\hat u(n)$ do you have in mind? $\endgroup$ Jun 15, 2018 at 17:26
  • $\begingroup$ 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... :) $\endgroup$ Jun 15, 2018 at 17:34
  • $\begingroup$ @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 $\hat u(n)=a_n$. $\endgroup$ Jun 15, 2018 at 17:39
  • $\begingroup$ 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... $\endgroup$ Jun 15, 2018 at 18:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .