Fractional Sobolev space $H^s_p(\mathbb R), s>0, 1<p<\infty$ is a space of tempered distributions $f$ that satisfy $F^{-1}((1+|\xi|^2)^{s/2} F(f)) \in L_p(\mathbb R)$. Here, $F$ denotes the Fourier transform. Such spaces are known to admit Littlewood-Paley characterisation, i.e. one can esitmate the norm $\|\phi\|_{H^s_p}$ by $$ \| ( \sum_{n\geq 0} 4^{sn} |\phi_n|^2 )^{1/2} \|_{L_p},$$ where $\phi_n(x) = F^{-1}(\psi_n(\xi)F(\phi))(x)$ for $\psi_n(x)$ being a smooth function that is equal to $1$ on the ring $2^{n-1}<|\xi|<2^{n}$ and $0$ outside a neighborhood of it.

Above is the definition of the Sobolev space for $\mathbb R$. It is well known that using local charts one can define Sobolev spaces on manifolds, and get the Littlewood-Paley characterisation as well.

It is also well known that one can define Fourier transform directly on the circle and get a Fourier series as a result, that are somewhat easier to work with.

Now the question: is it known that if I define fractional Sobolev spaces on the circle directly (without charts), it will have a Littlewood-Paley characterisation of such space? If yes, what would be a reference?


1 Answer 1



A sledge hammer with which you can hit this is Stein's Topics in Harmonic Analysis related to the Littlewood-Paley Theory, published by the Annals of Math Studies series of the PUP. The main thing you are looking for the Theorem 2, the "square function theorem" for decompositions of functions on compact Lie groups.

(Note that in this characterisation the two way inequality is only true of $f$ has mean zero.)

On the other hand, I am pretty sure that sledge hammer is not necessary for your particular problem. If you look at the proof of the Littlewood-Paley square function estimate on $\mathbb{R}^n$ using dyadic Martingales, I am pretty sure the exact proof applies to the $\mathbb{T}^n$ case with very minimal modifications. Maybe check Grafakos' Classical Fourier Analysis?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.