I am trying to solve an exercise from the book: "Classical and Multilinear Harmonic Analysis" from Muscalu and Schlag, but I am stuck on the following problem:
Given a function $f \in C^0(\mathbb{T}^1)$ that satisfies the following property: $$ \iint_{\mathbb{T}^1\times \mathbb{T}^1} \frac{\lvert f(x)-f(y)\rvert^2}{\sin^2(\pi(x-y))} \,dx\,dy < \infty $$
Then $ f \in H^{\frac{1}{2}}(\mathbb{T}^1)$.
I know that $\|f\|_{H^{\frac{1}{2}}} = \lvert\hat f(0)\rvert^2 + \sum_{n \neq 0} \lvert n\rvert\lvert\hat f(n)\rvert^2$. I was trying to bound the $n^{th}$ Fourier coefficient of $f$, but I did not manage to make any progress. I would like $\sum_{n \neq 0} \lvert n\rvert\lvert \hat f(n)\rvert^2$ to be convergent but I don't see how.
What I was able to show: \begin{eqnarray} \lvert\hat f(n)\rvert^2 &=& \left|\int_0^1 f(x) e^{-2 \pi i n x} \,dx \right|^2 \\ &\leq& \int_0^1 \lvert f(x) - f(y)\rvert^2 \,dx \; (\textrm{using the fact that $\int_0^1 e^{-2\pi inx}dx = 0$) } \\ &\leq& \iint_{\mathbb{T}^1\times \mathbb{T}^1} \frac{\lvert f(x)-f(y)\rvert^2}{\sin^2(\pi(x-y))} \,dx\,dy \\ &<& \infty \end{eqnarray}
Can someone help me? I would be very thankful.
Since $\lim_{t\to 0}\frac{\sin^2 t}{t^2}=1$ you can find $\delta>0$ such that $$\frac12\le \frac{\sin^2 t}{t^2}\le\frac32$$ for all $0<|t|\le\delta$. In turn $$\int\int_{\{|\pi (x-y)|\le\delta\}} \frac{|f(x)-f(y)|^2}{|x-y|^2}dxdy\le 2 \int\int_{\{|\pi (x-y)|\le\delta\}} \frac{|f(x)-f(y)|^2}{\sin^2(\pi(x-y))}dxdy.$$ On the other hand, $$\int\int_{\{|\pi (x-y)|>\delta\}} \frac{|f(x)-f(y)|^2}{|x-y|^2}dxdy\le \frac1{\delta^2}\int\int_{\{|\pi (x-y)|>\delta\}} |f(x)-f(y)|^2dxdy<\infty.$$ Thus, $$\int_{\mathbb{T}^1}\int_{\mathbb{T}^1} \frac{|f(x)-f(y)|^2}{|x-y|^2}dxdy\le \frac1{\delta^2}\int\int_{\{|\pi (x-y)|>\delta\}} |f(x)-f(y)|^2dxdy<\infty.$$ This is an equivalent seminorm in $H^{1/2}$.