Fourier transform of distributions on the circle

Question

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).

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.