Sobolev multiplication theorem

Question

I would like to know whether multiplication defines a bounded map $$H^{1/2} \otimes H^{1/2} \to H^{-1/2}$$ dimension of the domain is $3$.

I have checked two different sources and one said that it works but the other that this map is bounded as a map $$H^{1/2} \otimes H^{1/2} \to H^{s}$$ where $s < -\frac{1}{2}$. That is why I am confused.

Notation: $H^{1/2} = W^{1/2, 2} = L^2_{1/2}$

I would be satissfied with a source i can rely on.

Answer 1

See this notes by Terry Tao. Also, Sobolev multiplication below the borderline.

Answer 2

For $\Omega \subseteq \mathbb{R}^3$ there are Sobolev embeddings $$ H^{1/2} \subset L^3, \, L^{3/2} \subset H^{-1/2} $$ where the second one follows from the first one by duality. Thus $$ f, \, g \in H^{1/2} \subset L^3 \; \Rightarrow \; fg \in L^{3/2} \; (\text{by Cauchy-Schwarz}) \; \Rightarrow \; fg \in H^{-1/2} \, . $$