I am proving something, and I may need the following inequality: $$ (g,f)_{L^2} \le C||g||_{H^{-s}}||f||_{H^s} $$ where $(g,f)_{L^2} $ is inner product and $f$ and $g$ have higher enough regularity and $s \in \mathbb{R}$.
I think I can prove it and it is very easy. I don't know if it is right and where I can find a book referring to it (because I think this inequality would very useful and should be mentioned in some book). Anyway, I will put my proof here, I hope someone could check it and tell me where I can find it (there might be some other obvious reasons for it, if so, I hope you can tell me. I feel a little bit dizzy now due to the whole day study and work, perhaps I just miss some obvious facts).
I put all the constant into $C$.
Proof: $$ (g,f)_{L^2} =C \int_\mathbb{R}\hat{g}(\zeta)\hat{f}(\zeta)d\zeta=C \int_\mathbb{R}\hat{g}(\zeta)(1+|\zeta|^2)^{-s/2}\hat{f}(\zeta)(1+|\zeta|^2)^{s/2}d\zeta \le C(\int_\mathbb{R}|\hat{g}(\zeta)|^2(1+|\zeta|^2)^{-s} d\zeta)^{1/2} (\int_\mathbb{R}|\hat{f}(\zeta)|^2(1+|\zeta|^2)^{s} d\zeta )^{1/2} \le C||g||_{H^{-s}}||f||_{H^s} $$