Fractional Sobolev space $H^{1/2}(-\pi,\pi)$

Question

Let $H^{1/2}(-\pi,\pi)$ be the space of $L^2$ functions whose Fourier series coefficients $\{c_n\}_n$ satisfy the summability constraint $\sum_n |n| |c_n|^2 < \infty$. Are functions in $H^{1/2}(-\pi,\pi)$ continuous (in the sense that each admits a continuous representant)?

Answer 1

No. The function $$f(t)=\sum_{n=2}^\infty \frac{e^{in t}}{n \log n} $$ is in $H^{1/2}$ but its real part is unbounded near zero. Informally, it's because the series diverges when $t=0$. But a more careful proof is called for.

One approach: take small $t>0$, then choose $N=\lfloor 1/t\rfloor $ so that $\cos nt\ge \cos 1>0$ for $2\le n\le N$. The partial sum over $2\le n\le N$ is at least $\sum_2^N \frac{\cos 1}{n\log n} \ge c\log\log (1/t)$... and then somehow bound the tail using summation by parts. Looks tedious.

Another approach: if $f\in L^\infty$, then multiplication by $f$ is a bounded operator on $L^2$. Multiplication is convolution of Fourier coefficients. Convolving $1/(n\log n)$ with $$ \frac{1}{\sqrt{n\log n} (\log \log n)}$$ (which is in $L^2$, just barely) produces at least $$\begin{split} \frac{1}{\sqrt{n\log n} (\log \log n)} \sum_{k=2}^{n-2} \frac{1}{k\log k} & \approx \frac{1}{\sqrt{n\log n} (\log \log n)} \log \log n \\ & = \frac{1}{\sqrt{n\log n}} \end{split}$$ which is not in $L^2$.

Answer 2

$H^{1/2}_{per}[-\pi,\pi]$ contains discontinuous (even unbounded) functions - see the answer of Care Bear.

On the other hand, Sobolev inbedding provides that $$ H^{1/2+\varepsilon}_{per}[-\pi,\pi]\subset C_{per}[-\pi,\pi], $$ for every $\varepsilon>0$, and in fact the imbedding is compact.