Let $u,v\in\dot H^{1/4}(\mathbb R)$. Since by Sobolev embedding $\dot H^{1/4}(\mathbb R)\hookrightarrow L^4(\mathbb R)$, I know that the product $uv$ lies in $L^2(\mathbb R)$, that is, it holds the estimate $$ \|uv\|_{L^2}\leq C\|u\|_{\dot H^{1/4}}\|v\|_{\dot H^{1/4}}. $$ My question is, is it possible to find a better estimate on $uv$? Does the product lie in some Besov space of positive regularity, or any normed space that is strictly contained in $L^2$?
The Sobolev space $\dot H^s(\mathbb R)$ is the space of tempered distributions such that the Fourier transform is locally integrable and the following norm is finite: $$ \|u\|^2_{\dot H^s}=\left|\int_{\mathbb R}|\xi|^{2s}|\widehat u(\xi)|^2\,d\xi\right|, $$ where $\widehat u$ is the Fourier transform of $u$.
So, I have found an answer. One can use the Kato-Ponce estimates in $\mathbb R^n$: $$ \||\nabla|^s(fg)\|_{L^p} \lesssim \||\nabla|^sf\|_{L^{p_1}} \|g\|_{L^{p_2}}+\|f\|_{L^{q_1}} \||\nabla|^sg\|_{L^{q_2}}, $$ where $p\in[1,\infty]$, $p_j,q_j\in(1,\infty]$ and $s>0$, and where the indices satisfy $$ \frac{1}{p}=\frac{1}{p_1}+\frac{1}{p_2}=\frac{1}{q_1}+\frac{1}{q_2}. $$ One can choose $p=4/3$, $p_1=q_2=2$, $p_2=q_1=4$, $s=1/4$ and obtain (for $n=1$) $$ \||\partial_x|^{1/4}(fg)\|_{L^{4/3}}\lesssim \||\partial_x|^{1/4}f\|_{L^{2}} \|g\|_{L^{4}}+\|f\|_{L^{4}} \||\partial_x|^{1/4}g\|_{L^{2}}\lesssim \|f\|_{\dot H^{1/4}}\|g\|_{\dot H^{1/4}}. $$ This means that the product of two functions in $\dot H^{1/4}$ lies in the Triebel-Lizorkin space $\dot F_{4/3,2}^{1/4}(\mathbb R)$, and it holds the chain of inclusions $$ \dot F^{1/4}_{4/3,2}(\mathbb R)\hookrightarrow \dot B^{1/4}_{4/3,2}(\mathbb R)\hookrightarrow L^2(\mathbb R). $$ I am not sure this is the best one can achieve in the Besov/Triebel-Lizorkin scale, so please if anyone has a better answer leve a comment.
These lecture notes contain a proof of the K-P estimates (Theorem 7.10).