I want to verify that $$H^{1/2}(S^1,S^1)\subsetneq \operatorname{vmo}(S^1,S^1) \subsetneq \operatorname{bmo}(S^1,S^1)=L^1(S^1,S^1)$$ Where vmo and bmo are the spaces of vanishing and bounded mean oscillation like here and $H^{1/2}$ is the sobolev-space $W^{1/2,2}$.
The inclusions $H^{1/2}\subset\operatorname{vmo}\subset \operatorname{bmo}$ are clear for me, but why are holding the inequalities?
Thank you, Maxi
To show the first inclusion is strict, start with a Weierstrass function $f:S^1\to\mathbb R$ $$f(e^{i\theta})=\sum_{n=1}^\infty 2^{-n}\sin(4^n\theta).$$
The Fourier coefficients satisfy $\sum_p|\hat f(p)|<\infty$ but $\sum_p|p\hat f(p)|^2=\infty.$ This ensures that $f$ is continuous hence in $\mathrm{VMO}(S^1),$ but not in $H^{1/2}(S^1).$ To get the codomain in $S^1$ we can take $f+i\sqrt{1-f^2}$ which will still be continuous - this uses $|f|\leq 1.$
To show the second inclusion is strict take $$g(e^{i\theta})=\begin{cases} 1&(0\leq \theta<\pi)\\ -1&(-\pi\leq \theta<2\pi). \end{cases}$$
This is bounded hence in $\mathrm{BMO}(S^1,S^1),$ but the mean oscillation in an interval $(-\epsilon,\epsilon)$ is $1$ so $g$ is not in $\mathrm{VMO}(S^1,S^1).$